Dec 26

Related Rates: Defining Variables

By Tutor GuyNo Comments

 

Related rate problems will no longer seem so complicated once you master the art of defining rates as variables. And not just any variables, but variables in the form of derivatives! In other words, express the known and unknown rates as derivatives such as dV/dt or dA/dt. Remember that dA/dt means “the rate of change of the Area”. Your task is to find expressions for rates in the problem and “translate” them from English into Calculus. Here are some examples:

  • “The radius is increasing at a rate of 3.0 cm/s.”

\dfrac{dr}{dt}=3.0

  •  “If the volume is decreasing at a rate of 1.5 liters per minute…”

\dfrac{dV}{dt}=-1.5

  • “How fast is the ladder falling down the side of the building?”

\dfrac{dy}{dt}= \, ?

 Remember that speed at which something moves is the rate of change of its position. Also, this derivative expresses the unknown we are going to find in the problem.

  • “The temperature is dropping by 2.3° per hour.”

\dfrac{dT}{dt}=-2.3

 Note that when a value is falling or decreasing, the value of the derivative is negative.

Once you have defined all your rates as variables, your next step is to write the equation that expresses one variable in terms of the others. However, the equation has to be in terms of the quantities in the numerators, not in terms of the derivatives. Yes, this seems all wrong, but you will get the derivatives when you differentiate each side of the equation. For example, if your two rates are dV/dt and dr/dt, then you need to write an equation that expresses V in terms of r. When you differentiate this equation, you get the related rates equation and can plug in all the values you have.

Example: A hot air balloon in the shape of a sphere is being filled with gas at the rate of 20 cubic meters per minute. How fast is the radius changing when the volume is 288π cubic meters?

Solution: First, find all the rates mentioned in the problem:

\dfrac{dV}{dt}=20; \; \dfrac{dr}{dt}= \, ?

 Now, you need an equation that relates volume (V) to radius (r). Because the balloon is a sphere, use the formula for the volume of a sphere:

V= \dfrac{4}{3} \pi r^3

 Next, differentiate with respect to time:

\dfrac{dV}{dt}=4 \pi r^2 \dfrac{dr}{dt}

 This equation includes r. You need to plug in the value of V in the original equation to solve for r:

V= \dfrac{4}{3} \pi r^3 \rightarrow 288 \pi= \dfrac{4}{3} \pi r^3 \rightarrow 216=r^3 \rightarrow r=6 \, m

 Now plug in all the values to solve for the unknown rate:

\dfrac{dV}{dt}=4 \pi r^2 \dfrac{dr}{dt} \rightarrow 20=4 \pi (6)^2 \dfrac{dr}{dt} \rightarrow \dfrac{dr}{dt}= \dfrac{20}{4 \pi (36)} \approx 0.044 \, m/min

 (Look for other tips on solving related rate problems on this website if you need help with other aspects of related rate problems.)

Calculus
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