2011 calculus bc free response

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2011 calculus bc free response

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So h prime prime of 0 is going to be equal to cosine of 0 is 1. Created by Sal Khan. And so when you evaluate it at k, it's e to the 4k.

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If you're seeing this message, it means we're having trouble loading external resources on our website. AP Calculus BC Arc length for a curve. So let's draw some axes here to just make sure and you wouldn't obviously have to do this if you were doing it under time pressure during the actual AP exam but my point here to make sure we all are understanding what's going on. Well they tell us; x 0 is 0 so x is 0 and y 0 is -4 so we're at the point 0, I'll just say that's 21 right over there. So this is 21 and we figured this out in the last problem, I think it was And between these point who knows what the path is.

2011 calculus bc free response

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Now part c. Let me just write them out. And we can figure out the derivative of V with respect to k very easily because we have V as a function of k right over here. Do we have any 6th degree terms? So let's just say-- well, we'll put our formula for the Taylor series up here around x is equal to 0. We will get our actual volume. Sine of x squared is this thing. And then, if we want to take the integral, you might be able to do this just by inspection, but if you want to say, hey look, I have a little function here, I want its derivative sitting here, what we can do is we can multiply this expression by 4, and we can also divide it by 4. So this radius is e to the 2x power. And then you have the second derivative of h of x. Those are the first four non-zero terms of cosine of x.

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Plus 1 times x to the 4th over 4 factorial. If you're seeing this message, it means we're having trouble loading external resources on our website. So we did this first part. So this part right over here is pi e squared. Volume of a solid of rotation and Chain Rule for rates of change. The lowest degree term over here is this 1. So we can actually evaluate this integral analytically, it's not too bad. We do. So this radius is e to the 2x power. Because either-- let me rewrite it. The volume, V, found in part b, changes as k changes. This is a non-zero term. And then, if you have h prime of x, that's going to be equal to negative sine of x. Brian Liu.