Question

13. A plastic pond is being filled with water when a leak forms at the bottom of the pond, causing water to drain out.

h³
The volume of water, in in³, in the pool can be modeled by the function V (h) = where h is the height of
15
water inside the pond in inches. The rate that the height of the water is changing is given by
dh
dt
-2√h.
where h is measured in inches and t is measured in seconds. Find the rate that the volume of water in the pond
is changing with respect to time, when the height of the water is 4 inches.

Answer 1

The rate that the volume of water in the pond is changing with respect to time, when the height of the water is 4 inches, is -128/15 cubic inches per second.

Step-by-step explanation:

To find the rate that the volume of water in the pond is changing with respect to time, when the height of the water is 4 inches.

We need to find the derivative of the volume function V(h) with respect to time t and then evaluate it when h = 4 inches.

Given V(h) = (h^3) / 15 and dh/dt = -2√h, we can differentiate V(h) implicitly with respect to t using the chain rule.

dV/dt = dV/dh * dh/dt

= (3/15) * (h^2) * (-2√h)

= -2h^(5/2) / 15.

Substituting h = 4 into the expression for dV/dt,

dV/dt = -2(4^(5/2)) / 15

= -128 / 15.

Therefore, the rate that the volume of water in the pond is changing with respect to time, when the height of the water is 4 inches, is -128/15 cubic inches per second.