# Volume and Surface Area problems?

Problem 1) Water in the tank
A water tank in the shape of a right circular cone has a height of 10 feet. The top rim of the tank is a circle with a radius of 4 feet. If water is being pumped into the tank at the rate of 2 cubic feet per minute, what is the rate of change of the water depth, in feet per minute, when the depth is 5 feet?

### 2) Identify what you are looking for

$$\frac{dh}{dt}$$

### 3) Identify the Givens

• The height of the cone is 10 ft.
• The radius of the cone is 4ft
• $\frac{dv}{dt}=2\frac{ft^3}{min}$
• The height of the water at the instant we are ask to solve the problem is 5.

### 4) Identify the equation that relates the variables

$V=\frac{1}{3}\pi r^2h$

### 5**) Identify the equation that relates the variables

This step is specific to related rates problems that deal with cones.
We have to use similar triangles to eliminate one variable.

$\frac{10}{4}=\frac{h}{r} \\ r=\frac{4}{10}h \\ \color{Blue} {{r}=\frac{2}{5}h}$
Substitute in the volume equation:
$V=\frac{1}{3}\pi (\color{Blue}{\frac{2}{5}h)}^2h \\ \\ V=\frac{1}{3}\pi (\frac{4}{25}h^2)h \\ V=\frac{4}{75}\pi h^3$

### 6) Take the derivative of the equation

$V=\frac{4}{75}\pi h^3 \\ \frac{dv}{dt}=\frac{12}{75}\pi h^2\frac{dh}{dt}$

### 7) Substitute

$$2=\frac{12}{75}\pi (5)^2\frac{dh}{dt}$$

N/A

### 9) Solve

$$2=\frac{12}{75}\pi (5)^2\frac{dh}{dt} \\ 2=\frac{12}{75}\pi 25\frac{dh}{dt} \\ 2=\frac{12}{3}\pi \frac{dh}{dt} \\ 2=4\pi \frac{dh}{dt} \\ \frac{dh}{dt}=\frac{2}{4\pi} \\ \frac{dh}{dt}=\frac{1}{2\pi}$$
The rate of change of the water depth, when the depth is 5 feet is $\frac{1}{2\pi} \frac{ft}{min}$
Problem 2) Corn in a cone
Corn is poured through a chute at the rate of 10$\frac{ft^3}{min}$, and falls in a conical pile whose bottom radius is always half the height.
(a) How fast will the radius of the base change when the pile is 8 ft high?

### 2) Identify what you are looking for

$$\frac{dh}{dt}$$

### 3) Identify the Givens

• $\frac{dv}{dt}$
• For the cone: $r=\frac{1}{2}h$
• Radius of the water is 8ft

### 4) Identify the equation that relates the variables

$V=\frac{1}{3}\pi r^2h$

### 5**) Identify the equation that relates the variables

This step is specific to related rates problems that deal with cones.
We have to use similar triangles to eliminate one variable.

For the cone: $r=\frac{1}{2}h$
$\color{Blue}{h=2r }$
Substitute in the volume equation: $V=\frac{1}{3}\pi (r)^2(h) \\ V=\frac{1}{3}\pi (r)^2( \color{Blue}{2r}) \\ V=\frac{2}{3}\pi (r)^3$

### 6) Take the derivative of the equation

$$V=\frac{2}{3}\pi (r)^3 \\ \frac{dv}{dt}=2\pi (r)^2\frac{dr}{dt}$$

### 7) Substitute

$$10=2\pi (8)^2\frac{dr}{dt}$$

N/A

### 9) Solve

$$10=2\pi (8)^2\frac{dr}{dt}$$ $$10=2\pi (64)\frac{dr}{dt}$$ $$10=128 \pi \frac{dr}{dt}$$ $$\frac{dr}{dt}=\frac{10}{128 \pi}$$ The radius of the base change $\frac{10}{128 \pi} \frac{ft}{min}$ when the pile is 8 ft high?
The rate of change of the water depth, when the depth is 5 feet is $\frac{1}{2\pi} \frac{ft}{min}$

### 10) Extension

How fast is the circumference of the base increasing at this time? answer button The answer:
Formula: $$C=2\pi r$$ Derivative: $\frac{dC}{dt}=2 \pi \frac{dr}{dt}$ We know from part (a) we know that $\frac{dr}{dt}=\frac{10}{128 \pi}$ $\frac{dC}{dt}=2 \pi \big(\frac{10}{128 \pi}\big)$ $\frac{dC}{dt}=\frac{20}{128}$