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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But this essentially gives us everything we need to know. This is going to be this end of the solid when it's rotated around, and it's going to look something like this.

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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

2011 calculus bc free response

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So the derivative of sine of x is cosine of x. And then the seventh the derivative at 0 is going to be 0. Well, surface area of a circle is pi times the radius squared. Created by Sal Khan. And then cosine of x, we just figured out, is plus 1 minus x squared over 2 factorial plus x to the 4th over 4 factorial minus x to the 6th over 6 factorial. So this is going to be minus-- there's a negative sign out in front of both of them. So this is our volume, pi over 4 times e to the 4k minus 1. So that solid is going to look something like this. 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. Downvote Button navigates to signup page. So what's going to be the volume of this white disk right over here? Use this series and the series for sine of x squared, which we found right over here, sine of x squared-- actually, I think we wrote it, this is the series for sine of x squared when we simplified it. Let me take this disk right over here so that you can imagine this little bit of a cross section. Well, it's going to be the surface area of the disk times the depth.

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And then going straight through the center of the solid, you have your x-axis. And so when you evaluate it at k, it's e to the 4k. If the solid was slightly transparent, you would see the other side of this opening right over here, and the x-axis goes straight through the center. And then that side is going to look something like that. So between both of these, the lowest degree term right over here is this 1. So let's write the 1 over here. So where's our function again? The lowest degree term over here is this 1. And then you get another 0 for the third degree term. Pi over 4 is a constant, so we can actually take it out of the integral. Downvote Button navigates to signup page. At So plus x to the 4th over 4 factorial.