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.

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. 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.

2011 calculus bc free response

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. Search for courses, skills, and videos. AP Calculus BC About About this video Transcript. Lagrange error bound for Taylor Polynomial approximation. Created by Sal Khan. Want to join the conversation?

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And so the fifth derivative at 0 is going to be 0. So you have up there. So this radius is e to the 2x power. This is what we figured out in part A. And then if we want the volume of this entire disk, we then multiply the area times the depth. That would seem to have the benefit of applying generally to any solid of revolution, you just plug in R. And we're done. If you want a more rigorous discussion, try out Chris Tisdell on Youtube. Could you also solve this problem through radian integration along the angle of the z-axis? Well, the radius is the height between the x-axis and the function. The region R is rotated about the x-axis to form a solid. So think about it this way. And so if we say that h of x is equal to cosine of x, then h of 0, is going to be equal to cosine of 0, which is equal to 1.

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You could say it's minus x to the 6th over 3 factorial, and then you could distribute the negative sign if you want. So this is going to be minus-- there's a negative sign out in front of both of them. So that is x squared minus x to the sixth over 3 factorial-- I'm going to just write the whole thing, and then later we'll pick out the first four terms-- plus x to the 10th over 5 factorial minus x to the 14th over 7 factorial. So times dx. I multiplied and divided by 4, multiplied and divided by 4. So that's going to be our solid and you can imagine that maybe this right over here is our y-axis. All right. So we can actually evaluate this integral analytically, it's not too bad. So we could say that cosine of x is approximately equal to-- because we're just approximating it with the first four non-zero terms. Flag Button navigates to signup page. Those are the first four non-zero terms of cosine of x. So then the cycle starts again. And so the fifth derivative at 0 is going to be 0. And then you get another 0 for the third degree term. So this radius is e to the 2x power.