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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 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

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# 2015 AP Calculus BC 2d

Finding distance travelled.

## Want to join the conversation?

- No mention of why the absolute value of the integral wasn't needed to be entered into the calculator for the calculation.(2 votes)
- The absolute value (magnitude) of the velocity vector is equal to the speed.

|𝒗(𝑡)| = √(cos²(𝑡²) + 𝑒^𝑡)(2 votes)

- How could you figure out that integral by hand?(2 votes)

## Video transcript

- [Voiceover] Find the total distance traveled by the particle from time t equals zero to t equals one. Always remember they didn't say
find the total displacement, they said find the total distance traveled by the particle. So if something goes to the right by one, and then goes up by one, their distance is two. And actually, and then if they go back, if they go back, over here, just let me draw a straighter line, they then go back to the original starting point. And then this distance
right here would be what? The square root of,
the square root of two, the displacement would be zero. They got back to where they started. For the distance would be one plus one plus the square root of two. So how do we figure out the total distance in this scenario right over here? Let me erase this since it's
not relevant to the problem, it's just a, for remind ourselves we're talking about distance. Well distance is equal to rate times time. Or it's equal to speed times time. So in a very small amount of time, if you want to figure out the distance, well, you could take your speed, which is the magnus magnitude
of your velocity function. If you took your speed, that's your speed right over there, and if you multiply it by a
little small change in time, that's gonna give you your infinitesimally
small change in distance over that infinitesimally
small change in time. And if you wanted to change your, if you wanted to find the total distance over a non-infinitesimal change in time, well, then you could integrate. And you can integrate those
little changes in time in this case from t equals
zero to t equals one. This is going to be the
expression for the total distance. Well what is this going to be? Well this is going to
be equal to the interval from zero to one. In the last part of the problem, we already have an
expression for our speed. Our speed we saw was this
business right over here. So it's going to be
equal to the square root, give myself a little bit more space to work with, is
going to be equal to the square root of the
x-component of the velocity, cosign of t squared squared, let me put the little
outer parenthesis there, squared, plus the y-component
of our velocity squared, the rate of change of y with
respect to time squared. So plus e to the 0.5 t, that's the rate of change of y with respect to time, or
the y-component of velocity, we're gonna square that. This is our expression for our
speed as a function of time. And then you multiply that times dt and we're gonna integrate all this and this is gonna give us our total, this is gonna give us our total distance. And lucky for us we can use our calculator in this part of the AP exam,
so let's evaluate this. So let me turn it on. Let me clear out. And so we're just going to evaluate, let's go to math, and function integral right over there, that's for evaluating definite integrals, and we want to evaluate the square root, an open parenthesis, of, I think I'm gonna have
a lot of parenthesis here, but we'll see if we can do it. So the square root of, cosign of, and I'm gonna us x as
my variable integration cosign of x squared, so that's that parenthesis,
I need another parenthesis, squared, squared, plus e to this business squared. Well this we already solved before. This is the same thing
as e to the t, right? If we raise something to the 0.5 t, and then we raise that to the second power two times 0.5 t is just t. So I can do that if I like. I could just type in all of
this business if you like. So plus second e to the, my variable of integration here is x, e to the x, so close that, and then I close my square root. Did I do that right? This closes around the x squared, that closes with that,
yep ok that looks right. And then my variable of integration is x, and I'm integrating from zero to one. And now let's let the calculator
munch on it a little bit. And I get approximately 1.595. So this is approximately 1.595, the total distance traveled by the particle in time equals zero to t equals one which is kind of neat that
we can do these things, just from, you know, the
information that they gave us at the beginning of the problem.