Unit 4.5: Solving Related Rates

How can we solve Related Rates?


As shown in the problem on the right and in the previous unit, Related Rates are solved when given relation with variables and at least the change of one equation.

The general steps to solving an equation come in this order:

  • Write out all relevant information given, drawing diagrams if necessary

  • Identify the problem and find an equation relating the unknown and known variables

    • Note: this might not always be given, use the context of the problem to find anything you will use

  • Differentiate the equation with respect to time relating all relevant variables

  • Plug and solve all know values, remembering to keep units

Let’s use the example on the right to practice.

Related Rates problem

More Problems with walkthroughs

A spherical snowball melts so its volume decreases at 6 cm³/min. How fast is the radius changing when the radius is 4 cm? 

1) Let’s find all the relevant information:

  • Volume is changing at a rate of -6 cm³/min

  • Need a sphere

2) We need to identify the problem and write the equation:

  • We want to find the rate of change of radius when radius = 4

  • We know the equation relating volume and radius is:

3) Let’s Differentiate with respect to time:

4) We have gotten the equation and now its time to just plug in.

First, let’s find all the relevant information:

  • 2 people are 50 feet apart

  • θ is changing at a rate of 0.01 rad/min

  • They form a right triangle, but the drawing is given

Next, we need to identify the problem and write the equation:

  • We want to know the rate of change of the distance, which is the hypotenuse with length x, when θ = 0.5

  • We can write the equation comparing hypotenuse x, 50, and θ. The resulting equation is:

  • Rewrite the equation to solve in terms of x, we get the equation: which can be rewritten:

Now, let’s Differentiate with respect to time:

Finally, we have gotten the equation and now its time to just plug in.

  • We know that θ = 0.5 and

    • Note: unless specified, assume everything is in radians, not degrees. Check your calculator

  • By solving the equation we should get that

Styled Equation

cos(θ) = 50 x

Styled Equation

x = 50 sec(θ)

Derivative Formatting

dx dt =

Styled Equation

50 sec(θ) tan(θ)

Styled Derivative Equation

dθ dt

Styled Equation

x = 50 cos(θ)

Derivative Formatting

dθ dt = 0.01 rad/min

Derivative Formatting

dx dt = 31.125 ft/min

A 5.5-foot-tall person walks away from a 16-foot-tall streetlamp at a rate of 2.2 ft/sec. How fast is the tip of their shadow moving when the person is 10 feet from the lamp? 

1) Let’s find all the relevant information:

  • 16 foot tall lamp

  • Person is 5.5 feet

  • Person is moving away at a rate of 2.2 feet/sec

  • Draw a triangle

2) We need to identify the problem and write the equation:

  • We want to find the rate of the tip of the shadow when the person is 10 feet away

  • Write the equation in terms of x

3) let’s Differentiate with respect to time:

4) we have gotten the equation and now its time to just plug in.

But we need to remember what we are trying to solve for:

More Problems with walkthroughs

  • The water level is rising at 0.269 m/min. 

  • The top of the ladder is sliding down at 1.125 ft/sec.

  • The volume of the container is increasing at a rate of about 86.4 cm^3/sec

  • The radius is increasing at a rate of about 0.0796 cm/sec.

  • The side length of the ice cube decreases at a rate of 1/3 cm/min.

  • The water is rising at a rate of 0.716 cm/min.

  • The shadow is shortening on the wall at a rate of 0.781 m/s.

  • The volume between the two balloons is increasing at about 515.5 cm^3/sec.