Worksheet 1 (Limits)

/by

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

 

5. 

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* \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

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

 

 

 

 

 

 

Answers:

  1. DNE
  2.  -6/7
  3. -8
  4. -1/3
  5. 2
  6.  a=-3,3
  7.  9/2
  8. 1
  9.  -3/2
  10. log (9/8)

Adding and Subtracting Polynomials

/by

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

/by

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:
  1. Each real number \dpi{120} a_{i} is called  coefficient. The number \dpi{120} a_{0} not multiplied with any variable is called constant.
  2. Each product \dpi{120} a_{i}x^{i} is called term of polynomial.
  3. The highest power of variable that occur in a polynomial is called degree of polynomial.
  4. 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.

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

 

 

 

 

 

Answers:

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

 

 

 

 

 

 

Surface Area

/by

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.

  1.  y=x³ ; 0≤x≤2
  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:

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

Arc length

/by

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^{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:

  1.   1261/240 units
  2.   1.22 units

3.      4√10 π

4.    16 units

Fluid Pressure and Force

/by

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:

  1. 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.
  2. 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.
  3. Find the hydrostatic force  on a circular plate of radius 2 that is submerged 6 ft  in the water.
  4. 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

/by

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

/by

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:

  1.  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.
  2. Find critical points by setting first derivative =0
  3. Find f(x) at those critical values which gives the maximum or  minimum value.
  4. 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.

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:

  1. 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.
  2. 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.
  3. A right circular cylinder is inscribed in a right circular cone. Find the dimensions that maximize the volume of the cylinder.
  4. 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?
  5. Show that the height of a closed cylinder of given surface and maximum volume is equal to the diameter of its base.

 

 

 

Answers:

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

 

 

 

One sided Limits from graphs

/by

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.

 

 

Practice problems:

  1. Use the given graph of function to answer each question.

a)  \dpi{120} \lim_{x\rightarrow 0^{-}}f(x)                     b) \dpi{120} \lim_{x\rightarrow 3 ^{-}}f(x)             c)  \dpi{120} \lim_{x\rightarrow 3 }f(x)                   d)  f(3)= ?

e) \dpi{120} \lim_{x\rightarrow 0^{+}}f(x)                       f)  \dpi{120} \lim_{x\rightarrow 3 ^{+}}f(x)              g)  \dpi{120} \lim_{x\rightarrow 0 }f(x)                  h)  f(0) = ?

2.  Sketch the graph of a function with the following limits.

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

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

\dpi{120} \lim_{x\rightarrow 1}f(x)=2

 

 

Answers:

1.  a) -∞          b) 1     c) DNE        d) 3       e) 2         f) 2         g) DNE     h)  DNE

Graphing linear equation using Slope and Y Intercept

/by

How to graph a line using  slope and y intercept.

To graph a line using slope and y intercept , we should have the equation in slope intercept form(y=mx+b) . So we have to convert the linear equation in slope intercept form , if not given already.

Slope Intercept form:    y=mx+b

Here coefficient of x  is slope(m)  and constant term b is y intercept.

Steps to graph using slope and y intercept:

1. Identify y intercept (b) from given equation which can be written as (o,b) in ordered pair form. This gives you first point on y axis.

2. Identify slope(m). Positive slope gives you rising line and negative slope gives you falling line.

\dpi{120} \mathbf{Slope(m)= \frac{y}{x}=\frac{Rise}{Run}}

Numerator of slope tells you how many steps you need to move up or down whereas denominator tells us how many steps to need to move left or right.

For positive slope we always go up (rise) , for negative slope we always  go down.

3. Right direction of slope takes you to the second point. Joining these two points gives to graph of linear equation.

 

Example1: Find slope and y intercept of equation y = 5x-3 and then graph the equation.

Solution: Comparing the given equation with slope intercept form  y=mx+b, we get

Slope(m)=5

Y intercept (b)=-3

Rewriting  y intercept in ordered pair form , we get first point as (0,-3) which is always on y axis.

\dpi{120} Slope(m)=\frac{Rise}{Run}=\frac{5}{1}

From first point rise 5 steps and then run 1 step right.The point where we stop at, is the second point. Join these two points  and get the graph of line equation.

 

Example2: Find slope and y intercept of equation 3y-x=6 and then graph it.

Solution: Since this equation is not in slope intercept form, so we convert it in that form first.

3y-x =6

3y =x+6

\dpi{120} y=\frac{1}{3}x+\frac{6}{3}

\dpi{120} y=\frac{1}{3}x+2

Comparing it with y=mx+b, we get  y intercept  (0,2)  as  first point  on y axis.

\dpi{120} Slope(m)=\frac{Rise}{Run}=\frac{1}{3}

From first point rise 1 step and run 3 steps right. We reach at second point (3,3). Join these two points and get the graph of given line equation.

We can also first run by 3 steps right and then rise by one step reaching at the same point

 

Example3:Find slope and y intercept of the given equation and then graph the equation.

\dpi{120} {\color{Red} y=\frac{-5}{4}x +1}

Solution: Comparing the given equation with y=mx+b we get,

Slope(m)=-5/4

Y intercept = (0,1)

\dpi{120} Slope(m)=\frac{Rise}{Run}=\frac{-5}{4}

First plot y intercept (0,1) as first point on y axis. Since slope is negative so we go down by 5 steps instead of going up. Basically we should say Fall instead of Rise. After going down by 5 steps , move  right by 4 steps and we reach at second point (4,-4).

Join these two points and get the following graph.

 

 

 

Practice problems:

  1. Find slope and y intercept of the given equation and then graph the equation  y=2x+3
  2. Convert the given equation in slope intercept form. Find slope and y intercept  and then graph the equation 3x-6y=9
  3. Find slope and y intercept of the given equation and then graph the equation  y=  -2x/3 +1