Main content

### Course: AP®︎/College Calculus BC > Unit 11

Lesson 3: AP Calculus BC 2011- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2011 Calculus BC free response #1 (b & c)

Position and slope of particle's path at a given time. Created by Sal Khan.

## Want to join the conversation?

- Analytically, it is sqrt(pi/2)S(t.sqrt(2/pi)) as per wolfram alpha(1 vote)
- Fresnel Integrals are not covered in the AP Calculus BC curriculum (think first 2 semesters of college calculus) which is why those students would need to use the calculator on this portion of the FRQ.(4 votes)

- for4:30. couldn't we solve the integral like this:

∫sin (t^2) dt=1/2∫(1-cos(2t))dt and then use u-substitution?(2 votes)- You got the trig identity wrong. It must be sin^2(t) not sin(t^2).(2 votes)

- At1:54, you calculate the value for sin9/13, but I heard you do not need to simplify on the AP test, so would I be okay in leaving it as sin9/13 or did that policy change or not apply to the calculator part?(1 vote)
- It is always better to evaluate the answer where possible, so that means you would have to calculate the value for sin9/13.(2 votes)

- At4:30, Sal said it was hard to take the integral of sin(t)^2. Why?(1 vote)
- When you integrate sin(x^2)dx you have to do a u substitution to integrate the sine function, so you set u= to the inside function which is x^2, so now you have the integral from 0 to t of sin(u)dx where u=x^2, now you need to have your integral all in one variable which would look like sin(u)du, so you need to relate du to dx, du=2xdx and herein lies the problem because you cannot introduce another x variable to the integral expression.(1 vote)

## Video transcript

Part b-- find the slope
of the tangent line to the path of the particle
at time t is equal to 3. So the slope of the
tangent line is just going to be equal to
the rate of change of y with respect to x at that point. And this is the same
thing as dy/dt over dx/dt. And dealing with differentials
is a little bit strange, especially when you want
to deal with rigorously, but you can view them as very
little, small changes in y, small changes in,
t small changes in x, small changes in t. When you view it this way,
you could say, hey, look. If I just multiply both
sides, both the numerator and the denominator,
by dt over dt-- which, if you do view them as
the same small change-- then they would
cancel out, and you would get back to
the dy over dx. And the reason why
I wrote it this way is that they actually
give us these values. They give us dy/dt and
they give us dx/dt. So if you want to find the
slope at time equals 3, we just have to find dy/dt at
time equals 3 and dx/dt at time equals 3. So dy/dt at time
equals 3-- well, this is dy/dt right over here. It's just going to be sine
of 3 squared, or sine of 9. And dx/dt at time 3, that's
just going to be-- they tell us what dx/dt is
as a function of time. 4 times 3 plus 1. Well, that is equal to 13. And then we can just
get our calculator out and evaluate this. We did a lot of this work
in the first problem. But let's just actually
take the value. So we have sine of 9 divided
by 13 gets as 0.0317. And we're done. That's the slope
of the tangent line to the path of the particle
at time t is equal to 3. I think we have time. We could crank out
part c as well. Part c-- find the position
of the particle at time t is equal to 3. So they give us x prime of t,
and they give us y prime of t. This is x prime of t. This is y prime of t. We need to figure out
x of x of 3 and y of 3. So if we can, let's
write x of t and y of t and then try to
evaluate them at 3. So let's see. x of
t is going to be equal to the antiderivative
of this right over here. So the antiderivative of 4t
is going to be 2t squared. Take the derivative here. You'll just get 4t. Antiderivative of
1 is t, plus t, and then you're going to
have a plus a constant. Or essentially, we just
took the indefinite integral of this side, of 4t plus 1. So you get your constant here. Because if you take
the derivative, this constant will
obviously disappear. You'll lose that information. Lucky for us, they also gave
us this initial condition. x of 0 is 0, when t is 0,
this term is going to be 0. This term is going to be 0. And they're saying
the whole thing is going to be equal to 0. So c is going to be 0, as well. Let me do that explicitly
just so we don't get confused. So x of 0 is going to be equal
to 0 plus 0 plus c, which they're telling
us is equal to 0. So c is equal to 0. So that tells us
that x of t is going to be equal to 2t
squared plus t. So if you want to
find x of 3, you just evaluate it right over here. So if you want to
find x of 3, it's going to be 2 times 9 plus 3. 2 times 9 is 18, plus 3 is 21. So x of 3 is 21. Now let's try to do y of t. I'll do it in a different color. So let me scroll over to
the right a little bit. So let's say y of t is
equal to-- so if you try to take the
antiderivative of this, this is not simple to
take the antiderivative of this right over here. It's actually very hard
to do it analytically. Not even clear if you
can do it analytically. Lucky for us, this
part of the AP exam, we are allowed to
use calculators. So we could just say
this is equal to-- and this is straight from
the fundamental theorem of calculus-- y of t is equal
to the integral from 0 to t. And I'll just write it
as sine of x squared dx. I didn't want to use a t here
and then use a t, as well. That would make it confusing. So you integrate with
respect to x, and then one of the upper bound
is going to be t. And this is y of t. And of course, you're going
to have to put a constant over here, because if you took
the derivative of this, you would get y of t, and
you'll lose information from this constant. Now, we can figure out
what that constant is, because they tell
us what y of 0 is. So y of 0 is going to be equal
to the integral from 0 to 0. So this part right over
here is going to be 0. So it's going to be equal to
c, which they tell us y of 0 is also equal to negative 4. So we get y of t is equal
to the integral from 0 to t of sine of x
squared dx minus 4. So that's y as a function of t. And we already figured
out what x of 3 is. Now we have to figure
out what y of 3 is. y of 3 is going to be
equal to the integral from 0 to 3 of sine of x
squared dx and then minus 4. And, once again, lucky for us,
we can use calculators here. So let's get the
calculator out, and we want to go to the
catalog of functions. So I press 2nd
Custom here, which gives us this little
catalog thing. And I want to do the
definite integral. And it's actually FN INT. So let me go to the F's. FN INT. So I have to go a
little bit further down. A little bit. There you go. That's my function
that I want to use. This is a definite
integral, and then I want to evaluate the
definite integral of this right over here. So the function is going
to be sine of x squared-- that's what I want to
evaluate the definite integral of-- from 0 to 3. And my variable--
I need to see what the variable of
integration is first-- so my variable of
integration is x, and I'm going to go from 0 to 3. And this gives me-- let the
calculator think a little bit-- 0.77356, and then to
that-- so that's just this first part right over here. And then I need to subtract 4. So minus 4 gives negative
3.226, I'll go with. So this is equal
to negative 3.226. Did I write that down right? I have a bad memory. Negative 3.226, and we're done. So y of 3 is equal
to negative 3.226. So the position of
the particle at time t equals 3, if you
wanted a coordinate, it would be 21, negative 3.226.