Question

A woman is emptying her aquarium at a steady rate with a small pump. The water pumped to a 12-in.-diameter cylindrical bucket, and its depth is increasing at the rate of 4.0 in. per minute. Find the rate at which the aquarium water level is dropping if the aquarium measures 24 in. (wide) × 36 in. (long) × 18 in. (high).

Answer 1

Answer:

Therefore the rate at which water level is dropping is $\frac{11}{21}$ in per minute.

Step-by-step explanation:

Given that,

The diameter of cylindrical bucket = 12 in.

Depth is increasing at the rate of = 4.0 in per minutes.

i.e $\frac{dh_1}{dt}=4$

$h_1$ is depth of the bucket.

The volume of the bucket is V = $\pi r^2 h$

                                                 $=\pi \times 6^2\itimes h_1$

$\therefore V=36\pi h_1$

Differentiating with respect yo t,

$\frac{dV}{dt}=36\pi \frac{dh_1}{dt}$

Putting  $\frac{dh_1}{dt}=4$

$\therefore\frac{dV}{dt}=36\pi\times 4$

The rate of volume change of the bucket = The rate of volume change of the aquarium .

Given that,The aquarium measures 24 in × 36 in × 18 in.

When the water pumped out from the aquarium, the depth of the aquarium only changed.

Consider h be height of the aquarium.

The volume of the aquarium is V= ( 24× 36 ×h)

V= 24× 36 ×h

Differentiating with respect to t

$\frac{dV}{dt}=24\times 36 \times \frac{dh}{dt}$

Putting $\frac{dV}{dt}=36\pi\times 4$

$36\pi\times 4= 24\times 36\times \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{36\pi \times 4}{24\times 36}$

$\Rightarrow \frac{dh}{dt}=\frac{11}{21}$

Therefore the rate at which water level is dropping is $\frac{11}{21}$ in per minute.