Related Rates: Finding missing variables

By Tutor GuyNo Comments

 

Often, a related rate problem will require a value for a variable, but you don’t have the value. When this occurs, use the original equation (before you took the derivative) to find the unknown value.

Example: Two cars leave an intersection at the same time. Car A travels north at 60 feet per second. Car B travels east at 80 feet per second. How fast is the distance between them changing 1.0 minute after the cars leave the intersection?

Solution: First draw a picture.

 

 

 

 

 

 

It makes sense to let y be the distance car A has traveled from the intersection, and let x be the distance car B has traveled from the intersection. There are two rates given in the problem, and we need to find a third rate:

\dfrac{dy}{dt}=60; \; \dfrac{dx}{dt}=80; \; \dfrac{dr}{dt}= \, ?

 It should be obvious from the picture that the Pythagorean Theorem provides the equation we need:

x^2+y^2=r^2

 Differentiating gives:

2x \dfrac{dx}{dt}+2y \dfrac{dy}{dt}=2r \dfrac{dr}{dt} \rightarrow x \dfrac{dx}{dt}+y \dfrac{dy}{dt}=r \dfrac{dr}{dt}

 To solve for dr/dt, we need to know the value of x, y and r. We can calculate x and y because the cars have been traveling for 60 seconds: x = 4800 ft and y = 3600 ft. To find r, we need to plug these values back into our original equation:

x^2+y^2=r^2 \rightarrow (4800)^2+(3600)^2=r^2 \rightarrow r=6000

Then plug everything in to solve:

x \dfrac{dx}{dt}+y \dfrac{dy}{dt}=r \dfrac{dr}{dt} \rightarrow (4800)(80)+(3600)(60)=(6000) \dfrac{dr}{dt} \rightarrow \dfrac{dr}{dt}=100 \, ft/s

(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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