Solving Related Rates Problems

The following problems involve the concept of Related Rates. In short, Related Rates problems combine word problems together with Implicit Differentiation, an application of the Chain Rule. Recall that if $ y=f(x) $, then $ D \{y \} = \displaystyle{ dy \over dx } = f'(x)=y' $. For example, implicitly differentiating the equation $$ y^3+y^2= y+1$$ would be $$ D\{y^3+y^2\} = D\{ y+1 \} \ \ \ \ \longrightarrow $$ $$ 3y^2 \cdot y'+ 2y \cdot y' = y' + 0 $$ If $ x=f(t) $ and $ y=g(t) $, then $ D\{x\} = \displaystyle{ dx \over dt } = f'(t) $ and $ D\{y\} = \displaystyle{ dy \over dt } = g'(t) $ . For example, implicitly differentiating the equation $$ x^3+y^2= x+y+3$$ would be $$ D\{x^3+y^2\} = D\{ x+y+3 \} \ \ \ \ \longrightarrow $$ $$ 3x^2 \cdot \displaystyle{ dx \over dt } + 2y \cdot \displaystyle{ dy \over dt } = \displaystyle{ dx \over dt } + \displaystyle{ dy \over dt } + 0 $$

In all of the following Related Rates Problems, it will be assumed that each variable function $y$ is a function of time $t$. For that reason, I will always use Leibniz notation and not the ambiguous prime notation for derivatives, i.e., i will use $$ \displaystyle{ dy \over dt } \ \ \ \ instead \ of \ \ \ \ y' $$ Here is my strategy for approaching and solving Related Rates Problems:



EXAMPLE 1: Consider a right triangle which is changing shape in the following way. The horizontal leg is increasing at the rate of $ 5 \ in./min. $ and the vertical leg is decreasing at the rate of $ 6 \ in./min $. At what rate is the hypotenuse changing when the horizontal leg is $ 12 \ in. $ and the vertical leg is $ 9 \ in. $ ?

Draw a right triangle with legs labeled $x$ and $y$ and hypotenuse labeled $z$, and assume each edge is a function of time $t$.

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GIVEN: $ \ \ \ \displaystyle{ dx \over dt } = 5 \ in./min. \ $ and $ \ \displaystyle{ dy \over dt }= -6 \ in./min. $

FIND: $ \ \ \ \displaystyle{ dz \over dt } $ when $ x=12 \ in. $ and $ y=9 \ in $.

Use the Pythagorean Theorem to get the equation $$ x^2 + y^2 = z^2 $$

Now differentiate this equation with repect to time $t $ getting $$ D \{ x^2 + y^2\} = D \{z^2\} \ \ \ \longrightarrow $$ $$ 2x \displaystyle{ dx \over dt } + 2y \displaystyle{ dy \over dt } = 2z \displaystyle{ dz \over dt } \ \ \ \longrightarrow \ \ $$ (Multiply both sides of the equation by $1/2$.) $$ x \displaystyle{ dx \over dt } + y \displaystyle{ dy \over dt } = z \displaystyle{ dz \over dt } \ \ \ \ \ \ \ \ \ \ \ \ \ \ (DE) $$ Now let $ x=12 $ and $ y=9 $ and solve for $z$ using the Pythagorean Theorem.

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$$ 12^2+9^2= z^2 \ \ \ \longrightarrow \ \ \ z^2=225 \ \ \ \longrightarrow \ \ \ z=15 $$ Plug in all given rates and values to the equation $(DE)$ getting $$ (12)(5) + (9)(-6) = (15) \displaystyle{ dz \over dt } \ \ \ \longrightarrow $$ $$ 6 = 15 \displaystyle{ dz \over dt } \ \ \ \longrightarrow $$ $$ \displaystyle{ dz \over dt } = {6 \over 15} = { 2 \over 5} \ in/min. $$

In the list of Related Rates Problems which follows, most problems are average and a few are somewhat challenging.



Click HERE to return to the original list of various types of calculus problems.

Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu

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Duane Kouba ... October 24, 2019