2011 calculus bc free response

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If you're seeing this message, it means we're having trouble loading external resources on our website. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Donate Log in Sign up Search for courses, skills, and videos. AP Calculus BC About About this video Transcript. Volume of a solid of rotation and Chain Rule for rates of change. Created by Sal Khan.

2011 calculus bc free response

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Well, surface area of a circle is pi times the radius squared. But we have that negative out front, so it's negative cosine of x. We need to solve for dV dt.

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2011 calculus bc free response

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We can rewrite this as pi over 4 times 4e to the 4x. So the fourth derivative at 0 is going to be the same thing as the function evaluated at 0, which is 1. It'll look a little bit like a loudspeaker. So I'll do this over here because I want this to be my final answer. So between both of these, the lowest degree term right over here is this 1. Well, the radius is the height between the x-axis and the function. So then the cycle starts again. All of that times pi over 4. So I could write, this is also equal to the fourth derivative at 0. So let's just say-- well, we'll put our formula for the Taylor series up here around x is equal to 0. So that's the first four terms of that.

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If you want a more rigorous discussion, try out Chris Tisdell on Youtube. If you multiply something by 4 and divide by 4, you haven't changed its value. So what's going to be the volume of this white disk right over here? Do we have any third degree terms here? Posted 12 years ago. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. So that's going to be our solid and you can imagine that maybe this right over here is our y-axis. Do we have any fourth degree terms here? I dont think so. So this would be the volume of this little disk over here that has a very small infinitesimally wide depth.