MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 14 August 2014

Lecture 21

Related Rates

IMPORTANT CONCEPT: The derivative is a rate of change. $$\begin{array}{ccc}{\displaystyle \frac{df}{dx}{|}_{x=a}}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x},}\\ {\displaystyle f(a+\mathrm{\Delta }x)-f\left(a\right)}& =& {\displaystyle \text{change in}\hspace{0.17em}f,}\\ {\displaystyle \mathrm{\Delta }x}& =& {\displaystyle \text{change in}\hspace{0.17em}x\text{.}}\end{array}$$ $$\begin{array}{c} a a + Δ x x f ( a ) f ( a + Δ x ) y \end{array}$$ $\frac{df}{dx}{|}_{x=a}$ measures how $f$ is changing as $x$ changes.

A TV camera is 4000 feet from the base of a launch pad. A rocket is launched and has a speed of 600 ft/s when it is 3000 ft high. How fast is the distance between the camera and the rocket changing? $$\begin{array}{c} H D ⏟ 4000 θ \end{array}$$ $$\begin{array}{ccc}{\displaystyle H}& =& {\displaystyle \text{height of rocket,}}\\ {\displaystyle D}& =& {\displaystyle \text{distance between camera and rocket,}}\\ {\displaystyle \text{speed}}& =& {\displaystyle \text{change in height as time changes}=\frac{dH}{dt}\text{.}}\end{array}$$ We know $$\frac{dH}{dt}{|}_{H=3000}=600\hspace{0.17em}\text{ft/s.}$$ We want $$\frac{dD}{dt}{|}_{H=3000}\text{.}$$ From the picture ${4000}^2+H^2=D^2\text{.}$ So $$2H\frac{dH}{dt}=2D\frac{dD}{dt}\text{.}$$ So $$\frac{dD}{dt}=\frac{2H}{2D}\frac{dH}{dt}=\frac{H}{D}\frac{dH}{dt}\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \frac{dD}{dx}{|}_{H=3000}}& =& {\displaystyle \frac{3000}{\sqrt{{3000}^2+{4000}^2}}\frac{dH}{dt}{|}_{H=3000}=\frac{3000}{\sqrt{{5000}^2}}·600}\\ & =& {\displaystyle \frac{3000·600}{5000}=3·\frac{600}{5}=3·120=360\hspace{0.17em}\text{ft/s.}}\end{array}$$ How fast is the angle of the camera changing? We want $\frac{d\theta }{dt}{|}_{H=3000}\text{.}$ Since $\text{tan}\hspace{0.17em}\theta =\frac{H}{4000},$ $$\frac{d\theta }{dt}{\text{sec}}^2\theta =\frac{1}{4000}\frac{dH}{dt}\text{.}$$ Since $\text{sec}\hspace{0.17em}\theta =\frac{1}{\text{cos}\hspace{0.17em}\theta }=\frac{D}{4000},$ $$\frac{d\theta }{dt}\frac{D^2}{{4000}^2}=\frac{1}{4000}\frac{dH}{dt}\text{.}$$ So $$\frac{d\theta }{dt}=\frac{{4000}^2}{D^2}\frac{1}{4000}\frac{dH}{dt}=\frac{4000}{D^2}\frac{dH}{dt}\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \frac{d\theta }{dt}{|}_{H=3000}}& =& {\displaystyle \frac{4000}{({3000}^2+{4000}^2)}·\frac{dH}{dt}{|}_{H=3000}=\frac{4000}{{5000}^2}·600}\\ & =& {\displaystyle \frac{4·600}{5·1000}=\frac{4·120}{1000}=\frac{24}{50}\hspace{0.17em}\text{radians per sec.}}\end{array}$$

A runner runs around a circular track of radius 100m at a speed of 7 m/s. The runners friend is standing 200m from the center. How fast is the distance between them changing when their distance is 200m? $$\begin{array}{c} F 100 100m R ⏟ \end{array}$$ $$\text{speed of runner}=\text{change in runner's distance w.r.t. time}=\frac{dR}{dt}$$ We want $$\text{change in distance between friends w.r.t. time}=\frac{dF}{dt}\text{.}$$ Really we want $\frac{dF}{dt}{|}_{F=200}\text{.}$ $R=100·\theta$ if $\theta$ is the angle at the point $(x,y)$ at which the runner is at. $$\begin{array}{ccc}{\displaystyle F=\sqrt{{(200-x)}^2+y^2}}& =& {\displaystyle \sqrt{{200}^2-400x+x^2+y^2}}\\ & =& {\displaystyle \sqrt{{200}^2-400x+{100}^2}\text{.}}\end{array}$$ So $$F^2={200}^2+{100}^2-400x\text{and}2F\frac{dF}{dt}=-400\frac{dx}{dt}\text{.}$$ So $$\frac{dF}{dt}=-\frac{200}{F}\frac{dx}{dt}\text{.}$$ Now $x=100\text{cos}\hspace{0.17em}\theta$ and $7=\frac{dR}{dt}=100\frac{d\theta }{dt}\text{.}$ So $$\frac{dx}{dt}=-100\text{sin}\hspace{0.17em}\theta \frac{d\theta }{dt}=-100\text{sin}\hspace{0.17em}\theta \frac{7}{100}=-7\text{sin}\hspace{0.17em}\theta \text{.}$$ So $$\frac{dF}{dt}=\frac{(-200)}{F}(-7\text{sin}\hspace{0.17em}\theta )=\frac{1400\text{sin}\hspace{0.17em}\theta }{F}\text{.}$$ So $$\frac{dF}{dt}{|}_{F=200}=\frac{1400\text{sin}\hspace{0.17em}\theta }{F}{|}_{F=200}=\frac{1400}{200}\text{sin}\hspace{0.17em}\theta {|}_{F=200}=7\text{sin}\hspace{0.17em}\theta {|}_{F=200}\text{.}$$ When $F=200$ $$\begin{array}{c} 200 100 100 100 θ \end{array}\begin{array}{c} 200 200 50 θ \end{array}$$ so $$\text{sin}\hspace{0.17em}\theta =\frac{\sqrt{{200}^2-{500}^2}}{200}\text{.}$$ So $$\frac{dF}{dt}{|}_{F=200}=\frac{7\sqrt{{200}^2-{500}^2}}{200}=\frac{7\sqrt{40000-2500}}{200}=\frac{7\sqrt{37500}}{200}=\frac{7}{2}\sqrt{375}\hspace{0.17em}\text{m/s}\text{.}$$

Notes and References

These are a typed copy of Lecture 21 from a series of handwritten lecture notes for the class MATH 221 given on October 27, 2000.

page history