## Trending News

# Sand poured on the ground at the rate of 3.00 m^3/min . . .?

forms a conical pile whose height is one-third the diameter of its base. How fast is the altitude of the pile increasing when the radius of its base is 2.00 m?

This question is an application of derivatives. I really don't know where to start . . thanks for any help!

### 2 Answers

- Dr DLv 71 decade agoFavorite Answer
r = radius at any point t.

h = 2r/3 = height at any point t.

V = 1/3 * h * πr^2 = 2/9 * π r^3

dV/dt = dV/dr * dr/dt

= 2/9 * πr^2 * dr/dt

Given that dV/dt = 3 m^3 / min

dr/dt = 3 / (2/9 * πr^2)

when r = 2,

dr/dt = 27 / (8π) = 1.074 m/s

dh/dt = 2/3 * dr/dt = 0.7162 m/s

- CPUcateLv 61 decade ago
vol = V = 1/3 pi r^3 h

h = dia/3 . . . . or . . . . . dia = 3h = 2 r

r = 3/2 h. . . . . . . substitute to the above equation

V = 1/3 pi ( 3/2 h )^3 h = 9/8 h^4

dV/dt = 9/8 pi h^3 dh/dt . . . . differentiating

dV/dt = 9/8 pi h^3 dh/dt . . . . change h = dia / 3 = 2 r /3

dV/dt = 9/8 pi ( 2 r /3 )^3 dh/dt

dV/dt = 1/3 pi r ^3 dh/dt

dh/dt = 3 dV/dt / (pi r^3) = 3 ( 3 ) / ( pi 2^3 ) = 9 / 8 pi

dh/dt = 0.358 meter / min