Question

The radius of a right circular cone is increasing at a rate of 4 inches per second and its height is decreasing at a rate of 5 inches per second. At what rate is the volume of the cone changing when the radius is 10 inches and the height is 40 inches?

Answer 1

Answer:

$V' = 900\pi$ $in^{3} /sec$

Step-by-step explanation:

V = $\frac{1}{3} \pi r^{2}h$

$V' = \frac{2\pi rh *r'}{3} + \frac{\pi r^{2} *h'}{3}$ (using differentiation product rule)

plug known values in for r, h, r', and h'

$V' = \frac{2\pi (10)(40)(4)}{3} + \frac{\pi (10^{2} )(-5)}{3}$

$V' = \frac{3200\pi }{3} - \frac{500\pi }{3} \\V' = \frac{2700\pi }{3}$

$V' = 900\pi$ $in^{3} /sec$