Initial Value Problem with Trigonometric Substitution


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.