Rates Of Change

Rates Of Change

Intro 1

This is an application of differentiation that involves real life examples. It involves the chain rule with 2 derivatives.

Eg₁ pumping air into a ball. As the radius increases, the volume will increase and even the surface area of the volume will increase.

Eg₂ A drop of ink on a tissue paper, the radius of the circle will increase as well as the area of the circle.

Intro 2

The quantity  $\frac{dy}{dx}$  can be interrupted as the rate of change of y with respect to x.

However, a rate of change is normally considered as being given with respect to time (t).

Eg1 Rate of change of y w.r.t. time (t)  $=\frac{dy}{dt}$  .
Eg 2 Rate of change of V w.r.t time (t)  $=\frac{dV}{dt}$  .
Eg3 Rate of change of A w.r.t time(t)  $=\frac{dA}{dt}$  .

N.B Rate of change can be negative or positive. 
Negative : Rate of change is decreasing. 
Positive: Rate of change is increasing.
Do not forget the units of measure per time.

Derivatives

Volume increases with time  $=>\frac{dV}{dt}$  

Area increases with time  $=>\frac{dA}{dt}$  
Area increases as radius  $=>\frac{dA}{dr}$  
Area of Circle  $=\pi r^2\rightarrow\ \frac{dA}{dr}=2\pi r$  
Volume of Sphere   $=\frac{4}{3}\pi r^3\ \rightarrow\ \frac{dV}{dr}=4\pi r^2$  

To find:

 $\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}$  

 $\frac{dr}{dt}=\frac{dA}{dt}\times\frac{dr}{dA}=\frac{dA}{dt}\times\frac{1}{\frac{dA}{dr}}$  

 $\frac{dV}{dt}=\frac{d\ V}{dr}\times\frac{dr}{dt}$  

 $\frac{dV}{dr}=\frac{dV}{dr}\times\frac{dr}{dt}$  

Examples

1.The radius of a circle increases at a rate of 3cm/s. Find the rate of increase of the Area when r=5cm.
(in this example one derivative was given in the question and one need to be found)

 $\frac{dr}{dt}=3cm\$ /s

 $A=\pi r^2\rightarrow\ \frac{dA}{dr}=2\pi r$  

(The rate of increase of the Area is represented by  $\frac{dA}{dt}$  )
 $\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}$  
 $=2\pi r\left(3cm\ per\ s\right)$  
 $=6\pi r\ cm$  /s
when r = 5
 $=6\pi\left(5\right)cm^2\$  /s
 $=30\pi\ cm^2$  /s

2.The radius of a sphere increases at a rate of 2cm/s. Find the rate of increase of its volume when r=3cm.

 $\frac{dr}{dt}=2cm$  /s      $V=\frac{4}{3}\pi r^3\rightarrow\ \frac{dV}{dr}=4\pi r^2$  

 $\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}$  
 $=4\pi r^2\left(2m\ pers\right)$  
 $=8\pi r^2cm^3$  /s
when r=3cm
 $=8\pi\left(3\right)^2cm^3$  /s
 $=8\pi\left(9\right)cm^3$  /s
 $=72\pi cm^3$  /s

3. The area of a circle increases at a rate of  $2\pi cm\$  /s. Find the rate of increase of the radius when the radius is 6cm.

 $A=\pi r^2\rightarrow\frac{dA}{dr}=2\pi r$    $\frac{dA}{dt}=2\pi\ cm$  /s
 $\frac{dr}{dt}=\frac{dA}{dt}\times\frac{dr}{dA}$  
 $=\frac{dA}{dt}\times\frac{1}{\frac{dA}{dr}}$  
 $=2\pi cm\times\frac{1}{2\pi r}$  
 $=\frac{1}{r}cm$  /s
when r = 6cm 
 $=\frac{1}{6}cm$  /s