Question

solve for dh/dt given the following: V=1/3*pi*r^2*h, r=9, h=15, dV/dt=10pi

Answer 1

I substituted 9 into the volume equation to get:

1/3 * pi * 9^2 * h = V

27pi * h = V

Differentiated with respect to t:

27pi * h' = V'

Substituted V' in:

27pi * h' = 10pi

h'=10/27

Is this correct? I'm mainly concerned about whether or not I can substitute that 9 in in the beginning before differentiating.

Answer 2

The full question is:

A water tank has the shape of a circular cone pointed end points down with the radius of the top of 12 feet and height 20 feet. If water is pumped into the tank at the rate of 10pi ft^3/min find the rate at which the water level is rising when the water is 15 feet deep.

Answer 3

r = 3 h/ 5, v' = +10 ft³/min, h' at h = 15ft
So I'm right just plug h = 15 into my solution :)

Answer 4

Thus h' = 10π * 25 / 9π ( 15)²
= 250/ 9 ( 225)

Answer 5

= +.11 ft/ min

Answer 6

I'm kind of unclear about what your order of operations is in most steps, but did you get 10/81 without approximating?

Answer 7

@Chlorophyll

Answer 8

Yep,
= 250 / (225 * 9 )
= 10 / 81

Answer 9

Thanks