Uploaded by TheIntegralCALC on 02.11.2010

Transcript:

Hi everyone. Welcome back to integralcalc.com. We're going to be doing both an initial value

problem and a trigonometric substitution problem today. It’s going to be Initial Value Problem

Example 8 and Trigonometric Substitution Example 7. Both concepts are illustrated in this example.

The problem that we're going to be doing is dy/dt equals 1 divided by 25 plus 9t squared.

And then we're also given this initial condition here, y of 5/3 equals pi/30.

{dy/dt}~=~{1/{25~+~9t^2}},~~y(5/3)~=~pi/30 So the first thing that we're going to do

with this problem is go ahead and lay out a bunch of formulas. because this is a trigonometric

substitution problem, we have this identity here, a squared plus u squared and then x

equals a tan of theta and the identity, 1 plus tan squared theta equals sec squared

of theta. a^2~+~u^2

x~=~a~tan theta 1~+~{tan^2} theta~=~{sec^2} theta

If you're having trouble or you're unfamiliar with either trigonometric substitution or

initial value problems, you can look at both of those sections on my website for more information

and more examples. For now I’m going to pretend you have basic familiarity with both

so that we can just move forward with this. So we can identify here that our trigonometric

substitution identity is a squared plus u squared because a represents the constant

and u represents the variable here and we have our constant plus our variable. So we

have to identify the value of a and the value of u, which is going to be the square root

of both of these terms. a^2~=~25

u^2~=~{9t^2} So a is going to be 5, because we take the

square root of 25. a^2~=~{25}

sqrt{a^2}~=~sqrt{25} a~=~5

And u is going to be 3t, because we take the square root of 9t squared to get a equals

5 and u equals 3t. u^2~=~{9t^2}

sqrt{u^2}~=~sqrt{9t^2} u~=~3t

So given those two values, we can plug them into our x equals a tan of theta here.

x~=~a~tan theta And we do so by plugging in u for x and then

we have 3t equals 5 tan theta. 3t~=~5~tan theta

And then we have this identity for later because we're going to need it to make a substitution.

So then we have a couple things to do before we get started. We first need to solve this

equation here for t. We do that by dividing both sides by 3 and we get t equals 5/3 tan

theta. 3t~=~5~tan theta

{3t}/3~=~{5~tan theta}/3 t~=~{5/3}~tan theta

Then we need to find the derivative of t, dt by taking the derivative of this right-hand

side over here, which we know is 5/3 sec squared of theta d theta.

t~=~{5/3}~tan theta} dt~=~{5/3}~{sec^2} theta~d theta

So we have those two things and then the last thing we need to do is first solve for tan

theta which we pulled from this equation here by dividing both sides by 5, so we get tan

theta equals 3/5 t. 3t~=~5~tan theta

{3t}/5~=~{5~tan theta}/5 {3t}/5~=~tan theta

And then we use that equation to solve for theta. we take the inverse tan of both sides

which makes this tan over here on the right-hand side cancel and we're left with theta equals

tan to the negative 1 of 3/5 t. {3t}/5~=~tan theta

{tan^{-1}}({3t}/5)~=~ {tan^{-1}}(tan theta) {tan^{-1}}({3/5}t)~=~theta

So that's all the prep work that we have to do when we have a trigonometric substitution

problem. Now that we have all those things laid out, we can go ahead and start actually

solving the problem. So I've rewritten the problem down here without the initial condition,

dy/dt. {dy/dt}~=~{1/{25~+~9t^2}}

And then the first thing that we're going to do is multiply both sides by dt and then

integrate. {dy/dt}(dt)~=~{1/{25~+~9t^2}}(dt)

dy~=~{1/{25~+~9t^2}}(dt) int{}{}{dy~~~}~=~ int{}{}{1/{25~+~9t^2}}~dt

I’ve gone ahead and integrated both sides after I moved dt over to the right hand side

here. Now for the integration. The left-hand side, this integral of dy simply becomes y

equals. The integral of dy is y, the integral of just dt alone would be t. so we're left

with just y on the left-hand side. y~=~ int{}{}{1/{25~+~9t^2}}~dt

And then the integral on the right-hand side, we have to tackle with trigonometric substitution.

So the first thing that we do, you can see here on the original problem, we have t.

{dy/dt}~=~{1/{25~+~9t^2}} When we solved for t up here, we got t equals

5/3 tan theta. t~=~{5/3}~tan theta

So we plug that in for our t right here. y~=~ int{}{}{1/{25~+~9t^2}}~dt

y~=~ int{}{}{1/{25~+~9({5/3}~tan theta)^2}}~dt Don’t forget to keep this squared sign here

so that we don't lose it. It’s very important. And then we also plug in for dt here. We solved

for dt up here, we're got 5 sec squared theta d theta.

dt~=~{5/3}~{sec^2} theta~d theta y~=~ int{}{}{1/{25~+~9({5/3}~tan theta)^2}~~~}~({5/3}~{sec^2}

theta)~d theta So we go ahead and plug that all in ere. Once

we've plugged in for t and dt, we can go ahead and start simplifying, substituting all of

those things. The first thing I do is I bring this 5/3 out of the integral and we can do

that because everything inside the integral is multiplied together so that 5/3 is like

a coefficient on this entire term and coefficients can be brought out in front.

y~=~ {5/3}~int{}{}{1/{25~+~9({5/3}~tan theta)^2}~~~}~({sec^2} theta)~d theta

And then the other thing I do is I move the sec squared of theta just up here to the numerator

to simplify. y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25~+~9({5/3}~tan

theta)^2}}~d theta And then the last thing I did in this step

was, I simplified this 9 times the 5/3 squared here. So 5/3 squared would be 25/9.

y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25~+~9({25/9})~{tan^2} theta}}~d theta

The 9s would cancel and we'll be left with 25.

y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25~+~{25}~{tan^2} theta}}~d theta

So we have that 25 here and notice that I also have tan squared theta. Because I have

this squared sign here so not only do we have to square the 5/3, we also have to square

tan theta. So that simplifies the 25 tan squared theta.

In the next step, I go ahead and factor the denominator because we're trying to get to

a point where we can use our identity to solve here. I go ahead and factor out the 25 and

I’m left with 1 plus tan squared theta, which is equal to sec squared theta and which

is part of our trigonometric substitution formula.

y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25(1~+~{tan^2} theta)}}~d theta

1~+~{tan^2} theta~=~{sec^2} theta So substituting, we have sec squared of theta

here on the bottom. y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25({sec^2}

theta)}}~d theta The 25, I pulled out in front, 5/3 times 1/25

would be 1/15. y~=~ {5/3}~(1/25)~int{}{}{{{sec^2} theta }/{{sec^2}

theta}}~d theta y~=~ {1/15}~int{}{}{{{sec^2} theta }/{{sec^2}

theta}}~d theta So I simplified that coefficient and then

make our substitution and we're just left with sec squared of theta divided by sec squared

of theta, which means that those two will cancel and we'll just be left with 1/15 times

the integral of d theta. y~=~ {1/15}~int{}{}{~d theta~~~}

That’s the only thing left in our integral. so remember, before, we took the integral

of dy as y, if we took the integral of dt it would be t. well, on this case taking the

integral of d theta will simply give us theta. So we have 1/15 times theta and then of course

we add c as that constant of integration to account for the constant.

y~=~ {1/15}~int{}{}{~d theta~~~} y~=~{1/15} theta~+~c

So that c is also going to be what we need to use our initial condition when we reach

our problem to solve for c. we have solved for theta and our answer for theta was tan to the negative 1 of 3/5t so

we plug in this whole thing here for theta. {tan^{-1}}({3/5}t)~=~theta

y~=~{1/15}~{{tan^{-1}}({3/5}t)}~+~c So we plug that in and then the next thing

we do is we use our initial condition. y(5/3)~=~pi/30

This is saying here that y of t equals some value. So when 5/3 is equal to t, y is equal

to pi/3. So we go ahead and plug in pi/30 for y here. And then we go ahead and plug

in 5/3 down here for t. {pi/30}~=~{1/15}~{{tan^{-1}}({3/5}(5/3))}~+~c

And then we simplify that and then eventually we can solve for c. So first thing you're

going to notice in the next step is we're going to have 3/5 times 5/3. So that's going

to cancel and just give you 1. {pi/30}~=~{1/15}~{{tan^{-1}}(1)}~+~c

I'm going to move everything to the left side. I'm going to subtract this whole term here

so I get c on its own, because again we're trying to solve for c.

{pi/30}~-~{1/15}~{{tan^{-1}}(1)}~=~c So tan to the negative 1 or the inverse tan

of 1 is pi/4 so we go ahead and substitute that in and then I'm going to simplify.

{{tan^{-1}}(1)}~=~{pi/4} {pi/30}~-~{1/15}~(pi/4)~=~c

1/15 times pi/4 would of course be pi/60. {pi/30}~-~{pi/60}~=~c

So solving there, I multiply the first term with 2/2 so that I can get a common denominator

here of 60. So we get 2 pi/60 minus pi/60 which of course is pi/60.

{{2 pi}/60}~-~{pi/60}~=~c {pi/60}~=~c

So I’ve solved for c as pi/60 and now what were going to do is take this value and plug

it in right here for our c. {y}~=~{1/15}~{{tan^{-1}}(1)}~+~{pi/60}

And that's going to be our final answer. So our final answer is this 1/15, and then arc

tan is the same thing as the inverse tan. {tan^{-1}}~=~arctan

It’s a cleaner way of writing it instead of tan to the negative 1. They’re equivalent.

So arctan of 3/5 t and then of course plus, instead of c, pi/60 because we went ahead

and solved for c. {y}~=~{1/15}~{arctan~({3/5}t)}~+~{pi/60}

So with trigonometric substitution, that is how you solve for this initial value problem.

So hope that helped and I will see you guys next time. Bye.

problem and a trigonometric substitution problem today. It’s going to be Initial Value Problem

Example 8 and Trigonometric Substitution Example 7. Both concepts are illustrated in this example.

The problem that we're going to be doing is dy/dt equals 1 divided by 25 plus 9t squared.

And then we're also given this initial condition here, y of 5/3 equals pi/30.

{dy/dt}~=~{1/{25~+~9t^2}},~~y(5/3)~=~pi/30 So the first thing that we're going to do

with this problem is go ahead and lay out a bunch of formulas. because this is a trigonometric

substitution problem, we have this identity here, a squared plus u squared and then x

equals a tan of theta and the identity, 1 plus tan squared theta equals sec squared

of theta. a^2~+~u^2

x~=~a~tan theta 1~+~{tan^2} theta~=~{sec^2} theta

If you're having trouble or you're unfamiliar with either trigonometric substitution or

initial value problems, you can look at both of those sections on my website for more information

and more examples. For now I’m going to pretend you have basic familiarity with both

so that we can just move forward with this. So we can identify here that our trigonometric

substitution identity is a squared plus u squared because a represents the constant

and u represents the variable here and we have our constant plus our variable. So we

have to identify the value of a and the value of u, which is going to be the square root

of both of these terms. a^2~=~25

u^2~=~{9t^2} So a is going to be 5, because we take the

square root of 25. a^2~=~{25}

sqrt{a^2}~=~sqrt{25} a~=~5

And u is going to be 3t, because we take the square root of 9t squared to get a equals

5 and u equals 3t. u^2~=~{9t^2}

sqrt{u^2}~=~sqrt{9t^2} u~=~3t

So given those two values, we can plug them into our x equals a tan of theta here.

x~=~a~tan theta And we do so by plugging in u for x and then

we have 3t equals 5 tan theta. 3t~=~5~tan theta

And then we have this identity for later because we're going to need it to make a substitution.

So then we have a couple things to do before we get started. We first need to solve this

equation here for t. We do that by dividing both sides by 3 and we get t equals 5/3 tan

theta. 3t~=~5~tan theta

{3t}/3~=~{5~tan theta}/3 t~=~{5/3}~tan theta

Then we need to find the derivative of t, dt by taking the derivative of this right-hand

side over here, which we know is 5/3 sec squared of theta d theta.

t~=~{5/3}~tan theta} dt~=~{5/3}~{sec^2} theta~d theta

So we have those two things and then the last thing we need to do is first solve for tan

theta which we pulled from this equation here by dividing both sides by 5, so we get tan

theta equals 3/5 t. 3t~=~5~tan theta

{3t}/5~=~{5~tan theta}/5 {3t}/5~=~tan theta

And then we use that equation to solve for theta. we take the inverse tan of both sides

which makes this tan over here on the right-hand side cancel and we're left with theta equals

tan to the negative 1 of 3/5 t. {3t}/5~=~tan theta

{tan^{-1}}({3t}/5)~=~ {tan^{-1}}(tan theta) {tan^{-1}}({3/5}t)~=~theta

So that's all the prep work that we have to do when we have a trigonometric substitution

problem. Now that we have all those things laid out, we can go ahead and start actually

solving the problem. So I've rewritten the problem down here without the initial condition,

dy/dt. {dy/dt}~=~{1/{25~+~9t^2}}

And then the first thing that we're going to do is multiply both sides by dt and then

integrate. {dy/dt}(dt)~=~{1/{25~+~9t^2}}(dt)

dy~=~{1/{25~+~9t^2}}(dt) int{}{}{dy~~~}~=~ int{}{}{1/{25~+~9t^2}}~dt

I’ve gone ahead and integrated both sides after I moved dt over to the right hand side

here. Now for the integration. The left-hand side, this integral of dy simply becomes y

equals. The integral of dy is y, the integral of just dt alone would be t. so we're left

with just y on the left-hand side. y~=~ int{}{}{1/{25~+~9t^2}}~dt

And then the integral on the right-hand side, we have to tackle with trigonometric substitution.

So the first thing that we do, you can see here on the original problem, we have t.

{dy/dt}~=~{1/{25~+~9t^2}} When we solved for t up here, we got t equals

5/3 tan theta. t~=~{5/3}~tan theta

So we plug that in for our t right here. y~=~ int{}{}{1/{25~+~9t^2}}~dt

y~=~ int{}{}{1/{25~+~9({5/3}~tan theta)^2}}~dt Don’t forget to keep this squared sign here

so that we don't lose it. It’s very important. And then we also plug in for dt here. We solved

for dt up here, we're got 5 sec squared theta d theta.

dt~=~{5/3}~{sec^2} theta~d theta y~=~ int{}{}{1/{25~+~9({5/3}~tan theta)^2}~~~}~({5/3}~{sec^2}

theta)~d theta So we go ahead and plug that all in ere. Once

we've plugged in for t and dt, we can go ahead and start simplifying, substituting all of

those things. The first thing I do is I bring this 5/3 out of the integral and we can do

that because everything inside the integral is multiplied together so that 5/3 is like

a coefficient on this entire term and coefficients can be brought out in front.

y~=~ {5/3}~int{}{}{1/{25~+~9({5/3}~tan theta)^2}~~~}~({sec^2} theta)~d theta

And then the other thing I do is I move the sec squared of theta just up here to the numerator

to simplify. y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25~+~9({5/3}~tan

theta)^2}}~d theta And then the last thing I did in this step

was, I simplified this 9 times the 5/3 squared here. So 5/3 squared would be 25/9.

y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25~+~9({25/9})~{tan^2} theta}}~d theta

The 9s would cancel and we'll be left with 25.

y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25~+~{25}~{tan^2} theta}}~d theta

So we have that 25 here and notice that I also have tan squared theta. Because I have

this squared sign here so not only do we have to square the 5/3, we also have to square

tan theta. So that simplifies the 25 tan squared theta.

In the next step, I go ahead and factor the denominator because we're trying to get to

a point where we can use our identity to solve here. I go ahead and factor out the 25 and

I’m left with 1 plus tan squared theta, which is equal to sec squared theta and which

is part of our trigonometric substitution formula.

y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25(1~+~{tan^2} theta)}}~d theta

1~+~{tan^2} theta~=~{sec^2} theta So substituting, we have sec squared of theta

here on the bottom. y~=~ {5/3}~int{}{}{{{sec^2} theta }/{25({sec^2}

theta)}}~d theta The 25, I pulled out in front, 5/3 times 1/25

would be 1/15. y~=~ {5/3}~(1/25)~int{}{}{{{sec^2} theta }/{{sec^2}

theta}}~d theta y~=~ {1/15}~int{}{}{{{sec^2} theta }/{{sec^2}

theta}}~d theta So I simplified that coefficient and then

make our substitution and we're just left with sec squared of theta divided by sec squared

of theta, which means that those two will cancel and we'll just be left with 1/15 times

the integral of d theta. y~=~ {1/15}~int{}{}{~d theta~~~}

That’s the only thing left in our integral. so remember, before, we took the integral

of dy as y, if we took the integral of dt it would be t. well, on this case taking the

integral of d theta will simply give us theta. So we have 1/15 times theta and then of course

we add c as that constant of integration to account for the constant.

y~=~ {1/15}~int{}{}{~d theta~~~} y~=~{1/15} theta~+~c

So that c is also going to be what we need to use our initial condition when we reach

our problem to solve for c. we have solved for theta and our answer for theta was tan to the negative 1 of 3/5t so

we plug in this whole thing here for theta. {tan^{-1}}({3/5}t)~=~theta

y~=~{1/15}~{{tan^{-1}}({3/5}t)}~+~c So we plug that in and then the next thing

we do is we use our initial condition. y(5/3)~=~pi/30

This is saying here that y of t equals some value. So when 5/3 is equal to t, y is equal

to pi/3. So we go ahead and plug in pi/30 for y here. And then we go ahead and plug

in 5/3 down here for t. {pi/30}~=~{1/15}~{{tan^{-1}}({3/5}(5/3))}~+~c

And then we simplify that and then eventually we can solve for c. So first thing you're

going to notice in the next step is we're going to have 3/5 times 5/3. So that's going

to cancel and just give you 1. {pi/30}~=~{1/15}~{{tan^{-1}}(1)}~+~c

I'm going to move everything to the left side. I'm going to subtract this whole term here

so I get c on its own, because again we're trying to solve for c.

{pi/30}~-~{1/15}~{{tan^{-1}}(1)}~=~c So tan to the negative 1 or the inverse tan

of 1 is pi/4 so we go ahead and substitute that in and then I'm going to simplify.

{{tan^{-1}}(1)}~=~{pi/4} {pi/30}~-~{1/15}~(pi/4)~=~c

1/15 times pi/4 would of course be pi/60. {pi/30}~-~{pi/60}~=~c

So solving there, I multiply the first term with 2/2 so that I can get a common denominator

here of 60. So we get 2 pi/60 minus pi/60 which of course is pi/60.

{{2 pi}/60}~-~{pi/60}~=~c {pi/60}~=~c

So I’ve solved for c as pi/60 and now what were going to do is take this value and plug

it in right here for our c. {y}~=~{1/15}~{{tan^{-1}}(1)}~+~{pi/60}

And that's going to be our final answer. So our final answer is this 1/15, and then arc

tan is the same thing as the inverse tan. {tan^{-1}}~=~arctan

It’s a cleaner way of writing it instead of tan to the negative 1. They’re equivalent.

So arctan of 3/5 t and then of course plus, instead of c, pi/60 because we went ahead

and solved for c. {y}~=~{1/15}~{arctan~({3/5}t)}~+~{pi/60}

So with trigonometric substitution, that is how you solve for this initial value problem.

So hope that helped and I will see you guys next time. Bye.