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First order Linear Differential Equations

January 9, 2020/by admin_ct

What is first order Linear differential Equation?

A differential equation is linear if the dependent variable ‘y’ and its derivative appear only in first degree. The general form of a linear differential equation of first order is,

$\dpi{120} \mathbf{\frac{dy}{dx}+Py =Q}$

Where P and Q are functions of x(or constants).

For example,

$\dpi{120} \frac{dy}{dx}+2y=x^{3}$

$\dpi{120} \frac{dy}{dx}+2y= sinx$

are linear differential equations.

How to solve linear differential equations?

This type of differential equations are solved when they are multiplied by a factor, called integration factor because by multiplication of this factor the left hand side becomes exact differential of some function.

Integrating factor is given as,

$\dpi{120} I.F = e^{\int pdx}$

Its solution is written as,

$\dpi{120} \mathbf{y (I.F)= \int Q (I.F) dx +C}$

Following examples illustrates the procedure.

Example1:Solve the following Linear differential equation.

$\dpi{120} {\color{Red} \frac{dy}{dx}-\frac{y}{x}=2x^{2}, x>0}$

Solution:

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

here P= -1/x and Q=2x^2

$\dpi{120} I.F = e^{\int pdx}=e^{\int \left ( \frac{-1}{x} \right )dx}$

$\dpi{120} I.F =e^{-lnx}=e^{lnx^{-1}}=x^{-1}=\frac{1}{x}$

Plugin these values into the formula

$y (I.F)= \int Q (I.F) dx +C$

$y (\frac{1}{x})= \int 2x^{2} \left (\frac{1}{x} \right ) dx +C$

$y (\frac{1}{x})= \int 2x dx +C$

$y (\frac{1}{x})= x^{2} +C$

$y= x^{3} +Cx$

Example2: Solve the following initial value problem

${\color{Red} \frac{dy}{dx}+\frac{2x}{x^{2}+1}y=\frac{1}{(x^{2}+1)^{2}} ;y(0)=0}$

Solution: This is a linear differential equation where,

$P=\frac{2x}{x^{2}+1} ; Q=\frac{1}{(x^{2}+1)^{2}}$

Integrating factor

$\dpi{120} I.F = e^{\int pdx}=e^{\int \left ( \frac{2x}{x^{2}+1} \right )dx}$

$\dpi{120} I.F = e^{ln(x^{2}+1)}=x^{2}+1$

Using formula,

$y (I.F)= \int Q (I.F) dx +C$

$y (x^{2}+1)= \int \frac{1}{(x^{2}+1)^{2}}\left (x^{2}+1\right )dx +C$

$y (x^{2}+1)= \int \frac{1}{(x^{2}+1)}dx +C$

$y (x^{2}+1)= tan^{-1}(x) +C$

Using initial condition y(0)=0

0(0+1)=tan^-1(0)+C

0=C

$y (x^{2}+1)= tan^{-1}(x)$

$y = \frac{tan^{-1}(x)}{x^{2}+1}$

Differential Equations reducible to linear form:

Some differential equations are not separable or linear but they can be transformed to linear form after a suitable substitution. Here is an example.

Example3 : Solve the following differential equation.

${\color{Red} 2xe^{2y}\frac{dy}{dx}=3x^{4}+e^{2y}}$

Solution: This equation seems complex to solve but after reducing it would be an easy linear differential equation.

Let u=e^2y

$\frac{du}{dx}=2e^{2y}\frac{dy}{dx}$

Using this we get the original differential equation transformed to

$x\frac{du}{dx}= 3x^{4}+u$

rewriting this as general form of linear differential equation

$\frac{du}{dx}+\frac{-1}{x}u= 3x^{4}$

$P=\frac{-1}{x}; Q=3x^{4}$

$\dpi{120} I.F = e^{\int pdx}=e^{\int \left ( \frac{-1}{x} \right )dx}$

$\dpi{120} I.F =e^{-lnx}=e^{lnx^{-1}}=x^{-1}=\frac{1}{x}$

Plugin these values into the formula

$y (I.F)= \int Q (I.F) dx +C$

$u(\frac{1}{x})= \int 3x^{3} \left (\frac{1}{x} \right ) dx +C$

$u (\frac{1}{x})= \int 3x^{2} dx +C$

$u (\frac{1}{x})= x^{3} +C$

Plugin back u as e^2y

$e^{2y}= x^{4} +Cx$

$y=\frac{1}{2}ln(x^{4}+Cx)$

When y is independent variable instead of x.

Example4:Solve the following differential equation.

${\color{Red} ydx-(x+2y^{2})dy=0}$

Solution:

$ydx=(x+2y^{2})dy$

$\frac{dx}{dy}=\frac{(x+2y^{2})}{y}$

$\frac{dx}{dy}=\frac{x}{y}+2y$

$\frac{dx}{dy}+\frac{-1}{y}x=2y$

here P=-1/y and Q= 2y

$\dpi{120} I.F = e^{\int pdy}=e^{\int \left ( \frac{-1}{y} \right )dy}$

$\dpi{120} I.F =e^{-lny}=e^{lny^{-1}}=y^{-1}=\frac{1}{y}$

Plugin these values into the formula

$x (I.F)= \int Q (I.F) dy +C$

$x(\frac{1}{y})= \int 2y \left (\frac{1}{y} \right ) dy +C$

$x (\frac{1}{y})= \int 2 dy +C$

$x (\frac{1}{y})= 2y +C$

x=y(2y+C)

Example5:A tank initially contains 100 gal of water in which is dissolved 2 lb of salt. Salt-water solution containing 1 lb of salt for every 4 gal of solution enters the tank at a rate of 5 gal per minute. Solution leaves the tank at the same rate, allowing for a constant
solution volume in the tank.

a)Use an analytic method to determine the eventual salt content in the tank.

b) Find the concentration of salt when t —>∞. Check and see how reasonable is your answer.

Solution: The key idea to solve this kind of differential equations is that ‘The rate of change of salt concentration in a tank is equal to the rate at which salt enters the tank minus the rate at which salt leaves the tank’.

let x(t) represents the pounds of salt present in the tank at time t.

Then dx/dt= rate in- rate out

Rate in:

Concentration of the solution entering the tank=(1/4) lb/gal

Rate at which solution is entering = 5 gal/min

Rate at which salt enters the tank = 1/4 *5 = (5/4) lb/min

Rate out :

Assuming perfect mixing, the concentration of salt in the solution is found by dividing the amount of salt by the volume of solution.

so the concentration of solution leaving the tank = (x/100) lb/gal

Rate at which solution is leaving = 5 gal/min

Rate at which salt leaves the tank= (x/100)*5= 5x/100 =x/20 lb/min

Initially there is 2 lb of salt present in the tank so x(0)=2 lb

So differential equation can be modeled as,

$\frac{dx}{dt}=\frac{5}{4}-\frac{1}{20}x ; x(0)=2$

$\frac{dx}{dt}+\frac{1}{20}x=\frac{5}{4}$

here P= 1/20 , Q=5/4

$\dpi{120} I.F = e^{\int pdt}=e^{\int \left ( \frac{1}{20} \right )dt}$

$\dpi{120} I.F = e^{ \frac{t}{20} }$

Plugin the values into the formula

$x (I.F)= \int Q (I.F) dt +C$

$x (e^{\frac{t}{20}})= \int \frac{5}{4} (e^{\frac{t}{20}}) dt +C$

$x (e^{\frac{t}{20}})= \frac{5}{4} \left [20(e^{\frac{t}{20}}) \right ] +C$

$x (e^{\frac{t}{20}})= 25(e^{\frac{t}{20}}) +C$

$x = 25 +Ce^{\frac{-t}{20}}$

Using initial condition x(0)=2, we get

$2 = 25 +Ce^{0}$

C=-23

b) Concentration of salt at any time t is given as,

c(t)= x(t)/100

$c(t) = \frac{25 -23e^{\frac{-t}{20}}}{100}$

Eventual concentration can be found as,

$\lim_{t\rightarrow \infty }c(t) = \lim_{t\rightarrow \infty }\frac{25 -23e^{\frac{-t}{20}}}{100}= \frac{25-0}{100}=\frac{1}{4}$ lb/gal

This answer is quite reasonable as the concentration of solution entering the tank is same 0.25 lb/gal

Practice problems:

Solve the following linear differential equations\

1. $(1+x^{2})\frac{dy}{dx}+y=e^{tan^{-1}(x)}$

2. $\sqrt{1-x^{2}}\frac{dy}{dx}+y=1 ; y(0)=4$

3. $2xy\frac{dy}{dx}+2y^{2}= 3x-6$ Hint : u=y^2

4. $\sqrt{1-y^{2}}dx=(sin^{-1}y-x)dy ; y(0)=0$

5. A tank contains 100 gal of pure water. At time zero, sugar-water solution containing 0.2 lb of sugar per gal enters the tank at a rate of 3 gal per minute. Simultaneously, a drain is opened at the bottom of the tank allowing sugar-solution to leave the tank at
3 gal per minute. Assume that the solution in the tank is kept perfectly mixed at all times. Find the amount of sugar present in the tank at time t.

Answers:

$y=tan^{-1}(x)-1 +Ce^{-tan^{-1}(x)}$
$y=1+3e^{-sin^{-1}(x)}$
$y^{2}x^{2}=x^{3}-3x^{2}+C$
$x-sin^{-1}y+1=e^{sin^{-1}y}: y\in (-1,1)$
$S(t)= 20\left (1-e^{\frac{-3t}{100}} \right )$

Separable Differential Equations

January 6, 2020/by admin_ct

What is separable differential equation?

A differential Equation is separable if it can be written as f(y) dy= g(x) dx. Its solution can be found by integration both sides. A differential equation where we are able to move all y terms and dy to left and all x terms along with dx to right side, is called separable differential equation.

For example

$\frac{dy}{dx}= x siny$ is separable but

$\frac{dy}{dx}=x+y$ is not separable because x+y is not a product f(y) g(x) so can’t be separated.

Example1: Solve the given differential equation.

${\color{Red} \frac{dy}{dx}=\frac{2xy}{x+1}}$

Solution: Here we can easily separate the variables because right side is a product of x and y terms where f(y) =y and g(x)=2x/(x+1)

Separating the variables, we get

$\frac{1}{y} dy=\frac{2x}{x+1}dx$

Integrating both sides,

$\int \frac{1}{y} dy=\int \frac{2x}{x+1}dx$

$\int \frac{1}{y} dy=2\int \frac{x+1-1}{x+1}dx$

$\int \frac{1}{y} dy=2\int 1-\frac{1}{x+1}dx$

ln|y|=2[x-ln|x+1|]+C

Example2:Solve the differential equation
${\color{Red} \frac{dy}{dx}=\sqrt{16x^{2}y-4x^2y^2}}$ , y(2)=1

Solution: After simplifying the right side to separate x and y terms, we get

$\frac{dy}{dx}=\sqrt{4x^{2}(4y-y^2)}$

$\frac{dy}{dx}=2x\sqrt{4y-y^2}$

$\frac{1}{\sqrt{4y-y^2}}dy=2x dx$

$\int \frac{1}{\sqrt{4-4+4y-y^2}}dy=\int 2x dx$

$\int \frac{1}{\sqrt{2^2-(y-2)^2}}dy=\int 2x dx$

$sin^{-1}\left ( \frac{y-2}{2} \right )=x^{2}+C$

Using given initial condition,plugin x=2 and y=1 to get value of C

$sin^{-1}\left ( \frac{1-2}{2} \right )=2^{2}+C$

$sin^{-1}\left ( \frac{-1}{2} \right )-4=C$

$\frac{7\pi }{6}-4=C$

So the final solution is given as,

$sin^{-1}\left ( \frac{y-2}{2} \right )=x^{2}+ \frac{7\pi }{6}-4$

Example3: A parachutist falling toward Earth is subject to two forces: the parachutist weight (w = 32m) and the drag of the parachute. The drag of the parachute is proportional to the velocity of the parachute and in this case is equal to 8|v|. The parachutist weight is 128lb and initial velocity is zero. Find formulas for the parachutist’s velocity v(t) and distance x(t). Also find parachutists terminal velocity.

Solution: Since the parachutist falls down (the positive direction) velocity is always positive so |v| = v. The resultant force will be the force of the weight of the parachutist minus the force of the drag of the parachute.

so Force F= 128-8v ……. (i)

And the mass of parachute is, 128= 32m

therefore, m=4

Using Newton’s second law, Force F=ma

since a= dv/dt ,

so F=4 dv/dt …….. (ii)

Using equations (i) and (ii) we get,

$4\frac{dv}{dt}=128-8v$

$\int \frac{1}{32-2v}dv =\int dt$

$\frac{1}{2}\int \frac{1}{16-v}dv =\int dt$

$\frac{-1}{2}ln|16-v| = t+C$

$ln|16-v| = -2t-2C$

$ln|16-v| = -2t-K$

Using initial condition v(0)=0 , we get

-ln(16)=K

$ln|16-v| = -2t+ln(16)$

ln|16-v|-ln(16) =-2t

$ln\left ( \frac{16-v}{16} \right )=-2t$

$\left ( \frac{16-v}{16} \right )=e^{-2t}$

Solving for v , we get

$v(t)=16-16e^{-2t}$

Terminal velocity

$\lim_{t\rightarrow \infty }v(t)=\lim_{t\rightarrow \infty }(16-16e^{-2t})=16-0=16$ ft/sec

To know how far parachute has fallen, we find position function x(t) by taking integral of velocity function.

x(t)=∫v(t) dt

$x(t)=\int (16-16e^{-2t}) dt$

$x(t)=16t+8e^{-2t}+C$

using initial condition x(0)=0 we get

x(0)=0+8+C

-8=C

$x(t)=16t+8e^{-2t}-8$

Example4. Suppose that a cup of coffee is initially at a temperature of 105° F and is placed in a 75° F room. Newton’s law of cooling says that

${\color{Red} \frac{dT}{dt}=-k(T-75)}$

where $k$ is a constant of proportionality.

a) Suppose you measure that the coffee is cooling at one degree per minute at the time the coffee is brought into the room. Use the differential equation to determine the value of the constant $k$

b)Find all the solutions of this differential equation.

c)What happens to all the solutions as $t \to \infty$ ? Explain how this agrees with your intuition.

d) What is the temperature of the cup of coffee after 20 minutes?

e) How long does it take for the coffee to cool to 80°?

Solution:

a) Given dT/dt= -1 when T=105

Using these values into the differential equation, we get

-1=-k(105-75)

1/30 =k

b) Using this value of k we get the original DE as,

$\frac{dT}{dt}=-\frac{1}{30}(T-75)$

$\frac{dT}{T-75}=\frac{-1}{30}dt$

$\int \frac{dT}{T-75}=\frac{-1}{30}\int dt$

$ln|T-75|=-\frac{1}{30}t+c$

$T-75=e^{\frac{-1}{30}t+c}$

$T=75+e^{\frac{-1}{30}t}*A$

$T=75+Ae^{\frac{-t}{30}}$

Using initial condition T(0)=105, we get

$T(0)=75+Ae^{0}$

105-75=A => A=30

$T(t)=75+30e^{\frac{-t}{30}}$

$\lim_{t\rightarrow \infty }T(t)= \lim_{t\rightarrow \infty }[75+Ae^\frac{-t}{30}]=75+A(0)=75$

When t approaches to infinity , exponential part become 0 and the temp of coffee is same as temp of room.

d) After 20 minutes

$T(20)=75+30e^{\frac{-20}{30}}$

T(20)=90.4ºF

e) given T(t)=80

$80=75+30e^{\frac{-t}{30}}$

$\frac{5}{30}= e^{\frac{-t}{30}}$

ln(1/6)=-t/30

t=-30ln(1/6)

t=53.75 min

Equations reducible to Separable form:

Differential equations of the form dy/dx = f(ax+by+c) can be reduced to variable separable form by substituting ax+by+c= v as discussed in the following example.

Example5:Solve the following differential equation.

${\color{Red} \frac{dy}{dx}=(4x+y+1)^{2}}$

Solution: here we are not able to separate variables x and y so we use substitution.

Let v=4x+y+1

$\frac{dv}{dx}=4+\frac{dy}{dx}$

$\frac{dv}{dx}-4=\frac{dy}{dx}$

Using these values we get the original DE changed to

$\frac{dv}{dx}-4= v^{2}$

$\frac{dv}{dx}= v^{2}+4$

$\frac{1}{v^{2}+4}dv = dx$

$\int \frac{1}{v^{2}+4}dv = \int dx$

$\frac{1}{2}tan^{-1}\left ( \frac{v}{2} \right )=x+C$

Plugin back v to get final answer in terms of x and y

$\frac{1}{2}tan^{-1}\left ( \frac{4x+y+1}{2} \right )=x+C$

Practice problems:

Solve the following differential equations.

1. $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$

Solve the following initial value problem.

2. $(x+1)\frac{dy}{dx}=2e^{-y}-1 : y(0)=0$

3. $\frac{dy}{dx}=\frac{(x-y)+3}{2(x-y)+5}$

4. A freshly brewed cup of coffee has temperature 95◦C in a 20◦C room. When its temperature is 70◦C, it is cooling at a rate of 1◦C per minute. When does this occur?

Answers:

y-x=C(1+xy)
$y=ln\left ( 2-\frac{1}{x+1} \right )$
2(x-y)+ln(x-y+2)=x+C
20.3 min

Logistic Differential Equations

December 14, 2019/by admin_ct

What is Logistic Differential Equation(LDE)

Differential equations can be used to represent the size of population as it varies over time t. A logistic differential equation is an ordinary differential equation whose solution is a logistic function.

An exponential growth and decay model is the simplest model which fails to take into account such constraints that prevent indefinite growth but logistic function is more realistic model that includes other factors effecting population growth.

LDE(logistic differential equation) include two positive parameters –

i) Growth parameter (growth rate)k: This parameter plays a role similar to that of r in exponential differential equation.

ii) Carrying capacity (M):The carrying capacity of an organism in a given environment is defined to be the maximum population of that organism that the environment can sustain indefinitely.

Let M represent the carrying capacity for a particular organism in a given environment, and let k be a real number that represents the growth rate. The function p(t) represents the population of this organism as a function of time t and the constant represents the initial population (population at time t=0). Then the Logistic Differential equation is ,

\dpi{120} \mathbf{\frac{dP}{dt}= kP\left ( 1-\frac{P}{M} \right )}

A solution to this Logistic Differential equation is given as,

$\dpi{120} \mathbf{P(t)=\frac{M}{1+Ae^{-kt}}}$

Where

$\dpi{120} \mathbf{A=\frac{M-P_{0}}{P_{0}}}$

and

$\dpi{120} \mathbf{P_{0}}$ = initial population

Example1: Suppose that a population develops according to logistic differential equation

$\dpi{120} {\color{Red} \frac{dp}{dt}= 0.02p-0.0002p^{2}}$

where t is measured in weeks.

a) What is its carrying capacity and find value of k.

Answer: First we need to rewrite the equation as,

$\dpi{120} \frac{dp}{dt}= 0.02p\left ( 1- 0.01p \right )$

$\dpi{120} \frac{dp}{dt}= 0.02p\left ( 1-\frac{p}{100} \right )$

Comparing it with Logistic differential equation we get,

Carrying capacity(M) = 100

k= 0.02

b) Draw the direction field for this logistic equation and answer the questions.

Where are the slopes close to 0?

Answer: To get the points where slopes are 0 , we set dp/dt=0 and solve the equation for p.

0.02p(1+P/100)=0

0.02p =0 , 1-p/100=0

p=0 , 1= p/100 => p=100

So the slopes are 0 at p=0,100

Where are they largest?

Answer: Slopes are largest at p=M/2

So the slopes are largest at P= 100/2= 50

Which solutions are increasing?

Answer: Solutions are increasing when dp/dt >0

Therefore 0.02p(1- p/100)>0

0.02p >0 and 1-p/100>0

when p>0 and p < 100

So the solutions are increasing from 0 to 100

In interval form we can write it as (0,100)

Which solutions are decreasing?

Answer: Solutions are decreasing when (1-p/100) <0

As p, being population can’t be negative so 1-p/100<0

p>100

All solutions are decreasing when p>100

Which solutions have inflection points? At which population lever do they occur?

Answer: Since slope is largest at p=50 so all solution curves less than 50 will have inflection points at p=50. So the solution curves at p=10,20,30,40 will have inflection points at p=50.

What are the equilibrium solutions?

Answer: Equilibrium solutions are those where dp/dt =0 so answer is p=0,100 where slope is 0.

Conclusion: All the solutions approach p=100 as t increases. Solutions from 0 to 100 are increasing while solutions above p=100 are decreasing. Some solutions have inflection points while some don’t. Since slopes are largest at p=50 so the solution curves below 50 have inflection point at p=50. Increasing solutions move away from p=0 and all non zero solutions approach p=100 as t approach to infinity.

Example2: The Pacific halibut fishery has been modeled by the differential equation

${\color{Red} \frac{dY}{dt}=kY\left ( 1-\frac{Y}{M} \right )}$

Where Y(t) is the biomass ( the total mass of the members of the population) in kg at time t (measured in years) the carrying capacity is estimated to be ${\color{Red} M=9\times 10^{7}}$ and k= 0.75 per year.

a) If ${\color{Red} Y(0)=2\times 10^{7}}$ kg, find the biomass a year later. (Round the answer to two decimal places)

Solution: First we find A

$A=\frac{M-Y(0)}{Y(0)}= \frac{9\times 10^{7}-2\times 10^{7}}{2\times 10^{7}}$

$A=\frac{7\times 10^{7}}{2\times 10^{7}}= \frac{7}{2}=3.5$

Solution to the LDE is given as,

$Y(t)=\frac{M}{1+Ae^{-kt}}$

$Y(1)=\frac{9\times 10^{7}}{1+3.5e^{-0.75(1)}}= 3.39\times 10^{7}$

b) How long it will take for the biomass to reach ${\color{Red} 4\times 10^{7}}$ .(Round the answer to two decimal places)

Solution:

$4\times 10^{7}=\frac{9\times 10^{7}}{1+3.5e^{-0.75t}}$

$1+3.5 e^{-0.75t}=\frac{9\times 10^{7}}{4\times 10^{7}}$

$1+3.5 e^{-0.75t}= 2.25$

$3.5 e^{-0.75t}= 1.25$

$e^{-0.75t}= \frac{1.25}{3.5}$

$-0.75t = ln\left ( \frac{1.25}{3.5} \right )$

t=1.37 years

Example3: Suppose that a population grows according to a logistic model with carrying capacity 6000and k=0.0015 per year.

a) Write the logistic differential equation for this model.

Solution: Logistic differential equation formula is given as,

$\frac{dP}{dt}= kP\left ( 1-\frac{P}{M} \right )$

Plugin given values M= 6000 and k=0.0015 into this formula we get,

$\frac{dP}{dt}= 0.0015P\left ( 1-\frac{P}{6000} \right )$

b) If the initial population is 1000, write a formula for the population after t years. Use it to find the population after 50 years.

Solution: Solution to LDE is given as

$P(t)=\frac{M}{1+Ae^{-kt}}$

Given P(0)= 1000

$A=\frac{M-P(0)}{P(0)}= \frac{6000-1000}{1000}= 5$

So the formula for population after t years is given as,

$P(t)=\frac{6000}{1+5e^{-0.0015t}}$

And the population after 50 years is,

$P(50)=\frac{6000}{1+5e^{-0.0015(50)}} = 1064$

Example4: The population of wild pigs outside a small town is modeled by the function

${\color{Red} P(t)=\frac{220}{1+72e^{-0.8t}}}$

where t is measured in months.

a) What is ${\color{Red} \lim_{t\rightarrow \infty }P(t)}$ ?

Answer:

$\lim_{t\rightarrow \infty }\frac{220}{1+72e^{-0.8t}}= \frac{220}{1+72(0)}= 220$

b) What is the initial population of pigs?

Solution: To find initial population just plugin t=0

$P(0)=\frac{220}{1+72e^{-0.8(0)}}=\frac{220}{1+72(1)}$

$\frac{220}{73}= 3.013$

Initial population of pigs =3

c) When does the population reach 60 pigs?

Answer:

$60 =\frac{220}{1+72e^{-0.8t}}$

$1+72e^{-0.8t}=\frac{220}{60}$

$1+72e^{-0.8t}=3.667$

$72e^{-0.8t}=2.667$

$e^{-0.8t}= \frac{2.667}{72}$

-0.8t = ln(2.667/72)

t=4.119 ≈ 4.2 months

Practice problems:

Suppose that a population develops according to logistic differential equation

$\frac{dp}{dt}= 0.05p-0.0005p^{2}$

where t is measured in weeks.

a) What is its carrying capacity and find value of k.

b) Draw the direction field for this logistic equation and answer the questions.

Where are the slopes close to 0? Where are they largest?

Which solutions are increasing and which are decreasing?

Which solutions have inflection points? At which population lever do they occur?

What are the equilibrium solutions?

2. A population is modeled by the differential equation

$\frac{dP}{dt}=\frac{P}{8} \left (1-\frac{P}{12} \right )$

a) Find carrying capacity and growth constant.

b) Find the logistic model given P(0)=3

3. Suppose that a population grows according to a logistic model with initial population 1000 and carrying capacity 10,000. if the population grows to 2500 after one year, what will be the population after another three years?

4. Suppose that a particular population of bacteria follows the logistic formula

$P(t)=\frac{2000}{1.5+3.7e^{-0.2t}}$

Where P is in mg and t in hours.

a) Find Carrying capacity , growth constant and initial population.

b) Find the differential equation that p(t) solves.

Answers:

1(a) M=100 , k=0.05

b) slopes zero at p=0,100, largest at p=50

increasing (0,100) , decreasing p>100

Solutions below p=50 have inflection points and they have inflections points at p=50

equilibrium solutions p=0,100

2. M=12 , k= 1/8

$P(t)=\frac{12}{1+3e^{-\frac{1}{8}t}}$

3. 9000

4. a) M= 1333 mg , k=0.2 P(0)=384.6 mg

b) $\frac{dP}{dt}=0.2P\left (1-\frac{P}{1333} \right )$

Differential Calculus

September 26, 2019/by admin_ct

Derivative

Derivative at a point : Let f(x) be a real valued function defined on open interval (a,b) and let c∈(a,b) then f(x) is said to be differentiable at x=c iff

$\dpi{120} \mathbf{\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}}$

exist finitely.

This limit is called derivative of f(x) at x=c and is denoted by f ’(c) or $\dpi{120} \frac{\mathrm{d} (f(x))}{\mathrm{d} x}$ so

$\dpi{120} \mathbf{f'(c)= \lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}}$

Derivative represents slope of function. This is basically the change in y values (output values) with respect to change in x values (input values). It can also be represented by $\dpi{120} \frac{\Delta y}{\Delta x}$ or $\dpi{120} \frac{\mathrm{d} y}{\mathrm{d} x}$ . It also represent the average rate of change of function over an interval.

There are two ways to find derivative.

Derivative using difference quotient or limit definition method.
Derivative using standard rules and formulas.

Polar Coordinates and their Conversion

September 16, 2019/by admin_ct

Polar coordinates

Polar coordinate system is a plane with Pole(point O) and polar axis which is horizontal axis from point O. Any point P in this plane is assigned polar coordinates represented as P(r,θ ) as shown below.

we measure as positive when moving counterclockwise and negative when moving clockwise.

If r > 0, then P is on the terminal side of θ. If r < 0, then P is on the terminal side of π+θ.

Plotting Points in the Polar Coordinate System

Example1.Plot the following points with given polar coordinates

a) P(2, π/3) b) Q(-1, 3π/4 ) c) R(3,-45° )

Finding all Polar Coordinates of a point

Let point P have polar coordinates (r,θ). Any other polar coordinates of P must be of the form

(r, θ+2nπ ) Or

(-r, θ+(2n+1)π ) Where n is any integer.

The coordinates (r,θ ) , (r, θ+2π), and (-r, θ+π) all name the same point.

Example2.If the point P has Polar coordinates (1.5,-20° ), then find all other polar coordinates for P

Solution: using the formula given above

(1.5, -20+2nπ )

(-1.5, -20+(2n+1)π )

These are all the coordinates of point P. Using these formulas we can find first two coordinates using n=0,1

(1.5, -20+2π ) = (1.5 , 340°)

(-1.5,-20+π ) = (-1.5 , 160°)

Coordinate conversion:

Using the following equations Polar coordinates can be converted to rectangular coordinates and vice versa .

Let P have polar coordinates (r,θ ) and rectangular coordinates (x,y), then

x = r cosθ y = r sinθ

$\dpi{120} \mathbf{r^{2}=x^{2}+y^{2}}$ $\dpi{120} \mathbf{\theta =tan^{-1}\left ( \frac{y}{x} \right )}$

Example3Use an algebraic method to find the rectangular coordinates of the points with given polar coordinates.

(2, 270° )

Solution: we have r=2 and θ=270

x = r cosθ y =r sinθ

x = 2cos(270) y =2sin(270)

x = 2(0) = 0 y = 2(-1) = -2

So the rectangular coordinates are (0,-2)

Example4.Convert the rectangular coordinates (-1,1) to two polar coordinates algebraically.

Solution: Given that x=-1 and y=1

$\dpi{120} \mathbf{r=\sqrt{x^{2}+y^{2}}}$ $\dpi{120} \mathbf{\theta =tan^{-1}\left ( \frac{y}{x} \right )}$

$\dpi{120} \mathbf{r=\sqrt{(-1)^{2}+(1)^{2}}}$ $\dpi{120} \theta =tan^{-1}\left ( \frac{1}{-1} \right )$

$\dpi{120} r=\sqrt{2}$ $\dpi{120} \theta =tan^{-1}(-1)= \frac{3\pi }{4}$

So the first polar coordinates are (√2 , 3π/4 )

Other polar coordinate would be (-√2 , 3π/4 + π) = (-√2 , 7π/4)

which can also be written as (-√2 , -π/4 )

Converting equations:

Example 5. Convert the following polar equations to rectangular form and identify the graph.

a) r secθ = 3

r/cosθ = 3

r = 3cosθ

r*r= 3*r*cosθ

$\dpi{120} r^{2}=3x$

$\dpi{120} x^{2}+y^{2}=3x$

$\dpi{120} x^{2}-3x +y^{2}=0$

By completing squares,

$\dpi{120} x^{2}-3x + \frac{9}{4}+y^{2}=0 +\frac{9}{4}$

$\dpi{120} \left ( x-\frac{3}{2} \right )^{2}+\left ( y-0 \right )^{2}=\left ( \frac{3}{2} \right )^{2}$
Which is equation of a circle.

Center = (3/2 ,0)

Radius(r) = 3/2

b) r=2 sinθ-4cosθ

r*r = r(2 sinθ-4cosθ)

$\dpi{120} r^{2}= 2rsin\theta -4 rcos\theta$

$\dpi{120} x^{2}+y^{2}= 2y-4x$

$\dpi{120} x^{2}+4x+y^{2}-2y = 0$

$\dpi{120} x^{2}+4x+4+y^{2}-2y+1= 0+4+1$

$\dpi{120} \left ( x+2 \right )^{2}+\left ( y-1 \right )^{2}= 5$

$\dpi{120} \left ( x+2 \right )^{2}+\left ( y-1 \right )^{2}= \left (\sqrt{5} \right )^{2}$

Which is the equation of circle.

center = (-2,1)

radius(r)= √5

Example6: Convert the following rectangular equations to polar form.

a) 2x-3y =5

2*rcosθ -3*rsinθ = 5

r(2cos -3sinθ ) = 5

$\dpi{120} r = \frac{5}{2cos\theta -3sin\theta }$

b) $\dpi{120} \left ( x+3 \right )^{2}+\left ( y+3 \right )^{2}=18$

x^2+6x+9 +y^2 +6y+9 = 18

$\dpi{120} x^{2}+y^{2}+6x+6y+18=18$

$\dpi{120} x^{2}+y^{2}+6x+6y=0$

$\dpi{120} r^{2}+6rcos\theta +6rsin\theta =0$

r^2 = -r(6cosθ +6sinθ )

r = -(6cosθ +6sinθ )

Finding Distance using Polar Coordinates

Example7. The location, given in polar coordinates, of two planes approaching the Vicksburg airport are (4mi, 12° ) and (2mi,72° ) . Find the distance between the airplanes.

Solution: let the two planes be A and B as shown in figure given below.

To find the distance between A and B we use cosine rule. Angle between OA and OB is given as,

θ=72-12=60°

$\dpi{120} AB^{2}=OA^{2}+OB^{2}-2OA*OB cos(60)$

$\dpi{120} AB^{2}=4^{2}+2^{2}-2*4*2 cos(60)$

$\dpi{120} AB^{2}=20-16\left ( \frac{1}{2} \right ) =12$

AB = √12 = 3.46 miles

Distance between two planes is 3.46 miles.

Example8. A square with sides of length a and center at the origin has two sides parallel to the x-axis. Find polar coordinates of the vertices.

Solution: A square with sides of length a and center at origin would be drawn as below.

Rectangular coordinates of vertex A are (a/2 , a/2). We find its equivalent polar coordinates.

x = a/2 , y = a/2

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

$\dpi{120} r=\sqrt{\left ( \frac{a}{2} \right )^{2}+\left ( \frac{a}{2} \right )^{2}}=\sqrt{\frac{2a^{2}}{4}}= \frac{a}{\sqrt{2}}$

$\dpi{120} \theta =tan^{-1}\left ( \frac{\frac{a}{\sqrt{2}}}{\frac{a}{\sqrt{2}}} \right )= tan^{-1}(1) = \frac{\pi }{4}$

Polar coordinates of vertex

$\dpi{120} A = \left ( \frac{a}{\sqrt{2}},\frac{\pi }{4} \right )$

For all other vertices value of r remain same as sides of squares are equal just θ changes in each quadrant. In each quadrant it get increased by π/2

$\dpi{120} B= \left ( \frac{a}{\sqrt{2}},\frac{3\pi }{4} \right )$

$\dpi{120} C = \left ( \frac{a}{\sqrt{2}},\frac{5\pi }{4} \right )$

$\dpi{120} D = \left ( \frac{a}{\sqrt{2}},\frac{7\pi }{4} \right )$

Practice problems:

Find all the polar coordinates for given point P

(2, π/6 )

Convert the following rectangular equation to polar form.

$\dpi{120} \left ( x-3 \right )^{2}+y^{2}=9$

Convert the given polar equation to rectangular form r cscθ = 1

Radar detects two airplanes at the same altitude. Their polar coordinates are (8 mi,110°) and (5 mi, 15°). How far apart are the airplanes?

Answers :

$\dpi{120} \left ( 2, \frac{\pi }{6}+2n\pi \right ), \left ( -2,\frac{\pi }{6} +(2n+1)\pi \right )$
r=6cosθ
$\dpi{120} x^{2}+\left ( y-\frac{1}{2} \right )^{2}=\frac{1}{4}$
9.80 miles

Graphs of Polar equations

September 16, 2019/by admin_ct

How to Graph Polar equations.

Polar curves are just the special cases of parametric curves. Polar curves are graphed in the (x,y) plane, despite the fact that they are given in terms of r and θ. That is why the polar graph of r =4cosθ is a circle.

In polar mode, points are determined by a directed distance from the pole that changes as the angle sweeps around the pole.

Symmetry:

There exist three types of symmetry in polar equations.

The x axis (polar axis) as a line of symmetry.
The y axis (the line θ=π/2 ) as a line of symmetry.
The origin (the pole) as point of symmetry.

Example1. Use the symmetry tests to prove that the graph of r=4sin(3θ ) is symmetric about the y-axis.

Solution: Replace (r,θ ) with (-r, – θ) and we get,

-r = 4sin(3(-θ ))

-r = 4sin(-3θ )

-r = -4sin(3θ ) sine as odd function, sin(- θ) =-sinθ

r = 4sin(3θ )

After replacement we get same equation as original so this polar equation has symmetry about y axis.

Graphing a polar equation

Graphing polar equation is much similar to graphing rectangular equation in (x,y) coordinates. In polar equations we assume random values of and find r values corresponding to that and then plot those points on polar coordinate grid. Knowing symmetric tests is an advantage in graphing polar equations manually.

There are some basic shapes for polar equations. It is always helpful to have knowledge of these basic shapes before sketching polar equations.

1) Circles in polar form:

Polar equation	Description	Graph
r=a	Circle with center at pole and ‘a’ as diameter.
r=a cosθ	Circle with diameter ‘a’ along x axis with its left most edge at pole
r=a sinθ	Circle with diameter ‘a’ along y axis with its bottom most edge at pole

2) Limacons(snails)

r = a ± b sin θ, where a > 0 and b > 0

r = a ± b cos θ, where a > 0 and b > 0

The limacons containing sine will be above the horizontal axis if the sign between a and b is plus or below the horizontal axis if the sign is minus. If the limaçon contains the function cosine then the graph will be either to the right of the vertical axis if the sign is plus or to the left of the vertical axis if the sign is minus.

The ratio of (a/b) will determine the exact shape of the limacon as explained in this table.

r = a ± b sin θ, where a > 0 and b > 0

Ratio(a/b )	r = a+bsinθ (above horz. axis)	r = a- bsinθ (below horz. axis)
$\dpi{120} \mathbf{\frac{a}{b}< 1}$
$\dpi{120} \mathbf{\frac{a}{b}= 1}$
$\dpi{120} \mathbf{1< \frac{a}{b}< 2}$
$\dpi{120} \mathbf{\frac{a}{b}> 2}$

The graphs of limaçons with cosine would have similar shapes but along the horizontal axis.

Special case : When ratio $\dpi{120} \mathbf{\left | \frac{a}{b} \right |=1}$ then limacon is called cardioid.

3) Rose curves:

A rose curve is a graph that is produced from a polar equation in the form of

r = a sin(nθ) or

r = a cos(nθ),

where a ≠ 0 and n is an integer > 1

They are called rose curves because the loops that are formed resemble petals. The number of petals depend on the value of n. The length of the petals depand on the value of a.

If n is an even integer, then the rose will have 2n petals as shown in following examples.

r = asin(2θ ) r = acos(4θ )

If n is an odd integer then rose will have n petals as shown in following examples.

r = a cos(5θ ) r= a cos(3θ )

4) Lemniscates:

lemniscates has the shape of a figure-8 or a propeller. General polar equation of lemniscates is given as,

$\dpi{120} \mathbf{r^{2}=a^{2}sin(2\theta )}$ or

$\dpi{120} \mathbf{r^{2}=a^{2}cos(2\theta )}$ where a ≠ 0

A lemniscate containing the sine function will be symmetric to the pole while the lemniscate containing the cosine function will be symmetric to the polar axis, to θ =π/2 , and the pole.

$\dpi{120} \mathbf{r^{2}=a^{2}sin(2\theta )}$ $\dpi{120} \mathbf{r^{2}=a^{2}cos(2\theta )}$

Example2. Graph the polar equation r = 4cosθ .Use the knowledge of symmetric tests to complete the graph manually.

Solution:

Step1.First we construct a table of r,θ values. We assume random values of but we prefer to use standard angles. It is useful to first find those values of θ which makes values of r as maximum, minimum and 0.

Step2.Since cosθ is even function so we check asymmetry about x axis by replacing θ with –θ .

r = 4cos(-θ )

r = 4cosθ cos is even function so cos(-θ )= cos(θ)

That shows this polar equation will be symmetric about x axis. So we just get the table for values of θ up to π . After that values of r will repeat themselves.

θ	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
r	4	3.5	2.8	2	0	-2	-2.8	-3.5	-4

Now we plot these points on polar grid. We have to be careful while plotting negative r values. As explained in previous section, To get negative r values we add pi to specific θ values.

For example to plot point (-2, 2π/3 ) we get (-2, 2π/3+ π ) = (-2, 5π/3 ) as shown in graph below. Negative r values are obtained by rotating the point by 180°

(a)

(b)

Example3 Graph the polar equation r = 3cos(2θ ).

Solution:

Step1.recognizing the equation. The polar equation is in the form of a rose curve, r = a cos nθ. Since n is an even integer, the rose will have 2n petals.

2n = 2(2) = 4 petals

Step 2. Tests for symmetry:

Polar axis

θ = 2π

Pole

r = 3 cos 2θ

r = 3 cos 2(-θ)

r = 3 cos (-2θ)

r = 3 cos 2θ

-r = 3 cos 2(-θ)

r = -3 cos (-2θ)

r = –3 cos 2θ

r = 3 cos 2θ

-r = 3 cos 2θ

r = –3 cos 2θ

Passes symmetry test

Fails symmetry test

Step 3.Now we evaluate r at different values of .

θ	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
r	3	3/2	0	-3/2	-3	-3/2	0	3/2	3

These points will provide us with enough points to complete the rest of the graph using the symmetry of the rose curve and we get the graph as below.

Example4. Graph the polar equation r = 1 – 2 cos θ.

Solution:

Step1. Recognizing the equation-

The polar equation is in the form of a limaçon, r = a – b cos θ.

Find the ratio of a/b to determine the equation’s general shape.

a/b = 1/2 < 1

Since the ratio is less than 1, it will have both an inner and outer loop. The loops will be along the polar axis since the function is cosine and to the left since the sign between a and b is minus.

Step 2. Lets check the symmetry.

Polar axis	θ = 2π		Pole
r = 1 – 2 cos θ r = 1 – 2 cos (-θ) r = 1 – 2 cos θ	r = 1 – 2 cos θ -r = 1 – 2 cos (-θ) -r = 1 – 2 cos θ r = –1 + 2 cos θ	r = 1 – 2 cos θ -r = 1 – 2 cos θ r = –1 + 2 cos θ
Passes symmetry test	Fails symmetry test		Fails symmetry test

Step3. Find r at different value of .

θ	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
r	-1	– 0.73	-0.41	0	1	2	2.41	2.73	3

Using above points and symmetry we get the following graph.

Parallel and Perpendicular lines in space

September 15, 2019/by admin_ct

Parallel Lines in space

Two lines having vector equations as

$\dpi{120} \overrightarrow{r_{1}}=\overrightarrow{a_{1}}+t\overrightarrow{b_{1}}$ and

$\dpi{120} \overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b_{2}}$
are parallel if their direction vectors b1 and b2 are parallel

i.e $\dpi{120} \mathbf{\overrightarrow{b_{1}}=k \overrightarrow{b_{2}}}$ for some scalar k.

Example1. Find the vector equation of a line passing through a point (2,-1,3) and parallel to the line r = (i+j) + t(2i+j-2k).

Solution: We know that any line parallel to given line will have same direction vectors as of the given line.

So we have to find the vector equation of a line which is passing through (2,-1,3) and have direction vector as 〈2,1,-2〉.

r = 〈2,-1,3〉 + t 〈2,1,-2〉

r = 〈 2+2t, -1+t , 3-2t〉

Parametric equations of line are:

x=2+2t, y=-1+t, z=3-2t

and Cartesian(symmetric) equation is given as…

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

Line passing through a point and perpendicular to two given lines.

Suppose a line is passing through point P and perpendicular to two lines

$\dpi{120} \overrightarrow{r_{1}}=\overrightarrow{a_{1}}+t\overrightarrow{b_{1}}$ and

$\dpi{120} \overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b_{2}}$

then its equation would be given as,

$\dpi{150} \mathbf{\overrightarrow{r}= \overrightarrow{p}+t\left ( \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}} \right )}$

Example2. A line passes through point (2,-1,3) and is perpendicular to the line = (i+j-k) + t(2i-2j+k) and = (2i-j-3k) + t(i+2j+2k). Find the equation of this line.

Solution:

$\dpi{120} \overrightarrow{b_{1}}\times \overrightarrow{b_{2}}=\begin{vmatrix} i & j & k\\ 2& -2 &1 \\ 1& 2& 2 \end{vmatrix}$

= i(-4-2) –j(4-1)+k(4+2)

= -6i-3j+6k

So the vector equation of perpendicular line would be,

$\dpi{120} \mathbf{\overrightarrow{r}= \overrightarrow{p}+t\left ( \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}} \right )}$

$\dpi{120} \mathbf{\overrightarrow{r}= \left \langle 2,-1,3 \right \rangle+t\left ( -6 -3,6 \right )}$

Intersection of two lines:

Following algorithm is used to find intersection of two lines.

Write the general points lying on both lines using parameters s and t.
If the lines intersect , then they must have a common point. So set the corresponding parts of both lines equal to each other.
Solve the three equations for parameters s and t. If these values satisfy the third equation then the lines intersect otherwise not.

Example3.Show that the lines r=〈1,1,-1〉 + s〈3,-1,0〉 and r = 〈4,0,-1〉 + t 〈2,0,3〉 intersect each other. Also find their point of intersection.

Solution: The general points lying on first line is given as,

r=〈1,1,-1〉 + s〈3,-1,0〉 => (1+3s , 1-s, -1)

General points lying on second line is given as,

r = 〈4,0,-1〉 + t 〈2,0,3〉 => (4+2t,0,-1+3t)

If two lines intersect then they must have some common point. So,

1+3s = 4+2t 1-s = 0 -1=-1+3t

3s-2t = 3 s =1

Substituting s=1 into 3s-2t=3, we get…

3(1)-2t = 3

-2t = 0 => t=0

These values of s and t must satisfy the third equation.

Substituting t=0 into -1+3t = -1

We get, -1 = -1

So these lines intersect at s=1 and t=0. Using these values in line equations , we get coordinates of intersecting point as,

(1+3s , 1-s, -1) = (4,0,-1)

(4+2t,0,-1+3t) = (4,0,-1)

Example4. Find point of intersection of following two lines.

$\dpi{120} {\color{Red} L_{1}: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}}$

$\dpi{120} {\color{Red} L_{2}: \frac{x-4}{5}=\frac{y-1}{2}= z}$

Solution: General points lying on line $\dpi{120} L_{1}: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=s$ are given as,

(2s+1 , 3s+2 ,4s+3)

Points on line $\dpi{120} L_{2}: \frac{x-4}{5}=\frac{y-1}{2}= z=t$ are given as,

(5t+4,2t+1,t)

If two lines intersect then they must have a common point. So,

2s+1=5t+4 3s+2=2t+1 4s+3 = t

2s-5t = 3 3s-2t = -1 4s-t = -3

Solving first two equations, we get

s = -1 and t= -1

Substituting these values into third equation, we get

4s-t =-3 => 4(-1)-(-1) = -3

-4+1 = -3

-3 = -3

The values of s and t satisfy third equation too so these lines intersect and coordinates of intersecting point are,

s =-1, (2s+1 , 3s+2 ,4s+3)= (2(-1)+1 , 3(-1)+2 ,4(-1)+3)

= (-1,-1,-1)

t= -1 (5t+4,2t+1,t) = (5(-1)+4,2(-1)+1, -1)

= (-1,-1,-1)

Check whether given lines are parallel, perpendicular or Skew.

Here is a video example to understand this concept.

Example 5. Here are two cool lines:

${\color{Red} L_1: x= 6+4t, y=4-3t , z=-7+t}$

${\color{Red} L_2 : \frac{x+1}{8}=\frac{y-1}{-6}=\frac{z+5}{2}}$

Are these lines parallel, intersecting or skew ?

Perpendicular Distance of a point from a line.

Perpendicular is the shortest distance of a point from the given line. The idea behind perpendicular distance is the projection of perpendicular (drawn from point to the line) on the given line.

This distance can be found using two ways.

First method (Using dot product) :

Let $\dpi{120} \overrightarrow{r} = \overrightarrow{a} +t \overrightarrow{b}$ be the given line. And P be the point whose distance is required from the line.

Write the position vector of a general point on the given line and assume it as point L.
Obtain $\dpi{120} \overrightarrow{PL}$ which would be in form of parameter t.
Set dot product $\dpi{120} \overrightarrow{PL}.\overrightarrow{b}=0$ and solve for parameter t.
Plug in t value into $\dpi{120} \overrightarrow{PL}$ and find $\dpi{120} \left |\overrightarrow{PL} \right |$

Here $\dpi{120} \left |\overrightarrow{PL} \right |$ is the perpendicular (shortest distance) from point P to line L.

Example5: Find shortest distance(length of perpendicular) of point P(0,2,3) from the line

$\dpi{120} {\color{Red} \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}}$

Solution: let L be any point lying on this given line and general form of this point is

$\dpi{120} \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t$

Step 1) Vector form of this line is r = 〈-3,1,-4〉 + t 〈 5,2,3〉

x= 5t-3, y=2t+1 , z=3t-4

So point L, in general form is given as,

L = (5t-3,2t+1,3t-4)

Step 2)

$\dpi{120} \overrightarrow{PL}$ = L-P = 〈5t-3-0, 2t+1-2 , 3t-4-3〉

= 〈5t-3, 2t-1 , 3t-7〉

Step 3) Since PL is perpendicular to given line so,

$\dpi{120} \overrightarrow{PL}.\overrightarrow{b}=\left \langle 5t-3, 2t-1, 3t-7 \right \rangle . \left \langle 5,2,3 \right \rangle$

= 5(5t-3)+2(2t-1)+3(3t-7) = 0

Solving it, we get t =1

Step 4) Substituting t=1 into PL we get,

$\dpi{120} \overrightarrow{PL}= \left \langle 2,1,-4 \right \rangle$

$\dpi{120} \left |\overrightarrow{PL} \right |= \sqrt{2^{2}+(-1)^{2}+4^{2}}=\sqrt{21} units$

Second method (using cross product):

For this method we use following steps.

Get a point lying on the given line , Say it L.
Find vector LP .
Find cross product of vector LP and direction vector of line, b and use the formula

$\dpi{120} \mathbf{d= \frac{\left | \overrightarrow{LP }\times \overrightarrow{b}\right |}{\left |\overrightarrow{b} \right |}}$

Example5: Find shortest distance(length of perpendicular) of point P(0,2,3) from the line

$\dpi{120} {\color{Red} \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}}$

Solution: Using Cartesian form of line,

$\dpi{120} \frac{x-x_{1}}{b_{1}}=\frac{y-y_{1}}{b_{2}}=\frac{z-z_{1}}{b_{3}}$

(x1 ,y1 ,z1 ) is the point lying on this line and 〈b1,b2,b3〉 is the direction vector of this line.

Step 1) Let L = (-3,1,-4) is the point on this line and b = 〈5,2,3〉 is the direction vector of this line.

Step 2) LP = (0,2,3)-(-3,1,-4) = (3,1,7)

Step 3)

$\dpi{120} \overrightarrow{LP}\times \overrightarrow{b }= \begin{vmatrix} i & j & k\\ 3 & 1 & 7 \\ 5& 2& 3 \end{vmatrix}$

= i(3-14)-j(9-35)+k(6-5)

= -11i+26j+k

$\dpi{120} \left |\overrightarrow{LP}\times \overrightarrow{b} \right |=\sqrt{(-11)^{2}+26^{2}+1^{2}}= \sqrt{798}$

$\dpi{120} \left |\overrightarrow{b} \right |=\sqrt{5^{2}+2^{2}+3^{2}}=\sqrt{38}$

Distance $\dpi{120} d = \frac{\sqrt{798}}{\sqrt{38}}=\sqrt{21}units$

Practice problems:

Find the vector equation of a line passing through a point (2,-1,3) and parallel to the line r = (i-2j+k) + t(2i+3j-5k).
Show that the lines

$\dpi{120} L_{1}: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and

$\dpi{120} L_{2}: \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$

intersect each other. Also find their point of intersection.

Find the length of perpendicular drawn from the point (5,4,-1) to the line r= i + t(2i+9j+5k).

Answers:

r = (2i-j+3k) + t(2i+3j-5k)
(1/2 ,-1/2, -3/2)
√(2109/110)

Shortest Distance between two lines

September 15, 2019/by admin_ct

How to find shortest distance between two lines in space.

The shortest distance d between two lines

$\dpi{120} \mathbf{\overrightarrow{r_{1}}=\overrightarrow{a_{1}}+s\overrightarrow{b_{1}}}$ and

$\dpi{120} \mathbf{\overrightarrow{r_{2}}=\overrightarrow{a_{2}}+s\overrightarrow{b_{2}}}$

is given by the following formula.

$\dpi{150} \mathbf{d = \left |\frac{\left ( \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right ).\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )}{\left | \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right |} \right |}$

Example1: Find shortest distance between the lines

r =〈4,-1,0〉 + s〈1,2-3〉 and r =〈1,-1,2〉 + t〈2,4,-5〉 .

Solution: Here we have a1 = (4,-1,0) , a2 = (1,-1,2) and b1 = (1,2,-3) , b2 = (2,4,-5)

Step1) a2-a1 = (1,-1,2)-(4,-1,0) =

Step2)

$\dpi{120} \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}}=\begin{vmatrix} i & j & k\\ 1& 2& -3\\ 2 & 4 & -5 \end{vmatrix}$

= i(-10+12) –j(-5+6) +k(4-4)

= 2i-j+0k

Step3) $\dpi{120} \left |\overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right |=\sqrt{2^{2}+(-1)^{2}+0^{2}}=\sqrt{5}$

Step4) Plug in all the values into formula

$\dpi{120} d = \left |\frac{\left ( \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right ).\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )}{\left | \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right |} \right | =\left |\frac{\left \langle -3,0,2 \right \rangle.\left \langle 2,-1,0 \right \rangle}{\sqrt{5}} \right |$

$\dpi{120} =\left |\frac{-6+0+0}{\sqrt{5}} \right |= \frac{6}{\sqrt{5}}$

Example2.By computing the shortest distance determine whether the following pair of lines intersect or not.

$\dpi{120} {\color{Red} \frac{x-1}{2}=\frac{y+1}{3}=z }$ and $\dpi{120} {\color{Red} \frac{x+1}{5}=\frac{y-2}{1},z=2 }$

Solution: Given equation can be rewritten in proper form as,

$\dpi{120} \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1}$ and $\dpi{120} \frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}$

Here we have a1 = (1,-1,0) , a2 =(-1,2,2) , b1= (2,3,1) and b2=(5,1,0)

Step1) a2 -a1 = (-1,2,2)-(1,-1,0) = (-2,3,2)

Step2)

$\dpi{120} \overrightarrow{b_{1}}\times\overrightarrow{ b_{2}}=\begin{vmatrix} i & j & k\\ 2& 3& 1\\ 5 & 1 & 0 \end{vmatrix}$

= i(0-1) –j(0-5)+k(2-15)

= -i+5j-13k

Step3) $\dpi{120} \left (\overrightarrow{ a_{2}}-\overrightarrow{a_{1}} \right ).\left ( \overrightarrow{b_{1}}\times \overrightarrow{b_{2}} \right )$

= (-2,3,2).(-1,5,-13)

= 2+15-26= -9 ≠ 0

If the shortest distance between two lines is 0 , then the lines intersect each other.

Here we got that distance is not 0 so the two lines doesn’t intersect each other.

Shortest distance between two Parallel lines:

Lets assume two parallel lines with same direction vector b are given by the equations

$\dpi{120} \mathbf{\overrightarrow{r_{1}}=\overrightarrow{a_{1}}+s\overrightarrow{b}}$ and

$\dpi{120} \mathbf{\overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b}}$

Then the shortest distance between them is found by the following formula.

$\dpi{150} \mathbf{d = \left |\frac{\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times b}{\left | \overrightarrow{b} \right |} \right |}$

Example3: Find shortest distance between the lines

r = 〈1,2,3〉 + s 〈2,3,4〉 and r =〈2,4,5〉+ t 〈4,6,8〉 .

Solution: Rewriting second equation we get ,

r = 〈2,4,5〉 + 2t 〈2,3,4〉

We see that both lines have same direction vector , therefore both are parallel lines. Here we have a1 = (1,2,3) , a2 = (2,4,5) and b= (2,3,4).

Step1)

a2 -a1 = (2,4,5)-(1,2,3) = 〈1,2,2〉

Step2)

$\dpi{120} \left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times\overrightarrow{ b}=\begin{vmatrix} i & j & k\\ 1& 2& 2\\ 2 & 3 & 4 \end{vmatrix}$

= i(8-6) –j(4-4)+k(3-4)

= 2i-0j-k

Step3)

$\dpi{120} \left |\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times \overrightarrow{b} \right |=\sqrt{2^{2}+0+(-1)^{2}}=\sqrt{5}$

$\dpi{120} \left | b \right |=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}$

Step 4) Plug in the values into the formula.

$\dpi{120} \mathbf{d = \left |\frac{\left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )\times b}{\left | \overrightarrow{b} \right |} \right |}=\frac{\sqrt{5}}{\sqrt{29}}$

Practice problems:

Find shortest distance between the lines

r = 〈3,8,3〉 + s〈3,-1,1〉 and r = 〈-3,-7,-6〉 +t 〈-3,2,4〉

Find shortest distance between the lines

$\dpi{120} \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ , and $\dpi{120} \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{5}$

By computing the shortest distance determine whether the following pair of lines intersect or not.

$\dpi{120} \frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ , and $\dpi{120} \frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}$

Answers:

√270
1/√6
NO

Straight Line in Space

September 13, 2019/by admin_ct

Different forms of Straight line in space

Any straight line in space can be written in three forms – vector equation, Cartesian form and parametric form

Vector equation of a line passing through a fixed point with position vector $\dpi{120} \overrightarrow{a}$ and parallel to a given vector $\dpi{120} \overrightarrow{b}$ is given as,

$\dpi{150} \mathbf{\overrightarrow{r}=\overrightarrow{a}+t\overrightarrow{b}}$ where t is a scalar.

Here a is the point lying on the line and b is the direction of line.

Vector equation of a line passing through two points:

If a line is passing through two points $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ then its vector equation is written as,

$\dpi{150} \mathbf{\overrightarrow{r}=\overrightarrow{a}+t\left (\overrightarrow{b} -\overrightarrow{a} \right )}$

Cartesian(symmetric) equations of a line

Cartesian equations of a line passing through two given points (x1,y1,z1) and (x2,y2,z2) is given by

$\dpi{120} \mathbf{\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}}$

This is also called symmetric equations of line.

Parametric form of a line

Set of parametric equations of a line can be obtained from Cartesian form itself by setting each equation equal to any parameter say ‘t’.

$\dpi{120} \mathbf{\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}=t}$

Then parametric equations are given as,

$\dpi{120} \mathbf{\frac{x-x_{1}}{x_{2}-x_{1}}=t \Rightarrow x=x_{1}+t(x_{2}-x_{1})}$

$\dpi{120} \mathbf{\frac{y-y_{1}}{y_{2}-y_{1}}=t \Rightarrow y=y_{1}+t(y_{2}-y_{1})}$

$\dpi{120} \mathbf{\frac{z-z_{1}}{z_{2}-z_{1}}=t \Rightarrow z=z_{1}+t(z_{2}-z_{1})}$

Example1.Find the vector equation of the line passing through points A(3,4,-7) and B(1,-1,6). Also find the Cartesian equation and parametric equations of line.

Solution: Given position vectors of two points are,

a = 〈3,4-7〉 b =〈1,-1,6〉

Vector equation of line is given as,

$\dpi{120} \overrightarrow{r}=\overrightarrow{a}+t\left (\overrightarrow{b} -\overrightarrow{a} \right )$

$\dpi{120} \overrightarrow{r}=\left \langle 3,4,-7 \right \rangle +t\left (\left \langle 1,-1,6 \right \rangle-\left \langle 3,4,-7 \right \rangle \right )$

=〈3,4,-7〉 + t〈1-3,-1-4,6+7〉

= 〈3,4,-7〉+ t〈-2,-5,13〉

Cartesian(symmetric) form is given as,

$\dpi{120} \frac{x-3}{1-3}=\frac{y-4}{-1-4}=\frac{z-(-7)}{6+7}$

$\dpi{120} \frac{x-3}{-2}=\frac{y-4}{-5}=\frac{z+7}{13}$

Parametric equations are given as,

$\dpi{120} \frac{x-3}{-2}=\frac{y-4}{-5}=\frac{z+7}{13}= t$

x = -2t +3

y = -5t+4

z = 13t-7

Example2. Determine if the line that passes through the point (0,−3,8) and is parallel to the line given by x=10+3t, y=12t and z= −3−t passes through the xz-plane. If it does, then give the coordinates of that point.

Solution: Line is given in parametric form. Direction vector of given line are the coefficients of t 〈3,12,-1〉.

Any line parallel to this given line will have same direction vector. So the vector form of new parallel line passing through point (0,-3,8) is given as

$\dpi{120} \overrightarrow{r}=\left \langle 0,-3,8 \right \rangle +t\left \langle 3,12,-1 \right \rangle$

=〈 3t, -3+12t, 8-t〉

If this line passes through xz plane then y coordinate must be 0, therefore

-3+12t = 0

t =1/4

So, the line does pass through the xz-plane. To get the complete coordinates of the point all we need to do is plug t=1/4 into equations. We’ll use the vector form

$\dpi{120} \overrightarrow{r}=\left \langle 3t,-3+12t,8-t \right \rangle=\left \langle 3\left ( \frac{1}{4} \right ),-3+12\left ( \frac{1}{4} \right ),8-\frac{1}{4} \right \rangle$

$\dpi{120} =\left \langle \frac{3}{4},0,\frac{31}{4} \right \rangle$

This is the point where new line is passing through xz plane.

Example3.The Cartesian(symmetric) equations of a line are 6x-2= 3y+1=2z-2. Find the vector equation of the line.

Solution: In symmetric form of line , coefficients of x,y and z must be unity. So to get proper form of symmetric equations ,we make the coefficients of x,y and z as unity.

6x-2 =3y+1 =2z-2

6(x- 2/6 ) = 3(y+1/3 ) = 2(z-1)

Dividing all sides by 6,

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

This shows line is passing through point (1/3 ,-1/3 ,1) and has direction vector as (1,2,3). So vector form of this line is given as,

$\dpi{120} \overrightarrow{r}=\left \langle \frac{1}{3},\frac{-1}{3},1 \right \rangle +t\left \langle 1,2,3 \right \rangle$

Example4. Find the direction cosines of given line . Also find its vector equation.

$\dpi{120} {\color{Red} \frac{x-2}{2}=\frac{2y-5}{-3} ,z=-1}$

Solution: Given symmetric equation can be rewritten as,

$\dpi{120} \frac{x-2}{2}=\frac{y-\frac{5}{2}}{\frac{-3}{2}} = \frac{z+1}{0}$

This shows this line is passing through point (2,5/2 ,-1) and has direction ratio proportional to 2,-3/2,0.

So its direction cosines are,

$\dpi{120} \frac{2}{\sqrt{2^{2}+\left (\frac{-3}{2} \right )^{2}+0^{0}}},\frac{\frac{-3}{2}}{\sqrt{2^{2}+\left (\frac{-3}{2} \right )^{2}+0^{0}}},\frac{0}{\sqrt{2^{2}+\left (\frac{-3}{2} \right )^{2}+0^{0}}}$

$\dpi{120} =\frac{2}{\sqrt{\frac{25}{4}}},\frac{\frac{-3}{2}}{\sqrt{\frac{25}{4}}},\frac{0}{\sqrt{\frac{25}{4}}}\Rightarrow \frac{2}{\frac{5}{2}},\frac{\frac{-3}{2}}{\frac{5}{2}},\frac{0}{\frac{5}{2}}\Rightarrow \frac{4}{5},\frac{-3}{5},0$

The line is passing through point having position vector 〈2, 5/2, -1〉 and direction vector as 〈2, -3/2 ,0〉 so its vector equation is written as,

$\dpi{120} \overrightarrow{r}=\left \langle 2,\frac{5}{2},-1 \right \rangle +t\left \langle 2,\frac{-3}{2},0 \right \rangle$

$\dpi{120} \overrightarrow{r}=2i+\frac{5}{2}j-k +t\left ( 2i-\frac{3}{2}j+0k\right )$

Example5.If the points A(-1,3,2) , B(-4,2,-2), C(5,5,k) are collinear then find value of k.

Solution: The Cartesian equations of a line passing through points A(-1,3,2) and B(-4,2,-2) are given as,

$\dpi{120} \frac{x-(-1)}{-4+1}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}$

$\dpi{120} \frac{x+1}{-3}=\frac{y-3}{-1}=\frac{z-2}{-4}$

$\dpi{120} \frac{x+1}{3}=\frac{y-3}{1}=\frac{z-2}{4}$

Since points A,B and C are given collinear so point C (5,5,k) must satisfy this equation,

$\dpi{120} \frac{5+1}{3}=\frac{5-3}{1}=\frac{k-2}{4}$

2 =2 = (k-2)/4

After solving 2=(k-2)/4 we get k= 10

Angle between two lines

Let the vector equations of two lines is given as,

$\dpi{120} \overrightarrow{r_{1}}=\overrightarrow{a_{1}}+t\overrightarrow{b_{1}}$ and

$\dpi{120} \overrightarrow{r_{2}}=\overrightarrow{a_{2}}+t\overrightarrow{b_{2}}$

And θ is the angle between them then,

$\dpi{150} \mathbf{cos\theta =\frac{b_{1}.b_{2}}{\left | b_{1} \right |\left | b_{2} \right |}}$

where b1 and b2 are the direction vectors of two lines.

Example6. Find the angle between the lines

$\dpi{120} {\color{Red} \overrightarrow{r_{1}}=\left \langle 3,2,-4 \right \rangle+t\left \langle 3,-2,0 \right \rangle}$

Solution: The direction vectors of given lines are :

b1=〈3,-2,0〉 and b2 =〈1, 3/2, 2〉

$\dpi{120} \left | b_{1} \right |=\sqrt{3^{2}+(-2)^{2}+0^{2}}=\sqrt{13}$

$\dpi{120} \left | b_{2} \right |=\sqrt{1^{2}+(\frac{3}{2})^{2}+2^{2}}=\sqrt{\frac{29}{4}}$

b1.b2= 〈3,-2,0〉. 〈1, 3/2, 2〉 = 3-3+0 = 0

$\dpi{120} {cos\theta =\frac{b_{1}.b_{2}}{\left | b_{1} \right |\left | b_{2} \right |}}=\frac{0}{\sqrt{13}\sqrt{29/4}}$

=> θ = π/2

Perpendicularity of two lines

If two lines are perpendicular then dot product of their direction vectors would be 0 and vice versa.

Example7. Find the value of k if the given lines are perpendicular to each other.

$\dpi{120} {\color{Red} \frac{1-x}{3}=\frac{7y-14}{2k}=\frac{z-3}{2}}$ and $\dpi{120} {\color{Red} \frac{7-7x}{3k}=\frac{y-5}{1}=\frac{6-z}{5}}$

Solution: First we need to write these equations in proper symmetric form making coefficients of x,y and z as unity.

$\dpi{120} \frac{x-1}{-3}=\frac{y-2}{\frac{2k}{7}}=\frac{z-3}{2}$ and $\dpi{120} \frac{x-1}{\frac{-3k}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$

Direction vectors of these lines are

b1=〈-3, 2k/7 ,2 〉 and b2 =〈-3k/7 ,1,-5〉

Since lines are given perpendicular so dot product of their direction vectors would be 0.

$\dpi{120} \overrightarrow{b_{1}}.\overrightarrow{b_{2}}=0\Rightarrow \left \langle -3,\frac{2k}{7},2 \right \rangle.\left \langle \frac{-3k}{7},1,-5 \right \rangle=0$

$\dpi{120} \frac{9k}{7}+\frac{2k}{7}-10=0$

11k/7 = 10

k = 70/11

Practice problems:

Find the vector equation as well as Cartesian equation of the line passing through two points A(1,2,-1) and B(2,1,1).
Show that the points with position vectors 〈 -2,3,0〉 , 〈1,2,3〉 and 〈7,0,-1〉 are collinear.
Show that the following lines are perpendicular to each other.

$\dpi{120} \frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\dpi{120} \frac{x-1}{-3}=\frac{y-2}{\frac{2k}{7}}=\frac{z-3}{2}$

If the coordinates of points A,B,C and D are (1,2,3) ,(4,5,7), (-4,3,-6) and (2,9,2) respectively . then find the angle between lines AB and CD.

Answers:

1) $\dpi{120} \overrightarrow{r}=\left \langle 1,2,-1 \right \rangle+t\left \langle 1,-1,2 \right \rangle , \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}$

4) 0

Vector (Cross) Product

September 12, 2019/by admin_ct

Vector(Cross) Product

Let $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ be two non zero, non parallel vectors. Then the vector product $\dpi{120} \overrightarrow{a}X\overrightarrow{b}$ is defined as,

$\dpi{150} \mathbf{\overrightarrow{a}x\overrightarrow{b}=\left | a \right |\left | b \right |sin\theta}$

Where θ is the angle between vectors a and b.

Properties of vector product:

If cross product of two vectors is zero that means vectors are parallel.

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=0 \Rightarrow$ parallel vectors

Vector product is not commutative.

$\dpi{120} \overrightarrow{a}X\overrightarrow{b} =-\left ( \overrightarrow{b}X\overrightarrow{a} \right )$

Vector Product is distributive

$\dpi{120} \overrightarrow{a}X\left ( \overrightarrow{b}+\overrightarrow{c} \right )=\overrightarrow{a}X\overrightarrow{b}+\overrightarrow{a}X\overrightarrow{c}$

$\dpi{120} \left ( \overrightarrow{b}+\overrightarrow{c} \right )X\overrightarrow{a}= \overrightarrow{b}X\overrightarrow{a}+\overrightarrow{c}X\overrightarrow{a}$

Some important applications of vector product

Area of parallelogram with adjacent sides $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ is

$\dpi{120} \mathbf{\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

Area of a triangle with adjacent sides $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ is

$\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

Area of parallelogram with diagonals $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ is

$\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

Volume of parallelepiped is given as

$\dpi{120} \mathbf{\left | \overrightarrow{a}.\left ( \overrightarrow{b}X\overrightarrow{c} \right ) \right |}$

Three vectors are coplanar if their scalar triple product is 0 i.e $\dpi{120} \mathbf{ \overrightarrow{a}.\left ( \overrightarrow{b}X\overrightarrow{c} \right ) }=0$ or three vector are coplanar if volume of parallelepiped formed by them is 0.

Example1. Find $\dpi{120} {\color{Red} \left | \overrightarrow{a}X\overrightarrow{b} \right |}$ for given vectors $\dpi{120} {\color{Red} \overrightarrow{a}}$ =i+3j-2k and $\dpi{120} {\color{Red} \overrightarrow{b}}$ =-i+3k.

Solution:

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=\begin{vmatrix} i & j & k \\ 1& 3 &-2 \\ -1& 0 & 3 \end{vmatrix}$

= i(9-0) –j(3-2) +k(0+3)

= 9i-j+3k

$\dpi{120} \left | \overrightarrow{a}X\overrightarrow{b} \right |= \sqrt{(9)^{2}+(-1)^{2}+(3)^{2}} =\sqrt{91}$

Example2. A plane is defined by any three points that are in the plane. If a plane contains the points P=(1,0,0), Q=(1,1,1) and R=(2,−1,3) find a vector that is orthogonal to the plane.

Solution : To find orthogonal(normal) vector of a plane, we just find the cross product of two vectors lying in that plane.

So first we find two vectors using three points lying in the plane.

$\dpi{120} \overrightarrow{PQ}$ = (1,1,1) –(1,0,0)= 〈0,1,1〉

$\dpi{120} \overrightarrow{PR}$ =(2,−1,3)–(1,0,0)= 〈1,-1,3〉

Orthogonal vector is,

$\dpi{120} \overrightarrow{PQ}X\overrightarrow{PR}=\begin{vmatrix} i & j & k \\ 0& 1 &1 \\ 1& -1 & 3 \end{vmatrix}$

= i(3+1) –j(0-1) +k(0-1) = 4i +j –k

Example3. Find a unit vector of magnitude 9, perpendicular to both the vectors = i-2j+3k and = i+2j-k.

Solution:

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=\begin{vmatrix} i & j & k \\ 1& -2 &3 \\ 1& 2 & -1 \end{vmatrix}$

= i(2-6) –j(-1-3) +k(2+2)

= i(-4)-j(-4) +4k

= -4i+4j+4k

$\dpi{120} \left | \overrightarrow{a}X\overrightarrow{b} \right |= \sqrt{(-4)^{2}+(4)^{2}+(4)^{2}} =\sqrt{48}=4\sqrt{3}$

Required unit vector

$\dpi{120} =9\frac{\overrightarrow{a}X\overrightarrow{b}}{\left | \overrightarrow{a}X\overrightarrow{b} \right |}$ (multiply with given magnitude 9)

$\dpi{120} =9\frac{\left \langle -4,4,4 \right \rangle}{4\sqrt{3}}=\frac{3\sqrt{3}\left \langle -4,4,4 \right \rangle}{4}= 3\sqrt{3}\left \langle -1,1,1 \right \rangle$

Example4. Show that area of parallelogram having diagonals 〈3,1,-2〉 and 〈1,-3,4〉 is 5√3 .

Solution: lets assume these diagonals are represented by vectors $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ .

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=\begin{vmatrix} i & j & k \\ 3& 1 &-2 \\ 1& -3 & 4 \end{vmatrix}$

= i(4-6) -j(12+2) +k(-9-1)

= -2i-14j-10k

$\dpi{120} \left | \overrightarrow{a}X\overrightarrow{b} \right |= \sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}} =\sqrt{300}$

Area of parallelogram = $\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

= 1/2 (√300) = 5√3

Example5.Find area of triangle whose vertices are A(3,-1,2), B(1,-1,-3) and C(4,-3,1).

Solution: First we find vectors using given points.

$\dpi{120} \overrightarrow{AB}$ = (1,-1,-3) –(3,-1,2)= 〈-2,0,-5〉

$\dpi{120} \overrightarrow{AC}$ = (4,−3,1)–(3,-1,2)= 〈1,-2,-1〉

$\dpi{120} \overrightarrow{AB}X\overrightarrow{AC}=\begin{vmatrix} i & j & k \\ -2& 0 &-5 \\ 1& -2 & -1 \end{vmatrix}$

=i(0-10)-j(2+5)+k(4-0)

= -10i-7j+4k

$\dpi{120} \left | \overrightarrow{AB}X\overrightarrow{AC} \right |= \sqrt{(-10)^{2}+(-7)^{2}+(4)^{2}} =\sqrt{165}$

Area of triangle ABC

$\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{AB}X\overrightarrow{AC} \right |}= \frac{1}{2}\sqrt{165}$

Example6. Determine if three vectors lie in same plane (coplanar) or not.

a=〈1,4,-7〉 , b=〈2,-1,4〉 , c =〈0,-9,18〉

Solution:

$\dpi{120} \overrightarrow{a}.\left ( \overrightarrow{b}X\overrightarrow{c} \right )= \begin{vmatrix} 1 &4 &-7 \\ 2& -1 & 4\\ 0& -9 & 18 \end{vmatrix}$

= 1(-18+36)-4(36-0)-7(-18-0)

= 18 – 144 +126 = 0

Since scalar triple product is 0 i.e volume of parallelepiped is 0 so all three vectors lie in same plane.

Practice problems:

Find a unit vector of magnitude 3, perpendicular to both the vectors a= 3i+j-4k and b= 6i+5j-2k.
Find the area of parallelogram determined by vectors 〈3,1,-2〉 and 〈1,-3,4〉
Determine of three vectors lie in same plane (coplanar) or not.

a=〈2,3,1〉 , b=〈1,-1,0〉 , c =〈7,3,2〉

Let a=i+4j+2k , b=3i-2j+7k and c=2i-j+4k. Find a vector d which is perpendicular to both a and b and c.d =15

Answers:

2i-2j+k
(5√3)/2 units
Yes, coplanar
1/3 (2i-2j-k)

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What is Logistic Differential Equation(LDE)

Polar coordinates

Plotting Points in the Polar Coordinate System

Finding all Polar Coordinates of a point

Coordinate conversion:

Converting equations:

Finding Distance using Polar Coordinates

How to Graph Polar equations.

Symmetry:

Graphing a polar equation

1) Circles in polar form:

2) Limacons(snails)

3) Rose curves:

4) Lemniscates:

Parallel Lines in space

Line passing through a point and perpendicular to two given lines.

Intersection of two lines:

Check whether given lines are parallel, perpendicular or Skew.

Perpendicular Distance of a point from a line.

How to find shortest distance between two lines in space.

Shortest distance between two Parallel lines:

Different forms of Straight line in space

Angle between two lines

Perpendicularity of two lines

Vector(Cross) Product

Properties of vector product:

Some important applications of vector product