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 dydx\frac{dy}{dx} given x3+2xy+y3=exyx^3+2xy+y^3=e^{xy}

Differentiate both sides

ddx[x3+2xy+y3]=ddx[exy]\frac{d}{dx}[x^3+2xy+y^3]=\frac{d}{dx}[e^{xy}]

3x2+2y+2xdydx+3y2dydx=exy(y+xdydx)3x^2+2y+2x\frac{dy}{dx}+3y^2\frac{dy}{dx}=e^{xy}(y+x\frac{dy}{dx})

Isolate for dydx\frac{dy}{dx}

dydx(2x+3y2xexy=3x22y+yexy\frac{dy}{dx}(2x+3y^2-xe^{xy}=-3x^2-2y+ye^{xy}

dydx=3x22y+yexy2x+3y2xexy\frac{dy}{dx}=\frac{-3x^2-2y+ye^{xy}}{2x+3y^2-xe^{xy}}

Observe - To understand dydx\frac{dy}{dx} we only need a relationship among the variables xx and yy. We do not need an explicit function y=f(x)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=L= length of the string
  • x=x= horizontal distance to the kite

Relate the variables

L2=x2+102L^2=x^2+10^2

Differentiate both sides

ddt[L2]=ddt[x2+102]\frac{d}{dt}[L^2]=\frac{d}{dt}[x^2+10^2]

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

2LdLdt=2xdxdt2L\frac{dL}{dt}=2x\frac{dx}{dt}

Solve for the needed length LL

202+102=L220^2+10^2=L^2

L=500L=\sqrt{500}

Thus, dLdt=2xdxdt2L\frac{dL}{dt}=\frac{2\cdot x\cdot \frac{dx}{dt}}{2L}

=22022500=40500=\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 πm2/hr\pi m^2/hr. How fast is its perimeter increasing when it has area 10π/m210 \pi /m^2?

Draw a picture

Define the variables

  • A=A= area of pond (m2m^2)
  • R=R= radius of pond (mm)
  • p=p= perimeter of pond (mm)

Relate the rates

A=πR2,p=2πRA=\pi R^2, p=2 \pi R

Thus, A=π(p2π)2=14πp2A=\pi (\frac{p}{2 \pi})^2=\frac{1}{4 \pi}p^{2}

Differentiate both sides

Want: dpdt=?\frac{dp}{dt}=?

Have: A=10πm2,dAdt=πm2/hrA=10 \pi m^2, \frac{dA}{dt}=\pi m^2/hr

dAdt=ddt[14πp2]=24πp dpdt\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=10π=14πp2A=10 \pi=\frac{1}{4}\pi p^2

=40π2=p2=40\pi^2=p^2

p=40π2p=\sqrt{40\pi^2}

Thus, dpdt=dAdt/12πp=10π/12π40π2=10\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 dydx\frac{dy}{dx} if xey+yex=1xe^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

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

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

If dAdt\frac{dA}{dt} >0>0, then we get dBdt<0\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=3cmR= 3 cm and H=10cm H = 10cm and you pour your tea at 5πcm3/s5\pi cm^3/s, how fast does the level of the tea rise when the cup contains 15πcm315\pi cm^3 of tea?

Draw a picture

Introduce the variables

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

Relate the variables

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

dVdt=9πdhdt\frac{dV}{dt}=9\pi \frac{dh}{dt}

Thus, dhdt=dVdt9π=5π9π=59cm/s\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 dhdt\frac{dh}{dt} not depend on time?
  2. What shapes of cup have dhdt\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=x= distance to damp
  • L=L= length of shadow

Relate the variables

By similar triangles, L1.75=x+L3\frac{L}{1.75}=\frac{x+L}{3}

Differentiate both sides

ddt[L1.75]=ddt[x+L3]\frac{d}{dt}[\frac{L}{1.75}]=\frac{d}{dt}[\frac{x+L}{3}]

dLdt1.75=dxdt+dLdt3\frac{\frac{dL}{dt}}{1.75}=\frac{\frac{dx}{dt}+\frac{dL}{dt}}{3}

dLdt(11.7513)=dxdt3\frac{dL}{dt}(\frac{1}{1.75}-\frac{1}{3})=\frac{\frac{dx}{dt}}{3}

dLdt=dxdt3(11.7513)=2.1\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=45m4=45m and height H=6mH=6m. If the water is leaving at 50m3/min50m^3/min, how fast is the water level hh changing when h=5m?h=5m?

Draw a picture

Introduce variables

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

Relate the variables

v=13πr2hv=\frac{1}{3} \pi r^2h

By similar triangles : rh=456r=456h\frac{r}{h}=\frac{45}{6}\Leftrightarrow r=\frac{45}{6}h

Thus, v=13π(456h)2h=13(456)2πh3v=\frac{1}{3}\pi(\frac{45}{6}h)^2h=\frac{1}{3}(\frac{45}{6})^2\pi h ^3

Differentiate both sides

  • Know: dvdt=50m3/min\frac{dv}{dt}=50m^3/min
  • Want: dhdt=?\frac{dh}{dt}=?

ddt[v]=ddt[13(456)2πh3]\frac{d}{dt}[v]=\frac{d}{dt}[\frac{1}{3}(\frac{45}{6})^2\pi h^3]

dvdt=13(456)2π3h2dhdt\frac{dv}{dt}= \frac{1}{3}(\frac{45}{6})^2\pi \cdot 3h^2\cdot \frac{dh}{dt}

=(456)2πh2dhdt=(\frac{45}{6})^2\pi \cdot h^2\cdot \frac{dh}{dt}

Thus, dhdt=dvdt(456)2πh2=50(456)2π52=8225π\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}


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