Subtopic Archive - Page 2 of 10

Worksheet 2 (Differential Calculus)

October 12, 2020/by admin_ct

1. If f(x)=|x-3| then f'(1) is

a) 0

b) 1

c) -1

d) Non existent

Solution: Derivative of this absolute function can be found either by using definition of absolute value function or by using formula.

Lets first use definition of absolute value function.

$\dpi{120} f(x) =\begin{cases} & x-3 \text{ if } x\geq 3 \\ & -(x-3) \text{ if } x < 3 \end{cases}$

Since x=1 lie in second domain (x<3) so we use lower function.

f(x) = -(x-3)

f ‘(x) = -(1-0) =-1

Therefore f ‘(1) = -1. Option c is correct.

Other way: Using formula for derivative of absolute functions.

Let f(x) = |U| where U= g(x)

Then $\dpi{120} f'(x)=\frac{U}{|U|}*U'$ or $\dpi{120} f'(x)=\frac{g(x)}{|g'(x)|}*g'(x)$

$\dpi{120} f'(x)=\frac{(x-3)}{|x-3|}*(x-3)'$

$f'(1)=\frac{(1-3)}{|1-3|}*1 =\frac{-2}{|-2|} =\frac{-2}{2}=-1$

2. Given that $\dpi{120} {\color{Red} f(x) =x^{10}h(x)}$ and h(-1)=4 , h'(-1) = 7 . Then find f ‘(-1).

a) 4

b) -33

c)-40

d)-70

Solution: Since f(x) is a product of two functions x^10 and h(x) so we use product rule to find derivative f'(x)

$f'(x)=(x^{10})'h(x)+x^{10}h'(x)$

$f'(x)=(10x^{9})'h(x)+x^{10}h'(x)$

$f'(-1)=10(-1)^{9}h(-1)+(-1)^{10}h'(-1)$

Plugin given values of h(1) and h'(-1)

$f'(-1)= -10(4)+ (1)(7)$

f ‘(-1) = -33

Option b) -33 is correct answer.

3. If ${\color{Red} f'(x) =\frac{(x+2)^2)(x^2 -1)}{3}}$ and ${\color{Red} g(x)= f(\sqrt{x+1})}$ what is the value of g'(3) ?

a) 16

b) 1/4

c) 4

d) 16

Solution : First we find derivative of g(x) using chain rule.

$g'(x)= f'(\sqrt{x+1})* (\sqrt{x+1})'$

$g'(x)= f'(\sqrt{x+1})* \frac{1}{2\sqrt{x+1}}$

$g'(3)= f'(\sqrt{3+1})* \frac{1}{2\sqrt{3+1}}$

$g'(3)= f'(2)* \frac{1}{2*2}$

To get f'(2) we plugin x=2 into given expression for f'(x)

$f'(2) =\frac{(2+2)^2(2^2 -1)}{3}= 16$

Using f'(2)= 16 we get g'(3) as

g'(3) = 16 *(1/4) = 4

g'(3)= 4

So option c is the answer.

4. Functions f, g and h are twice differentiable functions with g(2)= h(2)= 4. The line ${\color{Red} y= 4+\frac{2}{3}(x-2)}$ is tangent to both the graphs of g at x=2 and graph of h at x=2.

a) find h'(2)

b) Let a be a function given as

${\color{Red} a(x)= 3x^3 h(x)}$ . Write an expression for a'(x) and a'(2).

c) The function h satisfies that ${\color{Red} h(x)= \frac{x^2-4}{1-(f(x))^3} ; x \neq 2}$ . It is known that ${\color{Red} \lim_{x\rightarrow 2}h(x)}$ can be evaluated using L Hospital’s rule. Use this limit to find f(2) and f ‘(2).

d) It is known that g(x) ≤ h(x) for 1<x<3. Let k be a function satisfying g(x)≤k(x)≤h(x) for 1<x<3. Is k continuous at x=2? Justify your answer.

Solution: It is known that slope of tangent line to the curve at given point is same the slope of curve at that point.

Slope of tangent line at x=2 is 2/3

and that would be the slope of curve h(x) at x=2. so h'(2) = 2/3.

b) $a(x)= 3x^3 h(x)$ . Since given function is product of two functions so using product rule,

$a'(x)= 3[(x^3)' h(x) + x^3 h'(x)]$

$a'(x)= 3[(3x^2)' h(x) + x^3 h'(x)]$

$a'(2)= 3[(3(2)^2)' h(2) + 2^3 h'(2)]$

$a'(2)= 3[12*4 + 8*2/3]$

a'(2) = 160

$\lim_{x\rightarrow 2}h(x)= \lim_{x\rightarrow 2} \frac{x^2-4}{1-(f(x))^3}$

It is given that this limit can be evaluated using L Hospital’s rule. Which means it would have intermediate form(0/0) when x approaches to 2. so we have

$\frac{2^2-4}{1-(f(2))^3} = \frac{0}{0}$

$1-(f(2))^3=0$

$(f(2))^3= 1 \Rightarrow f(2)= 1$

Since h(x) is continuous so we have,

$\lim_{x\rightarrow 2}h(x)=h(2)$

Using L Hospital’s rule to find its limit,

$\lim_{x\rightarrow 2}h(x)= \lim_{x\rightarrow 2} \frac{(x^2-4)'}{(1-(f(x))^3)'} = \frac{2x-0}{0-3(f(x))^2 f'(x)}$ = 4

$\frac{2x}{-3(f(x))^2 f'(x)} =4$

$\frac{2*2}{-3(f(2))^2 f'(2)} =4$

$\frac{4}{-3(1)^2 f'(2)} =4$

$\frac{1}{-3f'(2)} =1$

$-\frac{1}{3} = f'(2)$

d) Given that

$g(x)\leq k(x)\leq h(x) on 1<x<3$

Since g(x) and h(x) are given continuous and g(2)= h(2) = 4

$\lim_{x\rightarrow 2}g(x) =4$ and $\lim_{x\rightarrow 2}h(x) =4$ . Using sandwich theorem we can conclude that,

$\lim_{x\rightarrow 2}k(x) =4$ and k(2)= 4

Therefore K(x) is continuous at x=2.

5. Let ${\color{Red} g(x)= \int_{2}^{x}f(t) dt}$ for 0≤t≤7 where the graph of differentiable function f, is shown below.

a) Find g(3) , g'(3) and g”(3)

Using the given definition of g(x) we have,

$g(3)= \int_{2}^{3}f(t) dt$ which means the area under the curve f(t) from x=2 to x=3. This area can be found using geometric formulas.

Area under the curve for given x values is a trapezoid, so we can use trapezoid’s area formula.

A = 1/2 (sum of parallel sides)* distance between parallel sides.

g(3) = A= 1/2(2+4)*1 = 6/2 = 3

Using Fundamental Theorem of Calculus

g'(3) = f(3) = 2

And g”(3)= f ‘(3) which is slope of tangent line at x=3

slope = (4-0)/(2-4) = -2

so g”(3) = -2

b) Find average rate of change of g(x) on 0≤c≤3

Average rate = $\frac{g(3)-g(0)}{3-0}$

For that first we need to find g(0) , same way as we had found g(3).

$g(0)= \int_{2}^{0}f(t) dt$

Which can be rewritten using property of definite integral.

$g(0)= -\int_{0}^{2}f(t) dt$ . which represent the area under the curve from 0 to 2.

g(0) = – [(1/2)*2*4] = -4

Average rate = $\frac{g(3)-g(0)}{3-0} = \frac{3-(-4)}{3} = \frac{7}{3}$

c) For how many values of c, where 0<c<3, is g'(c) equal to average rate found in part b? Justify your answer.

We know that ,

$g(c)= \int_{2}^{c}f(t) dt$

then g'(c) = f(c) = 7/3

And the function f(t) gets the value 7/3 twice on the interval 0<c<3.

d) Find the Graph of gx coordinate of each point of inflection of the graph of g on the interval 0<t<7. Justify your answer.

We know that g'(x) = f(x) and g”(x)= f'(x).

Graph of g will have inflection points where f(x) changes its direction. We can observe from given graph of f that it changes direction from increasing to decreasing at x=2 and again from decreasing to increasing at x=5. Also slopes f ”(2) and f”(5) are undefined being sharp corners. So the two inflections points are at x=2,5.

6. For what values of a and c , is the piecewise function differentiable.

${\color{Red} f(x)=\begin{cases} & ax^2+sinx, \text{ if } x\leq \pi \\ & 2x-c , \text{ if } x > \pi \end{cases} }$

We know that functions $ax^2+sinx$ and 2x-c are continuous and differentiable in their respective domains.

$f'(x)=\begin{cases} & 2ax+cosx, \text{ if } x\leq \pi \\ & 2 , \text{ if } x > \pi \end{cases}$

We are given that piecewise function is differentiable at x=π too. That means left and right hand derivatives of given function would be same at x=π. Therefore,

$f'(\pi )=\begin{cases} & 2a \pi +cos(\pi ) , \text{ if } x\leq \pi \\ & 2 , \text{ if } x > \pi \end{cases}$

2aπ +cos(π) = 2

2aπ -1 = 2

2a π = 3

$a = \frac{3}{2\pi }$

Also piecewise function is continuous at pi so left and right hand limits would be same as f(π)

Left limit is,

$\lim_{x\rightarrow \pi^- }(ax^2+sinx)= a\pi ^2+sin(\pi )= a\pi ^2$ which is same as value of function , i.e f(π)

And right limit is ,

$\lim_{x\rightarrow \pi^+ }2x-c =2\pi -c$

For the function to be continuous. we set left and right limits equal ,

$a\pi ^2 =2\pi -c$

$c= 2\pi -a\pi ^2$

Plugin value of a = 3/2π , we get

$c=2\pi -\frac{3}{2\pi }*\pi ^2$

$c=2\pi -\frac{3\pi }{2}$

$c=\frac{\pi }{2}$

So the solution is

$a=\frac{3}{2\pi } , c=\frac{\pi }{2}$

7. When the height of a cylinder is 12 cm and the radius is 4cm, the circumference of the cylinder is increasing at a rate of ${\color{Red} \frac{\pi }{4}}$ cm/min, and the height of the cylinder is increasing four times faster than the radius. How fast is the volume of the cylinder changing?

Given: height of cylinder(h)= 12 cm

radius of cylinder(r) = 4 cm

Rate of increase of circumference (2πr)

$2\pi \frac{dr}{dt}=\frac{\pi }{4}$

$2 \frac{dr}{dt}=\frac{1 }{4} \Rightarrow \frac{dr}{dt}=\frac{1 }{8}$

Height of cylinder is increasing 4 times faster than rate of change of radius.

$\frac{dh}{dt} = 4*\frac{1}{8} =\frac{1}{2}$

To find : Rate of change of volume $\dpi{120} \frac{dV}{dt} =?$

Volume of cylinder is given as,

$\dpi{120} V= \pi r^2 h$

Differentiating both sides using product rule, with respect to time(t) ,

$\dpi{120} \frac{dV}{dt}=\pi \left ( 2r\frac{dr}{dt}*h+r^2\frac{dh}{dt} \right )$

Plugin value into this expression.

$\dpi{120} \frac{dV}{dt}=\pi \left ( 2(4)\frac{1}{8}*12+4^2*\frac{1}{2} \right )$

$\dpi{120} \frac{dV}{dt}=\pi \left ( 1*12+8 \right ) = 20\pi$

So the volume of cylinder is changing at the rate of 20π cubic cm/min.

8. Let f be a function that is differentiable on open interval (1,10), If, f(1) = -6 ,f(5)= 2 , and f(10) = -4. Which of the following must be true. Justify your answer.

i. f has atleast two zeros.

ii. The graph of f has atleast one horizontal tangent.

iii. f ‘(5) =0

iv. For some c, 1<c<10 , f(c) =1

v. For some c, 1<c<10 , f ‘(c)= 2/9

i, ii, iv and v must be true. Here is the reason for each.

i. f has atleast two zeros.

When we move from left to right on interval (1,10) , y values goes from negative to positive (-6 to 2 ) and then again from positive to negative (from 2 to -4) that means function’s graph will cross x axis atleast twice. Which proves f has atleast two zeros.

ii. Since function’s graph cross x axis twice so graph will turn atleast once and there is horizontal tangent at each turn. This proves there is atleast one turn and hence atleast one horizontal tangent.

iv. Since the given function is differentiable and hence continuous. This graph is going from -6 to 2 when x is changing from 1 to 5 , that means continuous graph will pass through y value 1, while going from -6 to 2. So there must be some x value c on given interval such that f(c)=1

v. Since function is given differentiable and hence continuous on interval (1,10) , it satisfy hypothesis of Mean Value Theorem. That means there exist a number c on interval (1,10) which satisfy conclusion of MVT.

Therefoere,

$\dpi{120} f'(c)=\frac{f(10)-f(1)}{10-1}$

$\dpi{120} f'(c)=\frac{-4+6}{9}= \frac{2}{9}$

which proves the statement “ For some c, 1<c<10 , f ‘(c)= 2/9″

9. Graph of velocity function of a particle is shown below, where t is measured in seconds. Find the intervals when particle is speeding up and slowing down.

First we observe the given graph. We see that particle start at t=0 , reach max velocity at t=1 .

We make the following observations by looking at the graph :-

Acceleration is positive as long as velocity graph is increasing and acceleration is negative as long as velocity graph is decreasing because acceleration is nothing but the slope of velocity.

Velocity is positive as long as graph is above x axis and velocity is negative as long as graph is below x axis.

Velocity is 0 at t=0 and t=2 seconds. It reaches to max value at t=1. So we make the following chart.

0<t<1
Velocity (V)
Acceleration (A)

0<t<1
+
+

1<t<2
+
–

2<t<3
–
–

Rules to follow :

Particle is speeding up when velocity and acceleration both have same signs. Particle is slowing down when they have opposite signs.

So the answer is :

Particle is speeding up on intervals : (0,1) and (2,3)

slowing down on interval : (1,2)

10 . A particle moves according to position function ${\color{Red} S(t)= \frac{t}{1+t^2}}$ on the interval t ≥0 where t is measured in seconds and S in feet.

a) Find when particle is at rest.

Particle is at rest when its velocity is 0. So we find velocity function by taking its derivative and then set velocity function =0 and solve for t.

$V(t)=S'(t)= \frac{(1+t^2)t' - t(1+t^2)'}{(1+t^2)^2}$

$V(t)= \frac{(1+t^2)1 - t(2t)}{(1+t^2)^2} =\frac{1+t^2 -2t^2}{(1+t^2)^2}$

$V(t) =\frac{1-t^2}{(1+t^2)^2}$

$V(t) =0 \Rightarrow \frac{1-t^2}{(1+t^2)^2}=0$

This expression would be 0 when its numerator is 0 so,

$1-t^2 =0$

t = -1,1

But t=-1 doesn’t lie in the given domain so we accept only t=1.

That means particle is at rest at t=1 second

b) Find the intervals for which particle is moving in positive(right) and negative(left) directions.

Particle is moving towards right when velocity is positive and moving towards left when velocity is negative. So we need to check the intervals where velocity is positive and where it is negative. Domain is given [0,∞) which is divided into two intervals by t=1 (when velocity is 0) So we get the two intervals as,

[0,1) and (1,∞)

For interval [0,1) We get V(0.5) > 0

Velocity is positive, that means particle is moving towards right on [0,1)

For interval (1,∞) , We get V(2) < 0

Velocity is negative, that means particle is moving towards left on (1,∞)

c) Find the total distance traveled in first 8 seconds.

To find total distance, we find distance traveled in positive direction as well as in negative direction.

Distance traveled in positive direction:

$\left | S(1)-S(0) \right |= \frac{1}{2} -0 = \frac{1}{2}$ ft.

Distance traveled in negative direction:

$\left | S(8)-S(1) \right |= \frac{1}{2} -\frac{8}{65}$

Total Distance :

$\frac{1}{2}+\frac{1}{2}-\frac{8}{65} = 1-\frac{8}{65} =\frac{57}{65}$ ft

d) When the particle is speeding up and slowing down?

To find when the particle is speeding up or slowing down, we need to find acceleration function. We know that acceleration is derivative of velocity function. so,

$a(t)= V'(t)=\frac{(1+t^2)^2(1-t^2)'-((1+t^2)^2)'(1-t^2)}{(1+t^2)^4}$

$a(t)= \frac{(1+t^2)^2(-2t)-2(1+t^2)(2t)(1-t^2)}{(1+t^2)^4}$

$a(t)= \frac{(1+t^2)[(-2t)(1+t^2)-2(2t)(1-t^2)]}{(1+t^2)^4}$

$a(t)= \frac{[(-2t)(1+t^2)-2(2t)(1-t^2)]}{(1+t^2)^3}=\frac{-2t-2t^3-4t+4t^3}{(1+t^2)^3}$

$a(t)= \frac{-6t+2t^3}{(1+t^2)^3}=\frac{2t(t^2-3)}{(1+t)^3}$

Setting a(t)=0 and solving for t, we get

$2t(t^2-3)=0$

$t=0, t=\pm \sqrt{3}$

But -√3 doesn’t lie in the given domain, so we have only t=0 and t=√3 .

Using t=√3 , we get the domain t≥0 divided into two intervals as [0,√3) and (√3,∞). Now we check each interval one by one.

For interval [0,√3) we get ,

$a(1)=\frac{2*1(1^2-3)}{(1+1)^3} = \frac{-1}{2} < 0$

Acceleration is negative during this interval.

For interval (√3,∞)

$a(2)=\frac{2*2(2^2-3)}{(1+2)^3} = \frac{4}{27} > 0$

Acceleration is positive during this interval.

Lets make a sign chart using velocity and acceleration results,

0<t<1
Velocity (v)
Acceleration(a)

0<t<1
+
–

1<t<√3
–
–

t>√3
–
+

Using the fact that particle will be speeding up when both velocity and acceleration have same signs (either both positive or both negative) and speeding down when both have different signs, we get the following results.

Speeding up on 1<t<√3

Speeding down on 0<t<1 and t>√3.

Practice Worksheet

Find derivative of given function: $f(x)= 2sin^{-1}(4x^3)$
Find second derivative of $y^2-x^2=9$
Find x values of the horizontal tangents of the curve $y= x^4 -18x^2-4$
Given the function $f(x)= ax-bx^2$ . Determine the values of a and b such that the line 3y-6x+6=0 is tangent to the function at x=3.
At time t, the position of a body moving along the x-axis is given as $x(t)= t^3 -21t^2 +144t$ . Find the body’s acceleration
each time the velocity is zero.
The graphs of f and g are shown below. Use them to evaluate

a) $\left ( fg \right )'(5)$ b) $\frac{d}{dx}\left ( \frac{g(x)}{f(x)} \right )_{x=5}$

7. A particle moves along a coordinate axis in the positive direction to the right. Its position at time $t$ is given by $s(t)=t^3-4t+2$ . Find $v(1)$ and $a(1)$ and use these values to answer the following questions.

a) Is the particle moving from left to right or from right to left at time $t=1$ ?

b) Is the particle speeding up or slowing down at time

8. The velocity-time graph of an object moving along a straight line is shown below. Find the intervals for which object is slowing down.

9. Find least value of a such that the function $f(x)= x^2+ax+1$ is increasing on [1,2].

10. A man is walking at the rate of 6.5 km/hr towards the foot of a tower 120 m high. At what rate is he approaching the top of tower when he is 50m away from tower?

Answers:

$\frac{24x^2}{\sqrt{1-16x^6}}$
$\frac{9}{y^3}$
x=-3, 0, 3
a= 2/3 , b=-2/9
a(6)= -6m/s^2 and a(8)=6m/s^2
a) 1/6 b)7/24
a) Right to left b) slowing down
(1,3)
-2
2.5 km/hr

Worksheet 1 (Limits)

August 29, 2020/by admin_ct

1.Find the limit and choose the correct option:

$\large {\color{Red} \lim_{x\rightarrow \infty }\sqrt{x^{2}-4x}-x}$

a) -∞

b) -2

c) -4

d) 0

Solution: $\lim_{x\rightarrow \infty }\frac{\sqrt{x^{2}-4x}-x}{1}*\frac{\sqrt{x^{2}-4x}+x}{\sqrt{x^{2}-4x}+x}$

$\lim_{x\rightarrow \infty }\frac{\left (\sqrt{x^{2}-4x} \right )^{2}-x^{2}}{\sqrt{x^{2}-4x}+x}$

$\lim_{x\rightarrow \infty }\frac{x^{2}-4x -x^{2}}{\sqrt{x^{2}-4x}+x}=\frac{-4x}{\sqrt{x^{2}-4x}+x}$

$\lim_{x\rightarrow \infty }\frac{-4x}{x(\sqrt{1-\frac{4}{x}}+1)} = \frac{-4}{\sqrt{1-\frac{4}{x}}+1}$

Applying limit as x tends to infinity , we get

$\frac{-4}{\sqrt{1-0}+1} = \frac{-4}{2} = -2$

Answer is option b) -2

2. $\large {\color{Red} \lim_{x\rightarrow \infty }\frac{sinx}{x}}$

a) 0

b) 1

c) ∞

d) DNE

Solution: we know that sinx ranges from -1 to 1

-1 ≤ sinx ≤ 1

$\frac{-1}{x}\leq \frac{sinx}{x}\leq \frac{1}{x}$

$\lim_{x\rightarrow \infty }\frac{-1}{x}\leq \lim_{x\rightarrow \infty }\frac{sinx}{x}\leq \lim_{x\rightarrow \infty }\frac{1}{x}$

Using Squeeze theorem of limits , we know that

$\lim_{x\rightarrow \infty }\frac{-1}{x} = 0 , \lim_{x\rightarrow \infty }\frac{1}{x}=0$

Therefore ,

$\lim_{x\rightarrow \infty }\frac{sinx}{x}= 0$

So the answer is option a) 0

3. $\large {\color{Red} \lim_{x\rightarrow 2^{-}}\frac{x^2 -2x}{|x^2-x-2|}}$

a) -∞

b) -2/3

c) 2/3

d) 0

Solution: To find limit , we define absolute value function first.

$|x^2 -x-2|=\begin{cases} (x-2)(x+1) & \text{ if } x\leq -1 \\ -(x-2)(x+1)& \text{ if } -1<x<2 \\ (x-2)(x+1))& \text{ if } x\geq 2 \end{cases}$

To find left hand limit at x=2 we use function |x^2 -x -2|= -(x-2)(x+1)

$\lim_{x\rightarrow 2^{-}}\frac{x^2 -2x}{|x^2-x-2|} = \lim_{x\rightarrow 2}\left [\frac{x(x-2)}{-(x-2)(x+1)} \right ]$

$= \lim_{x\rightarrow 2}\left [\frac{x}{-(x+1)} \right ]$

$=\frac{2}{-(2+1)} = -\frac{2}{3}$

Answer is option b) -2/3

4. If $\large {\color{Red} \lim_{x\rightarrow a}\frac{\tan^{-1}(2x)-\tan^{-1}(2a)}{x-a}=\frac{2}{3}}$ then a could equal

a) √2

b) 1

c) √2/2

d) 1/2

Solution: We know that this is limit definition formula for derivative of tan(2x) at x= a. Therefore,

$\frac{\mathrm{d} \tan^{-1}(2a)}{\mathrm{d} x} = \frac{2}{3}$

$\frac{2}{1+4a^{2}} = \frac{2}{3}$

Cross multiplying , we get

$6= 2(1+4a^2)$

$3= 1+4a^2$

$2=4a^2$

$\frac{1}{2}=a^2 => a= \frac{\sqrt{2}}{2}$

Answer is option c) √2/2

Graph of f is shown above. Which of the following is true.

I. $\lim_{x\rightarrow 2^{-}}f(x)=1$

II. $\lim_{x\rightarrow 2^{+}}f(x)= -2$

III. $\lim_{x\rightarrow 2}f(x)= -1$

a) I only

b) II only

c) III only

d) I and II only

e) I,II and III

Solution: From the given graph, we see that graph is approaching to 1 from left at x=2. Therefore left limit is 1.

From right graph is approaching to -2 at x=2 , therefore limit is -2.

So I and II are correct.

But a limit exist only when left and right limits are same. But here left and right limits are different so limit at x=2 doesn’t exist.

Hence III is not correct.

So the correct answer is option d) I and II only.

6. If f(x) is continuous for all x, the maximum number of horizontal asymptotes that the graph of f can have is:

a) 0

b) 1

c) 2

d) None

Solution: Horizontal asymptote is the end behavior of a graph. Since a continuous function have only two ends so we analyze the end behavior of graph by finding limits at – inf and +inf.

Therefore a continuous function can have maximum 2 horizontal asymptotes.

So the correct option is c) 2

7. If f(x) is continuous for all x, the maximum number of vertical asymptotes that the graph of f can have is:

a) 0

b) 1

c) 2

c) No maximum number

Solution: If a function is continuous for all x, then it won’t have any vertical asymptote because a vertical asymptote always breaks the graph in parts and make it discontinuous.

Therefor a function continuous for all x, can’t have any vertical asymptote.

So the answer is option a) 0

8. $\large {\color{Red} f(x)= \frac{x^3 -64}{x-4}}$ has point discontinuity. Assign a value to f(x) that removes discontinuity.

a) 4

b) 64

c) 48

d) -4

Solution: We can see this function has a hole (point discontinuity) at x=4

To remove point discontinuity we can redefine the given function. For that we factor it and simplify.

$\lim_{x\rightarrow 4} \frac{x^3 -64}{x-4}$

$\lim_{x\rightarrow 4} \frac{(x-4)(x^2 +4x+16)}{x-4}$

$\lim_{x\rightarrow 4} (x^2 +4x+16) = 48$

So this function can be redefined as,

$f(x)=\begin{cases} \frac{x^3 -64}{x-4} & \text{ if } x\neq 4\\ 48 & \text{ if } x= 4 \end{cases}$

Therefore correct answer is option c) 48

9. Find $\large {\color{Red} \lim_{x\rightarrow \infty }\frac{1-x}{cosx}}$

a) 2

b) 1

c) 0

d) DNE

Solution;When we plugin limit we get the result as indeterminate form.

Therefore apply L Hospital’s rule,

$\lim_{x\rightarrow \infty }\frac{(1-x)'}{(cosx)'}=\lim_{x\rightarrow \infty }\frac{-1}{-sinx}$

$\lim_{x\rightarrow \infty }\frac{1}{sinx}$

Since sin x is periodic function. It doesn’t approach to any value when x approach to – inf or + inf. Thus the limit does not exist.

So the limit of original function also does not exist.

So answer is option d) DNE

10. Find $\large {\color{Red} \lim_{x\rightarrow \infty }\frac{x^n}{e^x}}$

a) -∞

b) 0

c) ∞

d) DNE

Solution: Let $y = \lim_{x\rightarrow \infty }\frac{x^n}{e^x}$

Taking ln of both sides,

$lny = \lim_{x\rightarrow \infty }ln\left (\frac{x^n}{e^x} \right )$

$lny = \lim_{x\rightarrow \infty }ln\left (x^n \right )-ln\left ( e^x \right )$

$lny = \lim_{x\rightarrow \infty }nln\left (x \right )-x$

$lny = \lim_{x\rightarrow \infty }x\left ( \frac{n lnx}{x} -1\right )$

$lny = \lim_{x\rightarrow \infty }x* \lim_{x\rightarrow \infty } \left ( \frac{n lnx}{x} -1\right )$

$lny = \lim_{x\rightarrow \infty }x* n\lim_{x\rightarrow \infty } \left ( \frac{ lnx}{x} \right )-1$

Using L Hospital’s rule $\lim_{x\rightarrow \infty }\frac{lnx}{x} =0$

ln y= ∞ * (0-1)

lny = -∞

$y= e^{-\infty }= 0$

Therefore limit is 0 and correct option is b)

Practice worksheet

$\dpi{120} \lim_{x\rightarrow 5}\frac{|5-x|}{5-x}$
$\dpi{120} \lim_{x\rightarrow -\infty }\frac{-6+(\frac{2}{x})}{7-(\frac{1}{x^2})}$
$\dpi{120} \lim_{x\rightarrow 0^{-}}\frac{8x}{|x|}$
$\dpi{120} \lim_{x\rightarrow 0}\left [\frac{1}{x^{2}} -\frac{1}{sin^{2}x} \right ]$
$\dpi{150} \lim_{x\rightarrow \infty }e^{\frac{ln2}{1+lnx}}$
If $\dpi{120} \lim_{x\rightarrow a}\frac{x^{5}-a^{5}}{x-a}= 405$ then find all possible values of a.
$\dpi{120} \lim_{x\rightarrow 9}\frac{x^{\frac{3}{2}}- 27}{x-9}$
$\lim_{x\rightarrow 0}\frac{sin2x+sin3x}{2x+sin3x}$
$\dpi{120} \lim_{x\rightarrow \pi }\frac{1+sec^3x}{tan^2x}$ Hint : factor numerator and denominator
$\dpi{120} \lim_{x\rightarrow 0}\frac{3^{2x}-2^{3x}}{x}$

Answers:

DNE
-6/7
-8
-1/3
2
a=-3,3
9/2
1
-3/2
log (9/8)

Adding and Subtracting Polynomials

March 16, 2020/by admin_ct

Before learning how to add or subtract polynomials, we have to be familiar with the concept of like and unlike terms.

Like terms are the terms which have same variables raised to same exponents. For example 4x and 7x are like terms. 5x³ and -x³ are like terms. The terms which are not like , are unlike terms.

We can add or subtract only like terms.

Characteristics of Polynomial Functions

February 16, 2020/by admin_ct

What is Polynomial ?

As its name suggests , ‘Poly” means many and ‘mial’ can be thought of as ‘terms’. So any expression made up of many algebraic terms is called polynomial. These terms are made up of variables and constants. Variable is an alphabetical letter and constant is any natural number. Therefore when many algebraic terms made up of variables and constants, are combined with each other, a polynomial is formed.

A polynomial can be exactly defined as an expression which is a sum or difference of terms each consisting of a variable and constant where variable is raised to a non negative integer power.

When a polynomial is written so that its powers are decreasing , we say that it is in standard form. So polynomial is an expression that can be written in the form…

$\dpi{120} \mathbf{{\color{Red} P(x)=a_{n}x^{n}+.....+a_{2}x^{2}+a_{1}x+a_{0}}}$

Characteristics of a Polynomial:

Each real number $\dpi{120} a_{i}$ is called coefficient. The number $\dpi{120} a_{0}$ not multiplied with any variable is called constant.
Each product $\dpi{120} a_{i}x^{i}$ is called term of polynomial.
The highest power of variable that occur in a polynomial is called degree of polynomial.
The term with highest power is called leading term and the constant attached with it is called leading coefficient.

Example1: Write the following polynomials in standard form and then identify the leading term, leading coefficient and degree .

a. 5+2x

b. $\dpi{120} {\color{Red} 3+x^{3}-5x}$

c. $\dpi{120} {\color{Red} -3x^{5}+2x^4-8x+7}$

Answer:

a) standard form : 2x+5

leading term = 2x

leading coefficient = 2

degree = 1

b) standard form :

$\dpi{120} x^{3}-5x+3$

leading term = x^3

leading coefficient = 1

degree = 3

c) standard form: This is already in standard form.

$\dpi{120} -3x^{5}+2x^4-8x+7$

leading term = -3x^5

leading coefficient = -3

degree = 5

Types of Polynomial depending on the number of terms it contain:

There are mainly three types of polynomial depending on the number of terms it contain and they are,

i) Monomial : When a polynomial expression contain only one term, it is called Monomial

Example: 3x, $\dpi{120} x^{2},5x^{3}$

ii) Binomial : When a polynomial expression contain only two terms, it is called Binomial.

Example: 3x-1 , 7-x^2

iii) Trinomial : As its name indicates , ‘tri’ means three so when a polynomial contain three terms then it is called trinomial.

Example: $\dpi{120} x^{2}-2x+1$ , $\dpi{120} x^{4}-5x^{3}+3$

Types of Polynomial depending on the degree of expression:

Constant polynomial : when degree of en expression is 0 or it is just a constant then it is called constant polynomial.

Example: 8,-16 These are constant polynomials because these terms can be written as 8x^0 and -16x^0

Linear Polynomial: When degree of an expression is 1 then the expression is called linear polynomial.

Example: 3x, 4x-1, 7+3x

Quadratic Polynomial : When degree of polynomial is 2 then it is called quadratic polynomial.

Example: $\dpi{120} x^{2}-2x+1$ , $\dpi{120} 2-9x^{2}, 3x-7x^{2}$

Cubic Polynomial: When degree of polynomial is 3 then it ic called cubic polynomial.

Example: $\dpi{120} x^{3}-12x^{2}+5x-1$ ,

$\dpi{120} 5-4x^{3}$ , $\dpi{120} 3x+x^{3}$

Quartic Polynomial: When degree of polynomial is 4 then it is called Quartic Polynomial.

Example: $\dpi{120} x^{4}+x^{3}$

$\dpi{120} 2x^{4}-5x ,x^{4}-x^{2}$

When an expression is NOT a Polynomial ?

i) A polynomial can’t have negative exponents.

ii) A polynomial can’t have any term divided by a variable.

iii) A polynomial can’t have fractional exponents.

iv) A polynomial can’t have variable under sqrt sign.

Examples of expressions which are not Polynomial:

$\dpi{120} x^3+3x^{-2}+2x-1$

$\dpi{120} 5x^{2}-\frac{1}{x}+4$

$\dpi{120} x^{\frac{1}{3}}-2x+1$

$\dpi{120} x^{2}-5\sqrt{x}+1$

What is the degree of a Polynomial in two variables?

The degree of a polynomial in two variables is the sum of exponents of both variables which make the highest value.

Example:

$\dpi{120} x^{3}y- 6x^{2}y^{4}+4xy-5$
Here the degree is 6 which is the highest sum of exponents of both variables (x and y)

Example2: Identify the following polynomials depending on their degrees and number of terms.

1) ${\color{Red} x^{2}-6x+8}$

Answer: Quadratic , Trinomial

2) ${\color{Red} 7x^{3}-1}$

Answer: Cubic , Binomial

3) 2x+1

Answer: Linear , Binomial

4) 5

Answer: constant , Monomial

5) ${\color{Red} 5x^{3}}$

Answer: Cubic, Monomial

6) ${\color{Red} x^{3}-5x +1}$

Answer: Cubic, Trinomial

Practice problems:

Which of the following expressions are polynomials. Identify the leading coefficient and degree of the ones which are polynomials.

$4x^{2}-x^{6}+5x-3$
$4x^{3}-\sqrt{x}-1$
$2x^{3}-x^{-2}+1$
$x^{5}-x^{3}+5$
$5-3x$

Answers:

Yes, leading coefficient =-1, degree =6
No
No
Yes, leading coefficient = 1, degree=5
Yes, leading coefficient =-3 , degree =1

Surface Area

February 6, 2020/by admin_ct

How to find the surface area of solid of revolution ?

If we revolve a curve y=f(x) on the interval [a,b] about x axis, then we calculate area of resulting surface by breaking the curve into pieces as we did for arc length. A piece of curve of length dS at an average distance y from x axis traces out a surface that is well approximated by a slice of a cone whose area is approximated by 2πydS. In general surface area if given as,

$\dpi{120} \mathbf{S=\int_{a}^{b}2\pi (radius)(arc length)dx}$

Thus we find that area of entire surface of revolution is,

$\dpi{120} \mathbf{S=\int_{a}^{b}2\pi f(x)\sqrt{1+[f'(x)]^{2}}dx} ;\mathbf{a\leq x\leq b}$

If the curve is rotated about y axis then surface area is given as,

$\dpi{120} \mathbf{S=\int_{a}^{b}2\pi x\sqrt{1+[f'(x)]^{2}}dx};\mathbf{a\leq x\leq b}$

Above formula can also be rewritten in terms of y as,

$\dpi{120} \mathbf{S=\int_{a}^{b}2\pi f(y)\sqrt{1+[f'(y)]^{2}}dy};\mathbf{0\leq y\leq b}$

One thing to be noted about these formulas that variable of integral is always the opposite to one we are rotating the curve about.

Example1: The given curve is rotated about the y axis. Find the area of resulting surface.

$\dpi{120} {\color{Red} y=\frac{1}{4}x^{2}-\frac{1}{2}lnx;3\leq x\leq 4}$

Solution:

$\dpi{120} f(x)=\frac{1}{4}x^{2}-\frac{1}{2}lnx$

$\dpi{120} f'(x)=\frac{1}{4}2x-\frac{1}{2}*\frac{1}{x} =\frac{x}{2}-\frac{1}{2x}$

Plugin the values into the formula we get,

$\dpi{120} S=\int_{3}^{4}2\pi x\sqrt{1+\left (\frac{x}{2}-\frac{1}{2x} \right )^{2}}dx$

$\dpi{120} S=\int_{3}^{4}2\pi x\sqrt{1+\frac{x^{2}}{4}-\frac{1}{2} +\frac{1}{4x^{2}}}dx$

$\dpi{120} S=\int_{3}^{4}2\pi x\sqrt{\frac{x^{2}}{4}+\frac{1}{2} +\frac{1}{4x^{2}}}dx$

$\dpi{120} S=\int_{3}^{4}2\pi x\sqrt{\left (\frac{x}{2}+\frac{1}{2x} \right )^{2}}dx$

$\dpi{120} S=\int_{3}^{4}2\pi x\left (\frac{x}{2}+\frac{1}{2x} \right )dx$

$\dpi{120} S=2\pi \int_{3}^{4} \frac{x^{2}}{2}+\frac{1}{2} dx$

$\dpi{120} S=\pi \int_{3}^{4} x^{2}+1 dx$

$\dpi{120} S=\pi \left [ \frac{x^{3}}{3}+x \right ]_{3}^{4}$

$\dpi{120} S=\pi (\frac{37}{3}+1)=\frac{40\pi }{3}$

Example2: Find the area of the surface obtained by revolving the portion of the curve y=x³ from x=0 to x=1 about x axis.

Solution: Here we have f(x)= x³

so f'(x)= 3x²

Plugin the values into the formula, we get

$\dpi{120} S=\int_{a}^{b}2\pi f(x)\sqrt{1+[f'(x)]^{2}}dx$

$\dpi{120} S=\int_{0}^{1}2\pi x^{3}\sqrt{1+\left (3x^{2} \right )^{2}}dx$

$\dpi{120} S=2\pi \int_{0}^{1} x^{3}\sqrt{1+9x^{4}}dx$

Using substitution , let

$\dpi{120} u=1+9x^{4}$

du=36x³dx

du/36=x³dx

When x=0 , u=1+0= 1

When x=1, u=1+9=10

So new limits in terms of u are 1 to 10. we get the new integral in terms of u as,

$\dpi{120} S=2\pi \int_{1}^{10} \sqrt{u}*\frac{du}{36}=\frac{2\pi }{36}\int_{1}^{10} \sqrt{u}du$

$\dpi{120} S=\frac{\pi }{18}\left [ \frac{2}{3} u^{\frac{3}{2}}\right ]_{1}^{10}$

$\dpi{120} S=\frac{\pi }{27}\left [ (10)^{\frac{3}{2}-1} \right ] = 3.56$

Surface Area when revolved about a line other than coordinate axis.

Example3:Find the surface area of the surface formed by revolving the curve x = sin 2t, y = cos 3t, for 0 ≤ t ≤ π/3, about the line x=2.

Solution: First we sketch the curve given by these parametric equations. Arc length is easy to find , the only thing about which we need to be careful, is radius. Since all x values on the curve are less than 2 , so radius of revolution is

radius = 2-x => 2-sin2t

So the surface are is given as,

$\dpi{120} S=\int_{a}^{b}2\pi (radius)(arc length)dx$

$\dpi{120} S=\int_{0}^{\frac{\pi }{3}}2\pi (2-sin2t)\left ( \sqrt{(2cos2t)^{2}+(-3sin3t)^{2}} \right )dt$

$\dpi{120} S=2\pi\int_{0}^{\frac{\pi }{3}} (2-sin2t)\left ( \sqrt{4cos^{2}(2t)+ 9sin^{2}(3t)} \right )dt$

S≈20.1 sq. units

integral can be calculated numerically .

Example4: Find the area of the surface obtained by rotating the curve about x axis.

$\dpi{120} {\color{Red} x=\frac{1}{3}\left ( y^{2} +2\right )^{\frac{3}{2}};1\leq y\leq 2}$

Solution:

$\dpi{120} x=\frac{1}{3}\left ( y^{2} +2\right )^{\frac{3}{2}}$

$\dpi{120} x'=\frac{1}{3}*\frac{3}{2}\left ( y^{2} +2\right )^{\frac{3}{2}-1}(y^{2}+2)'$

$\dpi{120} x'=\frac{1}{2}\left ( y^{2} +2\right )^{\frac{1}{2}}(2y)= y\sqrt{y^{2}+2}$

Using the formula for Surface area we get,

$\dpi{120} S=\int_{1}^{2}2\pi y\sqrt{1+(y\sqrt{y^{2}+2})^{2}}dy$

$\dpi{120} S=\int_{1}^{2}2\pi y\sqrt{1+y^{4}+2y^{2}}dy$

$\dpi{120} S=\int_{1}^{2}2\pi y\sqrt{\left (1+y^{2} \right )^{2}}dy=\int_{1}^{2}2\pi y\left (1+y^{2} \right )dy$

$\dpi{120} S=2\pi \int_{1}^{2} y+y^{3} \right )dy$

$S=2\pi \left [ \frac{y^{2}}{2}+\frac{y^{4}}{4} \right ]_{1}^{2}$

$S=2\pi \left [ \left ( \frac{4}{2}+\frac{16}{4}\right )-\left (\frac{1}{2} +\frac{1}{4} \right ) \right ]=\frac{21\pi }{2}$

Example5: Find the surface area of resulting surface when given curve is rotated about y axis.

$\dpi{120} {\color{Red} x=\sqrt{a^{2}-y^{2}}; 0\leq y\leq \frac{a}{2}}$

Solution:

$\dpi{120} f(y)=\sqrt{a^{2}-y^{2}}$

$\dpi{120} f'(y)=\frac{1}{2}(a^{2}-y^{2})^{\frac{-1}{2}}(-2y)=\frac{-y}{\sqrt{a^{2}-y^{2}}}$

Using the formula for surface area, we get

$\dpi{120} S=\int_{a}^{b}2\pi f(y)\sqrt{1+[f'(y)]^{2}}dy$

$\dpi{120} S=\int_{0}^{\frac{a}{2}}2\pi \sqrt{a^{2}-y^{2}}\sqrt{1+\left ( \frac{-y}{\sqrt{a^{2}-y^{2}}} \right )^{2}}dy$

$\dpi{120} S=\int_{0}^{\frac{a}{2}}2\pi \sqrt{a^{2}-y^{2}}\sqrt{1+ \frac{y^{2}}{a^{2}-y^{2}} }dy$

$\dpi{120} S=\int_{0}^{\frac{a}{2}}2\pi \sqrt{a^{2}-y^{2}}\sqrt{\frac{a^{2}-y^{2}+y^{2}}{a^{2}-y^{2}} }dy$

$\dpi{120} S=\int_{0}^{\frac{a}{2}}2\pi \sqrt{a^{2}-y^{2}}\sqrt{\frac{a^{2}}{a^{2}-y^{2}} }dy$

$\dpi{120} S=\int_{0}^{\frac{a}{2}}2\pi \sqrt{a^{2}-y^{2}}\frac{a}{\sqrt{a^{2}-y^{2}}} dy$

$\dpi{120} S=2\pi \int_{0}^{\frac{a}{2}}a dy$

$\dpi{120} S=2\pi(a) \left [ y\right ]_{0}^{\frac{a}{2}}$

$\dpi{120} S=2\pi(a) \left [ \frac{a}{2}-0\right ]=\pi a^{2}$

Practice problems:

Find the area of the surface obtained by rotating the curve about x axis.

y=x³ ; 0≤x≤2
y=coshx ; 0≤x≤1

Find the area of the surface obtained by rotating the curve about y axis.

3. y= 1-x² ; 0≤x≤1

4. y=lnx ; 0≤x≤1

Answers:

$\dpi{120} \pi \frac{1}{21}\left ( 145\sqrt{145}-1 \right )$
$\dpi{120} \pi \left [ 1+\frac{1}{4}\left (e^{2} -e^{-2} \right ) \right ]$
π/6
$\dpi{120} \pi \left ( \sqrt{2} +ln(1+\sqrt{2})\right )$

Arc length

January 30, 2020/by admin_ct

How to find arc length of a curve ?

Length of a curve is called arc length. To find arc length of a curve defined by function f(x) over a certain interval [a,b] we use the following formula.

$\dpi{120} \mathbf{L=\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx}$

Example1: Find the length of the curve given by the following function from x=0 to x=1

$\dpi{120} {\color{Red} y=2x^{\frac{3}{2}}}$

Solution;

$\dpi{120} f(x)=2x^{\frac{3}{2}}$

$\dpi{120} f'(x)=2*\frac{3}{2}x^{\frac{1}{2}}= 3\sqrt{x}$

Plugin the values into the formula and we get the integral as,

$\dpi{120} L=\int_{0}^{1}\sqrt{1+(3\sqrt{x})^{2}}dx$

$\dpi{120} L=\int_{0}^{1}\sqrt{1+9x}dx$

To solve this integral using U substitution, we get

u = 1+9x

du= 9dx

du/9= dx , also the limits get changed to 1 to 10 in terms of u.

So we have the integral as,

$\dpi{120} L=\frac{1}{9}\int_{1}^{10}\sqrt{u}du$

$\dpi{120} L=\frac{1}{9}*\frac{2}{3}[u^{\frac{3}{2}}]_{1}^{10}$

$\dpi{120} L=\frac{2}{27}[(10)^{\frac{3}{2}}-1]$

L=2.27 units

Example2: Find the exact length of curve

$\dpi{120} {\color{Red} x=\frac{y^{4}}{8}+\frac{1}{4y^{2}} : 1\leq y\leq 2}$

Solution: Here we have the function in terms of y , so we use following formula

$\dpi{120} L=\int_{a}^{b}\sqrt{1+[f'(y)]^{2}}dy$

$\dpi{120} f(y)=\frac{y^{4}}{8}+\frac{1}{4y^{2}}$

$\dpi{120} f'(y)=\frac{4y^{3}}{8}+\frac{-2}{4y^{3}}=\frac{y^{3}}{2}-\frac{1}{2y^{3}}$

$\dpi{120} L=\int_{1}^{2}\sqrt{1+\left (\frac{y^{3}}{2}-\frac{1}{2y^{3}} \right )^{2}}dy$

$\dpi{120} L=\int_{1}^{2}\sqrt{1+\frac{y^{6}}{4}-\frac{1}{2}+\frac{1}{4y^{6}} }dy$

$\dpi{120} L=\int_{1}^{2}\sqrt{\frac{y^{6}}{4}+\frac{1}{2}+\frac{1}{4y^{6}} }dy$

$\dpi{120} L=\int_{1}^{2}\sqrt{\left (\frac{y^{3}}{2}+\frac{1}{2y^{3}} \right )^{2}}dy$

$\dpi{120} L=\int_{1}^{2}\frac{y^{3}}{2}+\frac{1}{2y^{3}}dy$

$\dpi{120} L=\left [\frac{y^{4}}{8}+\frac{1}{2}\frac{y^{-2}}{-2} \right ]_{1}^{2}=\left [\frac{y^{4}}{8}-\frac{1}{4y^{2}} \right ]_{1}^{2}$

$\dpi{120} L=\left (\frac{16}{8}-\frac{1}{16} \right )-\left ( \frac{1}{8}-\frac{1}{4} \right ) =\frac{33}{16}$ units

Example3: Compute the exact arc length of the curve over the given interval

$\dpi{120} {\color{Red} y=\frac{x^{2}}{2}-\frac{lnx}{4}; 2\leq x\leq 4}$

Solution: First find the derivative y’.

$\dpi{120} y'=\frac{2x}{2}-\frac{l}{4x}= x-\frac{1}{4x}$

Plugin value of y’ into the formula and we get,

$\dpi{120} L=\int_{2}^{4}\sqrt{1+\left ( x-\frac{1}{4x} \right )^{2}}dx$

$\dpi{120} L=\int_{2}^{4}\sqrt{1+ x^{2}-\frac{1}{2}+\frac{1}{16x^{2}} }dx$

$\dpi{120} L=\int_{2}^{4}\sqrt{ x^{2}+\frac{1}{2}+\frac{1}{16x^{2}} }dx$

$\dpi{120} L=\int_{2}^{4}\sqrt{\left ( x+\frac{1}{4x} \right )^{2}}dx=\int_{2}^{4} x+\frac{1}{4x} dx$

$\dpi{120} L=\left [ \frac{x^{2}}{2}+\frac{1}{4}ln|x| \right ]_{2}^{4}$

$\dpi{120} L=\left ( 8+\frac{1}{4}ln4 \right )-\left ( 2+\frac{1}{4}ln2 \right )$

$\dpi{120} L=6+\frac{1}{4}(ln4-ln2 )=6+\frac{1}{4}ln(\frac{4}{2})$ θ

$\dpi{120} L=6+\frac{1}{4}ln2$

Arc length of curves in Polar Coordinates

To find arc length of polar equations r(θ) , following formula is used.

$\dpi{120} \mathbf{L=\int_{a}^{b}\sqrt{r^{2}+\left ( \frac{dr}{d\theta } \right )^{2}}d\theta }$

Example4: Find the length of polar curve described by r=9+9cosθ.

Solution: $\dpi{120} r(\theta )=9+9cos\theta$

$\dpi{120} r'(\theta )= -9sin\theta$

Curve is traced out once from 0 to 2π. But curve over -π<x<0 is mirror image of upper half of curve over the interval 0<x<π. So we can integrate over 0 to π and multiply the result with 2.

Plugin the values of r and r’ into the formula and we get,

$\dpi{120} L=2\int_{0}^{\pi }\sqrt{(9+9cos\theta )^{2}+\left ( -9sin\theta \right )^{2}}d\theta$

$\dpi{120} L=2\int_{0}^{\pi }\sqrt{81{(1+cos\theta )^{2}+\left 81(sin\theta \right )^{2}}}d\theta$

$\dpi{120} L=2\int_{0}^{\pi }9\sqrt{{1+cos^{2}\theta +2cos\theta +sin^{2}\theta }}d\theta$

$\dpi{120} L=18\int_{0}^{\pi }\sqrt{2 +2cos\theta}d\theta=18\sqrt{2}\int_{0}^{\pi }\sqrt{1 +cos\theta}d\theta$

$\dpi{120} L=18\sqrt{2}\int_{0}^{\pi }\sqrt{2cos^{2}(\frac{\theta }{2})}d\theta$

$\dpi{120} L= 36\int_{0}^{\pi }cos(\frac{\theta }{2})d\theta$

$\dpi{120} L=36 \left [2 sin\left ( \frac{\theta }{2} \right )\right ]_{0}^{\pi }$

$\dpi{120} L=72 \left (sin\frac{\pi }{2}-sin\frac{0}{2} \right )=72(1-0)$

Arc length = 72 units

Example5: Find the exact length of spiraling polar curve given below on the interval 0≤θ≤2π

$\dpi{120} {\color{Red} r=3e^{4\theta }}$

Solution:

$\dpi{120} r'=3e^{4\theta }(4)=12e^{4\theta }$

Plugin the values of rand r’ into the formula and we get,

$\dpi{120} L=\int_{0}^{2\pi }\sqrt{\left ( 3e^{4\theta } \right )^{2}+\left ( 12e^{4\theta } \right )^{2}}d\theta$

$\dpi{120} L=\int_{0}^{2\pi }\sqrt{9e^{8\theta }+144e^{8\theta }}d\theta$

$\dpi{120} L=\int_{0}^{2\pi }\sqrt{153e^{8\theta }}d\theta =\sqrt{153}\int_{0}^{2\pi }\sqrt{e^{8\theta }}d\theta$

$\dpi{120} L=\sqrt{153}\int_{0}^{2\pi }e^{4\theta }d\theta$

$\dpi{120} L=\sqrt{153}\left [ \frac{e^{4\theta }}{4} \right ]_{0}^{2\pi }$

$\dpi{120} L=\frac{\sqrt{153}}{4}\left [ e^{4(2\pi) }-e^{0} \right ]$

$\dpi{120} L=\frac{\sqrt{153}}{4}\left [ e^{8\pi }-1 \right ]$

Arc length of curves in Parametric form

To find arc length of curves given by parametric equations , following formula is used

$\dpi{120} \mathbf{L=\int_{a}^{b}\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt } \right )^{2}}dt }$

Example6: Find the exact arc length of a curve defined by parametric equations

$\dpi{120} {\color{Red} x(t)=e^{-t}cost ; y(t)=e^{-t}sint ;for 0\leq t\leq \frac{\pi }{2}}$

Solution:

$\dpi{120} \frac{dx}{dt}=(e^{-t})'cost +e^{-t}(cost)'$

$\dpi{120} \frac{dx}{dt}=-e^{-t}cost -e^{-t}sint =-e^{-t}(cost+sint)$

$\dpi{120} \frac{dy}{dt}=(e^{-t})'sint +e^{-t}(sint)'$

$\dpi{120} \frac{dy}{dt}=-e^{-t}sint +e^{-t}cost =e^{-t}(cost-sint)$

Plugin the values into the formula

$\dpi{120} L=\int_{0}^{\frac{\pi }{2}}\sqrt{\left ( -e^{-t}(cost+sint) \right )^{2}+\left ( e^{-t}(cost-sint) \right )^{2}}dt$

$\dpi{120} L=\int_{0}^{\frac{\pi }{2}}\sqrt{\left ( e^{-2t}(cos^{2}t+2costsint+sin^{2}t) \right )+\left ( e^{-2t}(cos^{2}t-2costsint+sin^{2}t) \right )}dt$

$\dpi{120} L=\int_{0}^{\frac{\pi }{2}}\sqrt{\left ( e^{-2t}(2cos^{2}t+2sin^{2}t) \right )}dt$

$\dpi{120} L=\int_{0}^{\frac{\pi }{2}}\sqrt{ e^{-2t}(2) }dt =\sqrt{2}\int_{0}^{\frac{\pi }{2}}\sqrt{ e^{-2t} }dt$

$\dpi{120} L=\sqrt{2}\int_{0}^{\frac{\pi }{2}}\sqrt{( e^{-t})^{2} }dt=\sqrt{2}\int_{0}^{\frac{\pi }{2}} e^{-t} dt$

$\dpi{120} L=\sqrt{2}\left [-e^{-t} \right ]_{0}^{\frac{\pi }{2}}$

$\dpi{120} L=-\sqrt{2}\left [ e^{\frac{-\pi }{2}}-1 \right ]$

$\dpi{120} L=\sqrt{2}\left [ 1-e^{\frac{-\pi }{2}} \right ]$

Arc Length of vector functions

Formula for arc length of vector functions is just another way of writing arc length formula for parametric curves.

For plane curve

Let $\dpi{120} \overrightarrow{r}(t)= \left \langle x(t),y(t) \right \rangle$ be the vector equation for a plane curve , then its arc length on interval [a,b] can be calculated as,

$\dpi{120} \mathbf{L=\int_{a}^{b}\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt } \right )^{2}}dt }$

which can also be written as

$\dpi{120} \mathbf{L=\int_{a}^{b}\left \| r'(t) \right \|dt}$

For space curve

Let $\dpi{120} \overrightarrow{r}(t)= \left \langle x(t),y(t),z(t) \right \rangle$ be the vector equation for a space curve , then its arc length on interval [a,b] can be calculated as,

$\dpi{120} \mathbf{L=\int_{a}^{b}\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt } \right )^{2}+\left ( \frac{dz}{dt} \right )^{2}}dt }$

Which can also be written as,

$\dpi{120} \mathbf{L=\int_{a}^{b}\left \| r'(t) \right \|dt}$ where

$\dpi{120} \mathbf{\left \| r'(t) \right \|=\sqrt{\left ( x'(t) \right )^{2}+\left ( y'(t) \right )^{2}+\left ( z'(t) \right )^{2}}}$

Example7: Find the arc length of following vector valued function.

$\dpi{120} {\color{Red} \overrightarrow{r}(t)=\left \langle tcost,tsint,2t \right \rangle ;0\leq t\leq 2\pi }$

Solution: First we find r'(t)

$\dpi{120} \overrightarrow{r}'(t)=\left \langle cost-tsint, sint +tcost,2\right \rangle$

$\dpi{120} \left \| r'(t) \right \|=\sqrt{\left (cost-tsint\right )^{2}+\left ( sint+tcost \right )^{2}+\left ( 2\right )^{2}}$

$\dpi{120} \left \| r'(t) \right \|=\sqrt{\left cos^{2}t-2tcostsint+t^{2}sin^{2}t+ sin^{2}t+2tcostsint+t^{2}cos^{2}t +4}$

$\dpi{120} \left \| r'(t) \right \|=\sqrt{\left cos^{2}t+t^{2}sin^{2}t+ sin^{2}t+t^{2}cos^{2}t +4}$

$\dpi{120} \left \| r'(t) \right \|=\sqrt{\left cos^{2}t+sin^{2}t+t^{2}(sin^{2}t+ cos^{2}t) +4}$

$\dpi{120} \left \| r'(t) \right \|=\sqrt{\left1+t^{2}(1) +4}$

$\dpi{120} \left \| r'(t) \right \|=\sqrt{\left5+t^{2}}$

Using arc length formula we have

$\dpi{120} L=\int_{0}^{2\pi }\sqrt{t^{2}+5}dt$

Using table integral formula we get,

$\dpi{120} L=\left [ \frac{t}{2}\sqrt{t^{2}+5}+\frac{5}{2}ln|t+\sqrt{t^{2}+5}|\right ]_{0}^{2\pi }$

$\dpi{120} L=\left [ \frac{2\pi }{2}\sqrt{4\pi ^{2}+5}+\frac{5}{2}ln|2\pi +\sqrt{4\pi ^{2}+5}|\right ]-\left [ 0+\frac{5}{2}ln\sqrt{5} \right ]$

L= 25.343 units

Practice problems:

Find arc length of following functions on the given intervals

1. $\dpi{120} y=\frac{x^{5}}{6}+\frac{1}{10x^{3}} ;1\leq x\leq 2$

2. y= lnx ; 1≤x≤2

2. $\dpi{120} \overrightarrow{r}(t)=\left \langle 2t,3cos2t,3sin2t \right \rangle ;0\leq t\leq 2\pi$

3. $\dpi{120} r=2+2cos\theta ;0\leq \theta \leq 2\pi$

Answers:

1261/240 units
1.22 units

3. 4√10 π

4. 16 units

Fluid Pressure and Force

January 24, 2020/by admin_ct

Fluid Pressure

The pressure on an object at depth h in a liquid is given as,

P=wh

Where w= weight density of liquid per unit of volume.

Pascal’s principle states that pressure exerted by a fluid at a depth h is transmitted equally in all directions. Fluid force on a submerged horizontal surface of area A is ,

F= P*A

Weight density of water= 62.4 lb/cubic ft. OR 9800N/cubic m.

Rectangular plate submerged horizontally

If the plate is submerged horizontally then pressure is constant and can be calculated using the equation F=P*A. We don’t need calculus for this which is illustrated by the following example.

Example1: Find the fluid force on a rectangular metal sheet measuring 3ft by 4ft, that is submerged horizontally in 6 ft. of water in a tank.

Solution: P=wh

P=(62.4)(6)= 374.4 lb/ sq.ft.

F=P*A = (374.4)(12)= 4492.8 lbs

Rectangular plate submerged vertically

If plate is submerged vertically then pressure is not constant throughout, so force acting on it must be calculated through slicing which leads to integral. So the force exerted by a fluid of constant weight density w, against a submerged vertical plane from y=a to y=b is given as,

$\dpi{120} \mathbf{F=w\int_{a}^{b}h(y)L(y)dy}$

Where h(y) is depth of fluid at y and L(y) is the horizontal length of region at y.

Example2: Consider a swimming pool 5 m wide, 10 m long and 3m deep. If pool is completely filled with water find the force on

a) bottom of the pool.

b) One of the 5×3 ends of the pool.

Solution: a) As the bottom of pool act like a horizontal plate submerged in water , so pressure on bottom would be constant and can be calculated as,

Force (F)= P*A

F=9800(3)(5)(10)=1,470,000 N

b) Since 5×3 end of pool act like a vertical plate. lets calculate the force on one slice of width Δy which is y m deep down.

Pressure on slice= 9800y

Area of slice= 5Δy

Force

$\dpi{120} F=\int_{0}^{3}9800 y(5)dy$

$\dpi{120} F=49000\int_{0}^{3} ydy$

$\dpi{120} F=49000\left [ \frac{y^{2}}{2} \right ]_{0}^{3}$

$\dpi{120} F= 24500[9-0]=220,500N$

Example3: The vertical ends of a tank are semicircles of radius 10 ft. with diameter running along the top. If the tank is filled to the top with water, find the force on one end of the tank.

Solution:

lets assume a strip of width Δy at a dept of y ft. inside the tank.

Pressure along strip = 62.4 y

Area of strip = 2x Δy

Using Pythagoras we have ,

$\dpi{120} x^{2}+y^{2}=100$

$\dpi{120} x=\sqrt{100-y^{2}}$

Force on strip = Pressure x Area

$\dpi{120} F= 62.4 y*2\sqrt{100-y^{2}}\Delta y$

Total Force

$\dpi{120} F=\int_{0}^{10}62.4 y (2\sqrt{100-y^{2}})dy$

$\dpi{120} F=124.8\int_{0}^{10}y (\sqrt{100-y^{2}})dy$

This integral can be solved further using u substitution.

let 100-y^2= u

-2y dy= du

y dy=-du/2

Integral limits also get changed to 100 to 0 in terms of u

$\dpi{120} F= -62.4\int_{100}^{0}\sqrt{u}du$

$\dpi{120} F= -62.4\left [ \frac{2}{3} u^{\frac{3}{2}}\right ]_{100}^{0}$

$\dpi{120} F= -62.4\left (\frac{2}{3} \right )\left [ 0-1000 \right ]= 41600$ lbs

Examnple4: A vertical gate in a dam of a freshwater lake has the shape of an isosceles trapezoid 8 feet across the top and 6 feet across the bottom, with a height of 5 feet, as shown in the figure to the right. What is the fluid force on the gate when the top of the gate is 4 feet below the surface of the water?

Solution: First of all we find the equation of right side of trapezium. Assuming origin as reference point on the surface of water we get the coordinates at top right and bottom right points as (4,-4) and (3,-9).

$m=\frac{-9+4}{3-4}=5$

y+4=5(x-4)

y=5x-24

$x=\frac{y+24}{5}$

Lets assume a slice of width Δy which is y ft down the surface,

So depth of slice h(y)=-y

Length of slice=2x =2(y+25)/5

Force on slice

$F= 62.4 (-y)[\frac{2(y+24)}{5}]\Delta y$

Total Force

$F= 62.4 \int_{-9}^{-4}(-y)[\frac{2(y+24)}{5}]dy$

$F= -62.4\frac{2}{5} \int_{-9}^{-4}(y^{2}+24y) dy$

$F= -24.96\left [ \frac{y^{3}}{3}+12y^{2} \right ]_{-9}^{-4}$

F= 13,936 lbs

Example5: A circular observation window on a marine science ship has a radius of 1 foot, and the center of the window is 8 feet below seawater level, as shown in the figure to the right. What is the fluid force on the window?

Solution:

Lets assume a slice of width Δy which is 8-y ft below the water.

Pressure on slice=w*h =62.4 (8-y)

Window is a circle of radius 1 ft, so we can use circle equation with radius r=1.

$x^{2}+y^{2}=1$

$x=\sqrt{1-y^{2}}$

Length of slice=2x

Area of slice = length*width = 2xΔy

So the force on slice

$F=62.4(8-y)*2\sqrt{1-y^{2}}\Delta y$

Total Force

$F=\int_{-1}^{1}62.4(8-y)*2\sqrt{1-y^{2}}dy$

$F=124.8\int_{-1}^{1}(8-y)\sqrt{1-y^{2}}dy$

$F=124.8\int_{-1}^{1}8\sqrt{1-y^{2}}dy + 124.8\int_{-1}^{1}y\sqrt{1-y^{2}}dy$

Using formula from table of integrals for first part and u substitution for second part we get,

$F=124.8(8)\left [ \frac{y}{2}\sqrt{1-y^{2}}+\frac{1}{2}sin^{-1}(y) \right ]_{-1}^{1} + 124.8\left (\frac{1}{3} \right )\left [ u \right ]_{-1}^{1}$

$F=124.8(8)\left [ 0+\frac{\pi }{2} \right ]+0$

F= 1568.28 lbs

Practice problems:

A 4 ft high and 5 ft wide rectangular plate is submerged vertically in water so that top is 1 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it.
A semicircular plate with radius 10m is submerged vertically in water so that top is 2m above the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it.
Find the hydrostatic force on a circular plate of radius 2 that is submerged 6 ft in the water.
Determine the hydrostatic force on a triangular plate which is in the shape of isosceles triangle with equal sides as 4m each and unequal side as 6m completely submerged in water.

Answers:

1. 3750 lbs

2. $3.84\times 10^{6}$ N

3. 313920π lbs

4. 156960 N

Mass, Work and Energy

December 27, 2019/by admin_ct

Integration is used to calculate mass of a given object based on a density function. We can calculate mass of a one dimensional and two dimensional object using density function.

Calculating mass of one dimensional object:

Lets consider a rod thin enough to be treated as one dimensional object, oriented along x axis over the interval [a,b]. Let ρ(x) denote the linear density function, then total mass of rod is given by,

$\dpi{120} \mathbf{M=\int_{a}^{b}\rho (x) dx}$

Example1: Find the total mass of a 1 m rod whose linear density function is

$\dpi{120} {\color{Red} \rho (x)= 10(x+1)^{-3} kg/m}$ for 0≤x≤1

Solution: Plugin the values into the formula above,

$\dpi{120} M=\int_{0}^{1}\ 10(x+1)^{-3} dx$

$\dpi{120} M=10\left [ \frac{(x+1)^{-2}}{-2} \right ]_{0}^{1}$

$\dpi{120} M= -5\left [ \frac{1}{(x+1)^{2}} \right ]_{0}^{1}$

$\dpi{120} M=-5\left [ \frac{1}{4}-1\right ]=\frac{15}{4}$

So the total mass of the rod is 3.75 kg

Mass of a two dimensional object:

To find mass of a two dimensional disk of radius r, we use area density function, which is actually the radial density function. If ρ(x) is the radial density function of given disk of radius r then the formula to calculate mass is given as,

$\dpi{120} \mathbf{M=\int_{0}^{r} 2\pi x\rho (x) dx}$

Example2: The radial density function for a city’s population( in thousands of people per square mile) is

$\dpi{120} {\color{Red} \rho (x)= 20(1+x^{2})^{-\frac{1}{2}}}$ .How many people live within 5 miles of the city’s center?

Solution: Plugin the radial density function into the formula above and we get total Population as,

$\dpi{120} P=\int_{0}^{5} 2\pi x*20(1+x^{2})^{\frac{-1}{2}} dx$

$\dpi{120} P= 40 \pi \int_{0}^{5} x(1+x^{2})^{\frac{-1}{2}} dx$

Now we solve this integral using u substitution,

Let u=1+x^2

du = 2x dx

du/2 = x dx

and the limits get converted as,

For x=0 , u=1+0= 1

For x=5, u=1+5^2 = 26 , so we got new limits of u from 1 to 26

$\dpi{120} P= 40 \pi \int_{1}^{26} u^{\frac{-1}{2}} \frac{du}{2}$

$\dpi{120} P= 20 \pi \int_{1}^{26} u^{\frac{-1}{2}}du =20\pi \left [ 2u^{\frac{1}{2}} \right ]_{1}^{26}$

$\dpi{120} P=40\pi (\sqrt{26}-1)= 515.098$

Work and Energy:

Work is said to be done if a force F , working on an object displaces the body through some distance dx. Let F(x) represents the force at point x, then the work done over the interval [a,b] is given as,

$\dpi{120} \mathbf{W=\int_{a}^{b}F(x) dx}$

Springs : Hooke’s law states that force required to stretch or compress a spring x units from its natural length is F= kx where k is known as spring constant.

Example3: A certain spring requires a force of 80N to stretch from its natural length of 2 m to a length of 4 m. How much work is required to stretch it from rest to a length of 5 m ?

Solution; By Hooke’s law F= kx

80=k(4-2) => 40=k

Therefore F= 40x

$\dpi{120} W=\int_{0}^{5}40x dx$

$\dpi{120} W=20\left [x^2 \right ]_{0}^{5} = 500J$

Example4: It takes 100J of work to stretch a spring 0.5m from its equilibrium position. How much work is needed to stretch it an additional 0.75 m ?

Solution:

$\dpi{120} W=\int_{a}^{b}F(x) dx$ where F(x)= kx

$\dpi{120} 100=\int_{0}^{0.5}kx dx$

$\dpi{120} 100=\frac{k}{2}\left [ x^{2} \right ]_{0}^{0.5}$

200=0.25k => k= 800

Work done in stretching additional 0.75m

$\dpi{120} W=\int_{0.5}^{1.25}800 x dx$

$\dpi{120} W=\frac{800}{2}\left [ x^{2} \right ]_{0.5}^{1.25}$

W=400(1.3125) = 525J

Vertical Ropes and Cables:

Suppose that a cable or rope is used to raise a heavy object to some height h. The work done in raising the object is calculated using same formula W= F*distance. But finding the work done in raising the cable is trickier. There are two ways to think about breaking the problem down:

consider raising the entire cable through a small distance.
consider raising a small section of the cable all the way to the top.

We used first approach in springs. We use second approach for cables as clear in the next example.

Example4:A uniform cable 20 ft. long hangs from the top of a tall building. If the cable weights 3 pounds per foot, find the work done in raising the entire cable to the top.

Solution: Here weight of cable is the force exerted by the cable, so F=3 lb/ft

Lets assume cable is raised to a distance of y ft.

Then work done in raising the entire cable

$\dpi{120} W=\int_{0}^{20}(3y) dy$

$\dpi{120} W=\frac{3}{2}\left [ y^{2} \right ]_{0}^{20}$

W= 600 lb ft

Example5:Determine the work done in propelling a 5 tons satellite to a height of

(a)100 miles above the Earth and

(b) 300 miles above the Earth.

Solution: Because the weight of body varies inversely as the square of its distance from center of earth, force exerted by gravity is given as,

$\dpi{120} F(x)=\frac{C}{x^{2}}$

$\dpi{120} 5=\frac{C}{(4000)^{2}}$ (considering earth radius as 4000 miles and neglecting air resistance)

$\dpi{120} 8\times 10^{7}=C$

a) The work done in propelling 100 miles above the earth

$W=\int_{4000}^{4100}\frac{8\times 10^{7}}{x^{2}}dx$

$W=8\times 10^{7}\int_{4000}^{4100} x^{-2}dx$

$W=8\times 10^{7}\left [ \frac{-1}{x} \right ]_{4000}^{4100}$

$W=8\times 10^{7}\left [ \frac{-1}{4100} +\frac{1}{4000}\right ]$

$W=\frac{2\times 10^{4}}{41}$ miles tons

(To convert miles tons into ft lbs we multiply with 5280 and 2000 because 1 ton= 2000lbs, 1 miles= 5280 ft)

$W=5.15\times 10^{9}$ ft lbs

1 ft lb = 1.35582 J so we multiply with 1.35582 to get the work done in Joules.

$W=7\times 10^{9}$ J

b) Work done in propelling 300 miles above the earth

$W=\int_{4000}^{4300}\frac{8\times 10^{7}}{x^{2}}dx$

$W=8\times 10^{7}\int_{4000}^{4300} x^{-2}dx$

$W=8\times 10^{7}\left [ \frac{-1}{x} \right ]_{4000}^{4300}$

$W=8\times 10^{7}\left [ \frac{-1}{4300} +\frac{1}{4000}\right ]$

$W=\frac{6\times 10^{4}}{43}$ miles ton

$W=1.47\times 10^{10}$ ft lbs

$W=2\times 10^{9}$ J

Pumping liquids from Tanks :

The method of slicing the object into small pieces and moving each piece all the way to the top applies very nicely to situations where water or any liquid is being pumped from a tank. The work integral so obtained will depend on the shape and geometry of slices that occur in each problem.

Example6: A cylindrical tank of height 10 m and radius 5 m is filled to a height of 8 m with water, which weighs 9800N per cubic meter. Find the work done to pump all the water from the tank.

Solution;

Lets first consider a slice of thickness Δy at a height y. Since this slice is in the form of a disk with thickness Δy, so its volume is given as

$v=\pi r^{2}\Delta y$

Volume of slice = $\pi 5^{2}\Delta y =25\pi \Delta y$

Force exerted on the slice= 9800*25πΔy=245000πΔy N

Distance moved by slice =(10-y) m

Work done on slice= [245000π(10-y)Δy] Nm

Total work done to pump all the water

$W=\int_{0}^{8}245000\pi (10-y)dy$

$W=245000\pi \int_{0}^{8} (10-y )dy$

$W= 245000\pi \left [ 10y-\frac{y^{2}}{2} \right ]_{0}^{8}$

W= 245000π(80-32)= 36,945,130 J

Example7:A spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubic foot. Find the work required to pump oil out through a hole in the top of the tank.

Solution: Lets assume oil in spherical tank is to be subdivided into slices of thickness Δy and radius x. These slices are disks, so volume of each disk is

$v=\pi x^{2}\Delta y$

Force exerted on slice

$F=50*\pi x^{2}\Delta y$

Here radius x is variable so we need to get it as a function of y. This sphere can be though of a circle with radius 8 and center at (0,8)

$x^{2}+(y-8)^{2}=8$

$x^{2}=16y -y^{2}$

$F=50*\pi (16y-y^{2})\Delta y$

Distance moved by the slice=16-y

Work done in moving the disk through a distance of 16-y,

W=F*(16-y)

Work done in pumping out oil from the tank

$W=\int_{0}^{8}50\pi (16y-y^{2})(16-y)\Delta y$

$W=50\pi \int_{0}^{8} (256y-32y^{2}+y^{3}) dy$

$W=50\pi \left [ 128y^{2}-\frac{32}{3}y^{3}+\frac{y^{4}}{4} \right ]_{0}^{8}$

$W=50\pi \left (\frac{11264}{3} \right )\approx 589,782$ ft. lbs

Example8: A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m. How much work is required to pump the water out of the trough when it is full?

Solution: Lets consider water in semicircular trough is subdivided into slices of thickness Δy. These slices are in the form of rectangular cross sections having length as 3m. Assuming the center of semicircle as origin(0,0) , width of each slice is given as 2x where x is the distance of semicircular edges from central line.

Equation of the semicircle is

$x^{2}+y^{2}=(0.25)^{2}$

$x=\sqrt{(0.25)^{2}-y^{2}}$

Area of slice = length*width =3*2x

$A=3*2\sqrt{(0.25)^{2}-y^{2}}$

Considering each slice start at height y and exit the trough at 0, then distance moved by the slice is =0-y

Total work done in pumping out water from trough

$W=\int_{-0.25}^{0}1000*9.8 *(-y)6\sqrt{(0.25)^{2}-y^{2}} dy$

$W=9800\int_{-0.25}^{0}(-6y)\sqrt{(0.25)^{2}-y^{2}} dy$

let $u =(0.25)^{2}-y^{2}$

du = -2y dy

du/-2 = ydy

And new integral limits in terms of u would change to 0 to 0.0625

We get the new integral as,

$W=9800\left ( \frac{6}{2} \right )\int_{0}^{0.0625}\sqrt{u} dy$

$W=9800(3)\left [ \frac{2}{3}u^{\frac{3}{2}} \right ]_{0}^{0.0625}$

$W=9800(2)\left [ (0.25)^{\frac{3}{2}} -0\right ]=306.25 J$

Practice problems:

1.It takes 100 J of work to stretch a spring 0:5 m from its equilibrium position. How much work is needed to stretch it an additional 0.75 m?

2.A space module weighs 15 metric tons on the surface of Earth. How much work is done in propelling the module to a height of 800 miles above Earth, (Use 4000 miles as the radius of Earth. Do not consider the effect of air resistance or the weight of the propellant.)

3.A water tank is shaped like an inverted cone with height 6 m and base radius 1.5 m. If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank?

4. Find the mass of the thin bar over the interval 0 ≤ x ≤π with the density function
$\rho (x)=\frac{4x}{3+x^{2}}$ . Find the exact answer.

5. A 60-m long, 9.4 mm diameter rope hangs freely from a ledge. The linear density of the rope is 55g/m. How much work is needed to lift the entire rope to the ledge?

Answers:

1. 525J

2. 10,000 miles tons

3. 66,150π J

4. $2ln(\frac{3+\pi ^{2}}{3})$

5. 970.2 J

Optimization

December 20, 2019/by admin_ct

One important application of derivatives is optimization, which play a significant role in wide range of disciplines like physical Sciences, Biology and Economics.

Steps to solve an optimization problem:

First identify and find the objective function which we need to maximize or minimize. Sometimes objective function is not given directly and we need to formulate it with the help of information provided in the statement.
Find critical points by setting first derivative =0
Find f(x) at those critical values which gives the maximum or minimum value.
Second derivative can also be used to prove the critical point as maxima or minima of objective function.

Minimizing cost

Example1: Suppose we want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10 per sq. ft and the material used to build the sides cost $6 per sq. ft. If the box must have a volume of 50 cubic ft, determine the dimensions that will minimize the cost to construct the box.

Solution: let the width of base = x and height of box = y.

Then length of base = 3x

Volume of box =length*width*height

50 =3x*x*y

50/3x^2 = y

Here we must find height y in terms of x so that we can get objective function in one variable (x) only.

Objective is to minimize the cost of box, so lets formulate the cost function.

Cost= $10(top+bottom area)+$6( area of 4 sides)

C(x)= 10(3x^2 +3x^2)+6(xy+xy+3xy+3xy)

C(x)= 60x^2+48xy

C(x)= 60x^2 +48x(50/3x^2)

C(x)= 60x^2+800/x

This is the objective function in one variable which we need to minimize.

C'(x)= 120x -800/x^2

C'(x)=0 => 120x=800/x^2

x^3=800/120

$x=\sqrt[3]{\frac{20}{3}}$

C”(x)=120+1600/x^3

We can plugin this critical value in C”(x) and check that C”(x) = 360 > 0 at $x=\sqrt[3]{\frac{20}{3}}$

That means cost would be minimum at $x=\sqrt[3]{\frac{20}{3}}$

Width = $\sqrt[3]{\frac{20}{3}}$

length = $3 \sqrt[3]{\frac{20}{3}}$

To get height y, we plugin x value into y= 50/3x^2

Height = $\frac{5}{2} \sqrt[3]{\frac{20}{3}}$ (After rationalizing and simplifying )

Maximizing profit

Example2. You operate a tour service that offer following rates-

-$200 per person if 50 people(minimum number of people to book the tour) go on the tour.

-For each additional person upto a maximum of 80 people total, the rate per person is reduced by $2.

It costs $6000( fixed cost) plus $32 per person to conduct the tour. How many people does it take to maximize your profit.

Solution: We know that Profit = Revenue- Cost

Let x be the number of additional people on the tour.

Revenue = number of people* price

R(x)=(50+x)(200-2x)

Cost function= fixed cost+ cost per person

C(x)= 6000+32x

Profit P(x)=R(x)-C(x)

P(x)= (50+x)(200-2x)- (6000+32x)

=4000+68x-2x^2

P'(x)= 68-4x

P'(x)=0 => 68-4x=0

x=17

P”(x)=-4 <0 which shows that profit is maximum at x=17.

Number of additional people to maximize profit = 17

Total people on the tour = 50+17= 67 people.

Minimizing travel time

Example3 :Your task is to build a road joining a ranch to a highway that enables drivers to reach the city in the shortest time. How should this be done if the speed limit is 60 km/h on the road and 110 km/h on the highway? The perpendicular distance from the ranch to the highway is 30 km, and the city is 50 km down the highway.

Solution: Lets first sketch the given situation.

let P be the point which represent perpendicular distance of highway from ranch.

Let Q be the point where road from ranch joins the highway.

Let x represent the distance between points P and Q.

Objective : To minimize the travel time.

For that we use the formula, Time = Distance/Speed

Distance from ranch to point Q = $\sqrt{(30)^2+x^2}$

speed on road = 60 km/h

Distance from point Q to city = 50-x

speed on highway =110 km/h

$T(x)=\frac{\sqrt{(30)^2+x^2}}{60}+\frac{50-x}{110}$

$T'(x)=\frac{1}{60}\left ( \frac{x}{\sqrt{(30)^2+x^2}} \right )-\frac{1}{110}$

Solve for critical points,

$\frac{1}{60}\left ( \frac{x}{\sqrt{(30)^2+x^2}} \right )-\frac{1}{110}=0$

$\frac{1}{60}\left ( \frac{x}{\sqrt{(30)^2+x^2}} \right )=\frac{1}{110}$

Cross multiplying and simplifying ,

$110x=60\sqrt{(30)^2+x^2}$

$121x^{2}=36 (30^{2}+x^{2})$

$121x^{2}-36x^{2}= 32400$

$x^{2}=\frac{32400}{85}$

$x=\sqrt{\frac{32400}{85}}= 19.52$

This value can be checked by finding travel time T(x) at critical point x=19.52 and end points x=0,50

T(0)= 0.95 hrs

T(19.52)=0.87 hrs

T(50)=0.97 hrs

So travel time is minimized when road is joined to the highway 19.52 km away from point P.

Example4:An observer stands at a point P, one unit away from the track. Two runners start at the point S in the figure and run along the track. One runner run three times faster than the other. Find the maximum value of the observer’s angle of sight θ between the runners.

media_413_413a52e4-1187-45ea-8940-7935dc024704_phpOln4PB

Solution: Here objective is to maximize tanθ. Lets get a sketch for the given situation.

Runner B runs three times faster than runner A.

Distance covered by runner A at time t = x

Distance covered by runner B at the same time t= 3x

In smaller triangle angle α is given as tanα=x

In larger triangle angle β is given as tanβ= 3x

tanθ = tan(β-α)

$tan\theta =\frac{tan\beta -tan\alpha }{1+tan\alpha tan\beta }$

$tan\theta =\frac{3x-x }{1+ x*3x }$

$f(x)= \frac{2x}{1+3x^{2}}$

$f'(x)=\frac{2(1+3x^{2})-2x(6x)}{(1+3x^{2})^{2}}$

$f'(x)=\frac{2+6x^{2}-12x^{2}}{(1+3x^{2})^{2}} =\frac{2-6x^{2}}{(1+3x^{2})^{2}}$

$\frac{2-6x^{2}}{(1+3x^{2})^{2}}=0$

$2-6x^{2}=0$

$x = \frac{1}{\sqrt{3}}$

x = 1/√3

tanα = 1/√3 => α = π/6

3x = √3

tanβ = √3 => β =π/3

θ = β-α = π/6

So the maximum angle of sight would be π/6.

Example5: Find the point on the parabola ${\color{Red} y=1-x^2}$ at which the tangent line cuts from the first quadrant , the triangle with the smallest area.

Solution: First we find tangent line to the curve at a point x=a,

$f(x) =1-x^2$

f'(x)=-2x

Tangent line at x=a : y = f'(a) (x-a)+f(a)

y=-2a(x-a)+1-a²

y= -2ax+1+a²

Now we find the area formed by this tangent line along with x and y axis. For that we need x and y intercepts.

X intercept: Plugin y=0

$x=\frac{1+a^{2}}{2a}$

Y intercept : Plugin x=0

$y=1+a^{2}$

Area = $\frac{1}{2}*\frac{1+a^{2}}{2a}(1+a^{2})$

$A=\frac{1}{4a}(1+a^{2})^{2}=\frac{1+2a^{2}+a^{4}}{4a}$

$A=\frac{1}{4}\left ( a^{3}+2a +\frac{1}{a} \right )$

$A'=\frac{1}{4}(3a^{2}+2-\frac{1}{a^{2}})$

$A'=0 \Rightarrow \frac{3a^{4}+2a^{2}-1}{4a^{2}}=0$

$\frac{(3a^{2}-1)(a^2+1)}{4a^2}=0$

$(3a^{2}-1)(a^{2}+1)=0$

$(a^{2}+1)\neq 0 , so (3a^{2}-1)=0$

Therefore

$a=\pm \frac{1}{\sqrt{3}}$

For first quadrant we use positive value of a, so a=1/√3

Checking values of derivative A’ for different a values we get that

$A' < 0, for 0<a<\frac{1}{\sqrt{3}}$

$A' > 0, for \frac{1}{\sqrt{3}}<a<\infty$

Which shows that triangle would have minimum area at point a=1/√3

Therefore at the point

$(a,1-a^{2})=(\frac{1}{\sqrt{3}},\frac{2}{3})$

the tangent line cuts from the first quadrant , the triangle with smallest area.

Constructing the fuel tank : a story problem

“Great, I’ve got the fuel, but how do i get it to the spacecraft? Computer, how do I transport this fuel?” Your computer responds that the burning of fuel destroys the container that it is held in, and that a new container is constructed for each launch of the spacecraft. The shape of the container is a cylinder with a hemisphere attached to the bottom. Your computer is able to generate an image (not to scale) and equations for the volume and surface area of this shape.

You see a warning notice that supplies of the materials for constructing this container are low. In order to create the container with limited supplies you will need to provide the height and radius, correct to two decimal places, that minimize the surface area of container. What values do you provide?

Total volume of fuel : 15π

Equation for volume of this container: $V=\pi r^2(h+\frac{2}{3}r)$

Equation for surface area: $S= 2\pi rh+3\pi r^2$

For this problem, we need to minimize surface area using volume equation. Using given constant volume we find variable h in terms of r and then plugin expression for h into surface area equation. Using the concept of derivatives, then objective surface area equation is minimized.

Volume of container :

$V=\pi r^2(h+\frac{2}{3}r)$

$15 \pi =\pi r^2(h+\frac{2}{3}r)$

$15 = r^2(h+\frac{2}{3}r)$

$\frac{15}{r^2} = (h+\frac{2}{3}r)$

$\frac{15}{r^2}-\frac{2}{3}r = h$

Next, we find surface area equation in terms of variable r only using expression of h from volume equation.

$S= 2\pi rh+3\pi r^2$

$S= 2\pi r(\frac{15 }{r^2}-\frac{2}{3}r)+3\pi r^2$

$S= \frac{30\pi }{r}-\frac{4\pi r^2}{3}+3\pi r^2$

$S= \frac{30\pi }{r}+\frac{5\pi r^2}{3}$

Next, find derivative of this surface area equation with respect to r.

$S'(r)= 30\pi(\frac{1 }{r})'+\frac{5\pi}{3}(r^2)'$

$S'(r)= 30\pi (-\frac{1}{r^2})+\frac{5\pi }{3}(2r)$

$S'(r)=\frac{-30\pi }{r^2}+\frac{10\pi r }{3}$

Set the derivative equal to 0 and solve for r to get critical value.

$\frac{-30\pi }{r^2}+\frac{10\pi r }{3}=0$

$\frac{10\pi r }{3}=\frac{30\pi }{r^2}$

$10\pi r^3 = 90\pi$

$r^3=9$

$r=9^{\frac{1}{3}}=2.08$

We can check this r value for maximum or minimum, using second derivative test.

$S''(r)= 30\pi (-\frac{1}{r^2})'+\frac{5\pi }{3}(2r)'$

$S''(r)= 30\pi (\frac{2}{r^3})+\frac{5\pi }{3}(2)$

$S''(r)=\frac{60\pi }{r^3}+\frac{10\pi }{3}$

We observe that second derivative is going to be positive always for any positive value of r.

Which proves that surface area will be minimum at r= 2.08

Next, find value of h using r value we got earlier, into following expression for h.

$h= \frac{15}{(2.08)^2}-\frac{2}{3}(2.08)$

h= 2.08

Following values of radius and height will minimize the surface area of container:

r = 2.08

h= 2.08

Practice problems:

A rectangular garden of area 75 sq.ft is bounded on three sides by a fence costing $8 per feet and on the 4th side by a fence
costing $4 per feet. Find the dimensions that will minimize the total cost.
A lifeguard wishes to get to a person 100 ft downstream on the opposite shore of a 50 ft wide river, as fast as possible. What path should she take if she can swim 5 ft/sec and run 15 ft/sec.
A right circular cylinder is inscribed in a right circular cone. Find the dimensions that maximize the volume of the cylinder.
Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum total area?
Show that the height of a closed cylinder of given surface and maximum volume is equal to the diameter of its base.

Answers:

length = 10 ft, width = 7.5 ft
Swim straight to the point x=25/√2
h=H/3 , r=2R/3
All 4ft for the circle, none for square

One sided Limits from graphs

December 3, 2019/by admin_ct

How to find limits from graphs

Sometimes it is important to investigate and work on one sided limits because a limit exist only when left and right sided limits exist separately and are same. When we approach to x from lesser values (left side) then it is called left sided limit and when we approach to x from greater values (right side) then it is called Right sided Limit.

At holes left and right sided limits exist and are same but the function is not continuous there.
lets work on some examples where we find limits from the given graph.

Example1:The function f(x) in given graph is not defined at x=0,2,4. Investigate the one and two sided limits at these points.

Solution: At x=0,

Left hand limit $\dpi{120} \lim_{x\rightarrow 0^{-}}f(x) = DNE$

left side limit at x=0 doesn’t exist because function oscillates infinitely towards left of 0.

Right hand limit $\dpi{120} \lim_{x\rightarrow 0^{+}}f(x) = 2$

since both limits are not same so limit doesn’t exist at x=0

$\dpi{120} \lim_{x\rightarrow 0}f(x) = DNE$

At x= 2

left hand limit $\dpi{120} \lim_{x\rightarrow 2^{-}}f(x) = 3$

Right hand limit $\dpi{120} \lim_{x\rightarrow 2^{+}}f(x) = 1$

Both one sided limits exist separately but not equal so limit does not exist (DNE) at x=2

$\dpi{120} \lim_{x\rightarrow 2}f(x) = DNE$

At x=4

left hand limit $\dpi{120} \lim_{x\rightarrow 4^{-}}f(x) = 2$

Right hand limit $\dpi{120} \lim_{x\rightarrow 4^{+}}f(x) = 2$

Both one sided limits exist at x=4 and are equal , so limit exist at x=4

$\dpi{120} \lim_{x\rightarrow 4}f(x) = 2$

Infinite Limits:

Some functions f(x) tends to +∞ or -∞ when x approach to a value c, then $\dpi{120} \lim_{x\rightarrow c}f(x)$ does not exist and we say f(x) has an infinite limit. One sided limits can be infinite limits. More precisely we write it as,

$\dpi{120} \lim_{x\rightarrow c}f(x) = \infty$

$\dpi{120} \lim_{x\rightarrow c}f(x) = -\infty$

When f(x) approach to ∞ or -∞ from one or both sides then line x=c is called Vertical Asymptote.Limits does not exist at vertical asymptotes.

Here is an example to understand this concept graphically.

Example2: Investigate one sided limits graphically.

a) $\dpi{120} {\color{Red} \lim_{x\rightarrow 2}\frac{1}{x-2}}$

Observing the graph , we find that y values are approaching to -∞ as we move closer to x=2 from left side and y values are approaching to ∞ as we move closer to x=2 from right side.

$\dpi{120} \lim_{x\rightarrow 2^{-}}\frac{1}{x-2}= -\infty$

$\dpi{120} \lim_{x\rightarrow 2^{+}}\frac{1}{x-2}= \infty$

Both one sided infinite limits are not same. That means limits does not exist at x=2

b) $\dpi{120} {\color{Red} \lim_{x\rightarrow 0}\frac{1}{x^{2}}}$

Looking at the graph,we find that y values are approaching to ∞ when we move closer to x=0 from both left and right sides.

That means both one sided infinite limits are same. But infinite limit itself does not exist therefore limit doesn’t exist at x=0

c) $\dpi{120} {\color{Red} \lim_{x\rightarrow 0}ln(x)}$

Natural log (ln) is not defined for negative values so its domain is (0,∞). Looking at the graph we find that y values are approaching to -∞ when we move closer to 0 from right side. Only one right sided limit is possible, which doesn’t exist being infinite limit. So x=0 is a vertical asymptote and limit does not exist here.

Example3: Determine the one sided limits of the function f(x) at points x=1,3,5,6 in the following graph.

Solution; At x=1

Function is defined and continuous at x=1. Both left and right sided limits approach to same number(3), so limit exist here.

$\dpi{120} \lim_{x\rightarrow 1}f(x)=3$

At x=3

As we move closer to 3 from left , y values approach to -∞ whereas y value is 4 as we move towards 3 from right.

$\dpi{120} \lim_{x\rightarrow 3^{-}}f(x)= -\infty$

$\dpi{120} \lim_{x\rightarrow 3^{+}}f(x)= 4$

At x=5

$\dpi{120} \lim_{x\rightarrow 5^{-}}f(x)= 2$

$\dpi{120} \lim_{x\rightarrow 5^{+}}f(x)= -3$

At x=6

Both left and right sided limits are infinite limits where y values approach to ∞ from both left and right sides of x=6. therefore x=6 is vertical asymptote here.

$\dpi{120} \lim_{x\rightarrow 6^{-}}f(x)= \infty$

$\dpi{120} \lim_{x\rightarrow 6^{+}}f(x)= \infty$

Sketching graphs from given limits

Example 4: Sketch the graph of a function with the given limits.

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2^{+}}f(x)= f(2)=3}$

$\dpi{120} {\color{Red} \lim_{x\rightarrow 2^{-}}f(x)= -1}$

$\dpi{120} {\color{Red} \lim_{x\rightarrow 4}f(x)= 2\neq f(4)}$

Solution:

From first two limits, we conclude that both left and right limits are different when x is approaching to 2. That means graph is approaching to 3 when x values are approaching to 2 from right side. Also function value is 3 at x=2.

Graph is approaching to -1 when x values are approaching to 2 from left.

Limits exist at x=4 and is equal to 2, that means both left and right sided limits are approaching to 2 when x values are approaching to 4 from left and right sides. But function value is not same as limit at x=4 which means there would be a hole at x=4. Gathering all the information above , we get the following graph.

Example 5: Sketch the graph of a function with the given limits.

$\dpi{120} {\color{Red} \lim_{x\rightarrow 1}f(x)= \infty }$

$\dpi{120} {\color{Red} \lim_{x\rightarrow 3 ^{-}}f(x)= 0}$

$\dpi{120} {\color{Red} \lim_{x\rightarrow 3 ^{+}}f(x)= -\infty }$

Solution:

From first limit we note that there is vertical asymptote at x=1 and graph is approaching to infinity from both left and right sides of x=1.

left and right limits are not same at x=3. From left of x=3, graph is approaching to 0 and from right of x=3 graph is approaching to – infinity. Again there is vertical asymptote at x=3.

Converting all the above information into a graph, we get the following graph.