# Chapter 21: Related Rates

We now introduce the primary method by which calculus enters in to applications.

Recall, the technique of implicit differentiation.

To find the derivative ﻿$\frac{dy}{dx}$﻿ given ﻿$x^3+2xy+y^3=e^{xy}$﻿

Differentiate both sides

﻿$\frac{d}{dx}[x^3+2xy+y^3]=\frac{d}{dx}[e^{xy}]$﻿

﻿$3x^2+2y+2x\frac{dy}{dx}+3y^2\frac{dy}{dx}=e^{xy}(y+x\frac{dy}{dx})$﻿

Isolate for ﻿$\frac{dy}{dx}$﻿

﻿$\frac{dy}{dx}(2x+3y^2-xe^{xy}=-3x^2-2y+ye^{xy}$﻿

﻿$\frac{dy}{dx}=\frac{-3x^2-2y+ye^{xy}}{2x+3y^2-xe^{xy}}$﻿

Observe - To understand ﻿$\frac{dy}{dx}$﻿ we only need a relationship among the variables ﻿$x$﻿ and ﻿$y$﻿. We do not need an explicit function ﻿$y=f(x)$﻿.

#### Ex.

A kite is 10m off the ground and 20m to the right. If the wind is blowing it at 2m/s to the right then how fast is the string unwinding?

Draw a picture Define the variables

• ﻿$L=$﻿ length of the string
• ﻿$x=$﻿ horizontal distance to the kite

Relate the variables

﻿$L^2=x^2+10^2$﻿

Differentiate both sides

﻿$\frac{d}{dt}[L^2]=\frac{d}{dt}[x^2+10^2]$﻿

• Want: ﻿$\frac{dL}{dt}=?$﻿
• Have: ﻿$\frac{dx}{dt}=2, \ x=20$﻿

﻿$2L\frac{dL}{dt}=2x\frac{dx}{dt}$﻿

Solve for the needed length ﻿$L$﻿

﻿$20^2+10^2=L^2$﻿

﻿$L=\sqrt{500}$﻿

Thus, ﻿$\frac{dL}{dt}=\frac{2\cdot x\cdot \frac{dx}{dt}}{2L}$﻿

﻿$=\frac{2\cdot 20\cdot 2}{2\cdot \sqrt{500}}=\frac{40}{\sqrt{500}}$﻿

Idea - How does the rate of change of one quantity effect the rate of change of another?

#### ex.

Suppose the surface area of a puddle is increasing at a rate of ﻿$\pi m^2/hr$﻿. How fast is its perimeter increasing when it has area ﻿$10 \pi /m^2$﻿?

Draw a picture Define the variables

• ﻿$A=$﻿ area of pond (﻿$m^2$﻿)
• ﻿$R=$﻿ radius of pond (﻿$m$﻿)
• ﻿$p=$﻿ perimeter of pond (﻿$m$﻿)

Relate the rates

﻿$A=\pi R^2, p=2 \pi R$﻿

Thus, ﻿$A=\pi (\frac{p}{2 \pi})^2=\frac{1}{4 \pi}p^{2}$﻿

Differentiate both sides

Want: ﻿$\frac{dp}{dt}=?$﻿

Have: ﻿$A=10 \pi m^2, \frac{dA}{dt}=\pi m^2/hr$﻿

﻿$\frac{dA}{dt}=\frac{d}{dt}[\frac{1}{4 \pi}p^2]=\frac{2}{4 \pi}p \ \frac{dp}{dt}$﻿

Solve for the needed perimeter

﻿$A=10 \pi=\frac{1}{4}\pi p^2$﻿

﻿$=40\pi^2=p^2$﻿

﻿$p=\sqrt{40\pi^2}$﻿

Thus, ﻿$\frac{dp}{dt}=\frac{dA}{dt}/\frac{1}{2\pi}p=10\pi/\frac{1}{2\pi}\sqrt{40\pi^2}=\sqrt{10}$﻿

#### ex.

Find ﻿$\frac{dy}{dx}$﻿ if ﻿$xe^y+ye^x=1$﻿.

Differentiate both sides #### Ex.

Suppose two products apples (A) and bananas (B) satisfy A+B =1000. How will an increase in apples sales effect bananas?

Differentiate both sides with respect to t

﻿$\frac{d}{dt}[A+B]=\frac{d}{dt}$﻿

﻿$\frac{dA}{dt}+\frac{dB}{dt}=0\rightarrow\frac{dA}{dt}=-\frac{dB}{dt}$﻿

If ﻿$\frac{dA}{dt}$﻿ ﻿$>0$﻿, then we get ﻿$\frac{dB}{dt}<0$﻿. Banana sales will decrease.

#### Ex.

Suppose you are pouring yourself a nice cup of tea. If your cup is a cylinder with dimensions ﻿$R= 3 cm$﻿ and ﻿$H = 10cm$﻿ and you pour your tea at ﻿$5\pi cm^3/s$﻿, how fast does the level of the tea rise when the cup contains ﻿$15\pi cm^3$﻿ of tea?

Draw a picture Introduce the variables

• ﻿$h=$﻿ height of tea level
• ﻿$V=$﻿ volume of tea

Relate the variables

﻿$\frac{d}{dt}[V]=\frac{d}{dt}[9\pi h]$﻿

﻿$\frac{dV}{dt}=9\pi \frac{dh}{dt}$﻿

Thus, ﻿$\frac{dh}{dt}=\frac{\frac{dV}{dt}}{9 \pi}=\frac{5\pi}{9\pi}=\frac{5}{9}cm/s$﻿ at all time.

#### Questions

1. Why does ﻿$\frac{dh}{dt}$﻿ not depend on time?
2. What shapes of cup have ﻿$\frac{dh}{dt}$﻿ non-constant?

#### Ex.

Suppose it is night and you are walking away from a street lamp at a speed of 1.5m/s. If the lamp is 3. tall and you are 1.75 m tall, how fast is the length of your shadow growing when you are 2m away from the lamp?

Draw a picture Introduce variables

• ﻿$x=$﻿ distance to damp
• ﻿$L=$﻿ length of shadow

Relate the variables

By similar triangles, ﻿$\frac{L}{1.75}=\frac{x+L}{3}$﻿

Differentiate both sides

﻿$\frac{d}{dt}[\frac{L}{1.75}]=\frac{d}{dt}[\frac{x+L}{3}]$﻿

﻿$\frac{\frac{dL}{dt}}{1.75}=\frac{\frac{dx}{dt}+\frac{dL}{dt}}{3}$﻿

﻿$\frac{dL}{dt}(\frac{1}{1.75}-\frac{1}{3})=\frac{\frac{dx}{dt}}{3}$﻿

﻿$\frac{dL}{dt}=\frac{\frac{dx}{dt}}{3(\frac{1}{1.75}-\frac{1}{3})}=2.1$﻿

#### ex.

Water is being drained from a conical reservoir with radius ﻿$4=45m$﻿ and height ﻿$H=6m$﻿. If the water is leaving at ﻿$50m^3/min$﻿, how fast is the water level ﻿$h$﻿ changing when ﻿$h=5m?$﻿

Draw a picture Introduce variables

• ﻿$v=$﻿ volume of water
• ﻿$h=$﻿ height of water
• ﻿$r=$﻿ radius of water

Relate the variables

﻿$v=\frac{1}{3} \pi r^2h$﻿

By similar triangles : ﻿$\frac{r}{h}=\frac{45}{6}\Leftrightarrow r=\frac{45}{6}h$﻿

Thus, ﻿$v=\frac{1}{3}\pi(\frac{45}{6}h)^2h=\frac{1}{3}(\frac{45}{6})^2\pi h ^3$﻿

Differentiate both sides

• Know: ﻿$\frac{dv}{dt}=50m^3/min$﻿
• Want: ﻿$\frac{dh}{dt}=?$﻿

﻿$\frac{d}{dt}[v]=\frac{d}{dt}[\frac{1}{3}(\frac{45}{6})^2\pi h^3]$﻿

﻿$\frac{dv}{dt}= \frac{1}{3}(\frac{45}{6})^2\pi \cdot 3h^2\cdot \frac{dh}{dt}$﻿

﻿$=(\frac{45}{6})^2\pi \cdot h^2\cdot \frac{dh}{dt}$﻿

Thus, ﻿$\frac{dh}{dt}=\frac{\frac{dv}{dt}}{(\frac{45}{6})^2\cdot \pi \cdot h^2}=\frac{50}{(\frac{45}{6})^2\cdot \pi \cdot 5^2}=\frac{8}{225\pi}$﻿