# Related Rates

## What are related rates in Calculus? |

Many things change with time. Our goal is to find the rate at which some quantity is changing by relating the quantity to other quantities whose rates of change are known.

# Steps for Solving Related Rates Problems

## General Steps

- Draw a picture
- Identify what quantity you are looking for
- Identify the given information . The rates $\frac{ d(variable)}{dt}$ can often be identified by the problem stating changing, increasing or decreasing. If the rate is decreasing be sure to make the quantity negative.
- Identify the equation that relates the variables..
- Take the derivative of the equation that relates the variables with respect to "t" (remember to make use of the derivative rules) before you plug the given information in.
- Substitute the given rates and values into the derivative equation
- Use Equation in Step 4 to find other variables we're ultimately looking for.
- Solve for the desired quantity

**Problem 1**When a circular shield of bronze is heated over a fire its radius increases at the rate of $\frac{1}{5}$ cm/sec. At what rate is the shield's area increasing when the radius is 50 cm?

### 1) Draw A Picture

### 2) Identify what you are looking for

What rate the area is increasing - $\frac{dA}{dt}$

### 3) Identify the Givens

The radius is 50 cm (r = 50 )

Radius is increasing at the

Radius is increasing at the

**rate**of ($\frac{dr}{dt}=\frac{1}{5}$)### 4) Identify the equation that relates the variables

Since we are working with a circle $A=\pi r^2$

### 5) Take the derivative of the equation

$\frac{dA}{dt}=(2)(\pi)(r)(\frac{dr}{dt})$

### 6) Substitute

$
\frac{dA}{dt}=(2)(\pi)(r)(\frac{dr}{dt})
\\
\frac{dA}{dt}=(2)(\pi)(50)(\frac{1}{5})$

### 7) Use Equation in Step 4 to find other variables we're ultimately looking for

N/A

### 8) Solve

$\frac{dA}{dt}=20 \pi $ cm/s

**Problem 2**A spherical snowball with an outer layer of ice melts so that the volume of the snowball decreases at a rate of 2 $\frac{cm^3}{min}$. How fast is the radius changing when diameter of the snowball is 10 cm?

### 1) Draw A Picture

### 2) Identify what you are looking for

What rate the area is increasing - $\frac{dA}{dt}$

### 3) Identify the Givens

The radius is 50 cm (r = 50 )

Radius is increasing at the

Radius is increasing at the

**rate**of ($\frac{dr}{dt}=\frac{1}{5}$)### 4) Identify the equation that relates the variables

Since we are working with a circle $A=\pi r^2$

### 5) Take the derivative of the equation

$\frac{dA}{dt}=(2)(\pi)(r)(\frac{dr}{dt})$

### 6) Substitute

$
\frac{dA}{dt}=(2)(\pi)(r)(\frac{dr}{dt})
\\
\frac{dA}{dt}=(2)(\pi)(50)(\frac{1}{5})$

### 7) Use Equation in Step 4 to find other variables we're ultimately looking for

N/A

### 8) Solve

$\frac{dA}{dt}=20 \pi $ cm/s

## Related Rates Problems

Triangle Problems (the classic ladder problem ) | |

Sphere Problems (volume and surface area) | |

Cylinder Problems (volume changing, water draining etc..) | |

Cone Related Problems |

Calculator Calculus Homework

Answers