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

$\frac{d}{dt}[A+B]=\frac{d}{dt}[1000]$

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

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