A water tank is formed by revolving the curve $y = 3 x ^ { 4 }$ about the $y$-axis. Water drains from the tank through a small hole in the bottom of the tank. At what constant rate does the water level, $y$, fall? (Use Torricelli's Law: $\mathrm { dV } / \mathrm { dt } = - \mathrm { m } \sqrt { \mathrm { y } }$.)
A) $\frac { d y } { d t } = \frac { - m \pi } { \sqrt { 3 } }$
B) $\frac { d y } { d t } = \frac { - \sqrt { 3 } } { m \pi }$
C) $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac { - \pi } { \mathrm { m } \sqrt { 3 } }$
D) $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac { - \mathrm { m } \sqrt { 3 } } { \pi }$

Question

Quizplus · Accepted Answer

First, we need to find the volume of the tank at any given height, which is the function we get when we revolve the curve around the y-axis. This is given by:

$$V(y) = \pi \int_{0}^{y} (3x^4)^2 dx = \pi \int_{0}^{y} 81x^8 dx = \frac{27}{5} \pi y^5$$

Now we can use Torricelli's law, which relates the rate at which the water level is falling to the rate at which the volume is decreasing:

$$\frac{dV}{dt} = -Am\sqrt{y} \frac{dy}{dt}$$

where A is the area of the hole and m is a constant. Since the hole is assumed to be small, we can assume that the cross-sectional area of the water tank does not change significantly as the water level decreases, so A is constant.

Taking the derivative of the volume function with respect to time, we get:

$$\frac{dV}{dt} = \frac{d}{dt} \left(\frac{27}{5} \pi y^5\right) = \frac{27}{5} \pi \frac{dy}{dt} 5y^4 = \frac{135}{5} \pi y^4 \frac{dy}{dt} = 27 \pi y^4 \frac{dy}{dt}$$

Substituting into Torricelli's law, we get:

$$27 \pi y^4 \frac{dy}{dt} = -Am\sqrt{y} \frac{dy}{dt}$$

Solving for dy/dt, we get:

$$\frac{dy}{dt} = \frac{-Am\sqrt{y}}{27 \pi y^4}$$

Substituting A = pi r^2 and r = 0 (since the hole is at the bottom), we get:

$$\frac{dy}{dt} = \frac{-m\sqrt{y}}{27 \pi}$$

Simplifying, we get:

$$\frac{dy}{dt} = \frac{-m}{27 \pi} \sqrt{y}$$

This matches the form of answer choice D.

A Water Tank Is Formed by Revolving the Curve y=3x4y = 3 x ^ { 4 }y=3x4

A Water Tank Is Formed by Revolving the Curve $y = 3 x ^ { 4 }$