AP Calculus BC exams: 2008 1 c&d parts c and d of problem 1 in the 2008 AP Calculus BC free response
AP Calculus BC exams: 2008 1 c&d
- So we were doing part c of the first problem on the Calculus
- BC exam, the free response part, and so what, I'll re-read
- it, it says the region r, this is the region r,
- is the base of a solid.
- I redrew the region r here, but with a little perspective, so
- we can hopefully visualize it in three dimensions.
- So it says for this solid, each cross section perpendicular
- to the x-axis, right?
- There's two ways you can do a cross-section, you could do
- them, you could cut this way, but that would be
- parallel to the x-axis.
- We wanted to cut perpendicular to the x-axis, or parallel
- to the y-axis, right?
- So if we cut it along the line like that.
- So they say, each cross-section perpendicular to the
- x-axis is a square.
- So I drew a couple of squares here.
- That's one.
- So here, this would be the base, and since we know the
- cross-section is the square, the height has to be the
- same length as the base.
- Same thing here.
- So here, the height will be really high, because this is
- kind of maybe our maximum point in terms of the base width.
- But it's still pretty wide, then it gets narrow again.
- So how do we figure out the volume of this solid?
- Which is kind of hard to visualize.
- And it's in some ways you easier to visualize it than
- it is to draw, so you have to give me some credit.
- But anyway, what we do is, we take the area of each of these
- squares, and I drew one of them here.
- You take you each of the area of each of these squares,
- multiply them by a super small change in x, and that we know,
- from everything we've learned in calculus hopefully, that
- that super small change in x, I'm trying to draw a little
- perspective, is dx.
- So if we multiply dx times the area of this square, that is
- the volume of this kind of part the entire solid.
- And if we were to sum up all of these infinitely thin solids,
- we would get the volume for the whole thing.
- So how do we do that?
- Well let's write our integral expression.
- So what is the area of each of these squares?
- Each of these cross-sections, what is the area?
- Well the base is going to be the difference between
- our two functions, right?
- This top function right here, that was sin of pi x.
- And this bottom function, right here, that is y is equal to
- x to the third minus 4x.
- So the base of the functions, the base of this distance right
- here is going to be the difference between the top
- function and the bottom function.
- So each base is going to be sin of pi x, minus this function,
- so minus x to the third, plus 4x, right switch the sign.
- Minus x to the third, plus 4x.
- So far this might look pretty similar to part a, but
- what's the twist here?
- We want the area.
- The area of each of the squares, not
- just this distance.
- So what's the area?
- It's going to be this distance squared, so we have to
- square this entire thing.
- So that's the area of each of these squares, and then
- we have to multiply them times a little bit of dx.
- And that gives us the volume for each of these, I guess,
- parts of the entire solid.
- And then what are our boundaries of integration?
- Well it's the same thing as in part a.
- This is 0, this is 2.
- So the boundaries are pretty straightforward.
- So we essentially just have to evaluate this now.
- And just like in part b, I've first tried to evaluate this
- analytically, and you end up getting a very, very,
- very hairy integral.
- Which you can do analytically, you but you have to know some
- power reduction formulas in trigonometry, you have to do
- some integration by parts, it would probably use up all of
- the time on the AP Exam.
- So since they said that a graphing calculator is required
- for some parts of the problem, I say why not use our
- graphing calculator?
- Because the graphing calculator is very good at numerically
- evaluating definite integrals like this.
- So let's get out my TI-85 emulator again.
- Here we go.
- I want you see the key strokes so that you, ok turned
- on, exit out of this.
- So we're going to use the calculus function here,
- so second, calculus.
- And this function right here, this is definite integral, a
- very useful function to use.
- Press F5, definite integral.
- And then we just type in the expression.
- So the expression, let me move it down a little bit.
- So it's open parentheses sin of where is pi?
- Pi is -- I haven't used one of these calculators in a long,
- long time -- oh there it is.
- Second pi x sin of pi x minus x to the third power, plus 4x.
- All of that squared.
- And then this definite integral function, you have to tell it
- which is the independent variable, or kind of, you
- know, what variable are we integrating across, and
- that's the variable x.
- And then you just tell it the boundaries of integration,
- and we're done.
- So integrate from 0, from x is equal to 0, to x is equal to 2.
- If I haven't made a mistake, I can hit enter, and let
- the calculator do the rest of the work.
- Let's see what it ends up with.
- OK, 9. -- that's the answer, that is the volume of this
- solid -- it's 9.9783 So you could write that this
- is equal to 9.9783.
- And I'm pretty sure they want you to use a calculator,
- because frankly, computing the integral, that's kind of just
- chug math, you know, very mechanical math, although it's
- pretty sophisticated, but it would take you forever.
- But this is kind of, I think, what they wanted you to do, set
- up the integral, recognize that each of the squares, the area
- of each of the squares is just going to be this distance, the
- distance between the functions squared.
- And then you integrate that from 0 to 2.
- Let's see how much time I have left.
- I have a couple of minutes.
- So let's do part d.
- The region r models, let me, paste, oh, I didn't want to
- do that, edit, undo, edit, paste, there you go.
- OK, so I wanted to make this a little smaller.
- So what does part d say?
- The region r models the surface of a small pond,
- so that's the surface now.
- At all points r at a distance x from the y-axis, the depth of
- the water is given by h of x is equal to 3-x.
- So essentially 3-x is the depth, right?
- So at this point of this pond, the depth is just 3, right 3-0.
- And at this point, the depth is 3-2 which is 1.
- So essentially the pond is going to get shallower and
- shallower or as we go further to the right.
- You could almost imagine it, let me see if I
- can draw it again.
- So this is the sin function with some perspective.
- This is the polynomial function below it.
- This is, it's probably drawn in, that's the x-axis.
- This is the y-axis.
- And so here, the depth of the pond is given by the function
- h of x is equal to 3-x.
- So over here, the depth is 3, so if I were to draw, so if I
- were to go straight down the depth is, you know
- maybe it's 3.
- And the pond essentially gets shallower and shallower
- as we go to the right.
- So how do we figure out the volume of this?
- Over here, what is the depth?
- It's going to be 1, right, 3-2.
- This is x is equal to 2.
- So here the depth is going to be 1.
- So if we take the cross-section just along the x-axis, the
- depth is going to look something like this, but then
- this is the top of it.
- I know that's kind of hard to visualize.
- But anyway, how do we figure out what the
- volume of this lake is?
- But actually, I realize I'm pushing nine minutes, so I
- will continue this in the next video.
- See you soon.
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