logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

The radius of a circle is increasing uniformly at the rate of $3\; cm/s.$ Find the rate at which the area of the circle is increasing when the radius is $10\; cm$.

$\begin{array}{1 1}(A)\;60\pi cm^2/s \\ (B)\;30\pi cm^2/s \\(C)\;20\pi cm^2/s \\(D)\;40\pi cm^2/s \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
  • $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1:
Given : $\large\frac{dr}{dt}$$=3cm/s$
Radius =$10cm.$
Area of the circle =$\pi r^2$
$A=\pi r^2$
Differentiating on both sides w.r.t $t$
$\large\frac{dA}{dt}$$=2\pi r\large\frac{dr}{dt}$
Step 2:
Substituting the values for $r$ and $\large\frac{dr}{dt}$ we get,
$\large\frac{dA}{dt}=$$2\times \pi\times 10\times 3$
$\quad\quad=60\pi cm^2/s$
Hence the rate at which the area of the circle is increasing is $60\pi cm^2/s$
answered Jul 5, 2013 by sreemathi.v
edited Jul 5, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...