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 2011 calculus bc free response of Khan Academy, please enable JavaScript in your browser. Donate Log in Sign up Search for courses, skills, and videos. AP Calculus BC

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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That is our y-axis. And then the next non-zero term is this negative 1 over here.

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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. Search for courses, skills, and videos. AP Calculus BC About About this video Transcript. Position and slope of particle's path at a given time. Created by Sal Khan.

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.

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So plus this x squared minus x squared over 2 factorial. Brian Liu. And then, if you have h prime of x, that's going to be equal to negative sine of x. 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. 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. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If I take the derivative of this, I get cosine of x. That's approximately equal to that. We have this term-- let me pick a nice color here-- we have this term, because we have a 1 coefficient. So you could say that f of x is going to be approximately equal to-- because we're just approximating it with this polynomial right over here. So we're centering it around x equals 0 right over here.

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.

And now we have the same denominator, because this is the same thing as 6 factorial. So it's negative x squared over 2 factorial. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. And then going straight through the center of the solid, you have your x-axis. Log in. All of that times pi over 4. Let's take a disk right over here. And this is going to be h prime of x. And so we can say that this expression right over here is the same as this expression right over here, from pi over 4 times the integral from 0 to k of 4e to the 4x dx. You have that negative out front. So, from cosine of x-- so you have this right over here is cosine of x. Let's find the first four non-zero terms of the Taylor series for cosine of x about x equals 0.