Question

The position function for an object is s(t)=3t^2+2t+5 where s is measured in meters and t is measured in seconds. Find the velocity of the object when t=2.

If the radius of a right circular cone equals its height, how fast is the volume changing with respect to changes in r when r=2 feet?

Answer 1

Given
\(s(t)=3t^2+2t+5\)
then
\(s(2)=3(2)^2+2(2)+5 = ?\)

Given r=h,
\(\large V=\frac{\pi r^3}{3}\)
\(\large \frac{dV}{dr}=V'(r)=\pi r^2\)
Calculate \(V'(2)\) for rate of change of volume at r=2.

Answer 2

thank you

Answer 3

You're welcome! :)