Calculus (Derivatives of Inverse Functions)
Suppose f(x) = sin(pi cos(x)). On any interval where the inverse function y = f^-1(x) exists, the derivative of f^-1(x) with respect to x is: I've come as far as y = arccos ((arcsin(x))/pi), but I am not certain this is right.
Thank you. From now on, I will include my work for reference. As you can see, I do know (to a degree) what I am doing. However, my practice instructions aren't always very clear on how to carry out these problems so I look for a second opinion to back me up. Again, thank you.
The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3. Here is what I have so far: Radius = 1/(x^3) Area of Cross Section = pi(1/(x^3))^2 Simplified: pi(1/(x^6)) Volume = (definite integral from 1 to 3) pi(1/(x^6)) dx = pi( -1 / 5(3)^5) - pi(-1 / 5(1)^5) = pi (-1 ...
Calculus (Solid of Revolution)
Here is what I have so far: Radius = 1/(x^3) Area of Cross Section = pi(1/(x^3))^2 Simplified: pi(1/(x^6)) Volume = (definite integral from 1 to 3) pi(1/(x^6)) = pi( -1 / 5(3)^5) - pi(-1 / 5(1)^5) = pi (-1 / 1215) - pi (-1 / 5) = pi(242 / 1215) = 0.625732858 Sincerely, Mooch
Calculus (Volume of Solids)
This is precisely why I posted, I thought that the wording of this practice problem might make sense to someone else, because it completely confused me. After a lot of thinking, I figured that the solid of revolution was a hemisphere. It was created by rotating the quarter cir...
awesome! Thank you. OK how about four fives to get 4??
tried 22-2x2 =18 22-2+2=18 how do I get to 14
show how to use four two's to get the quanity 14