1. ## related rates

oil from a leaking oil tanker radiates outward in the form of a circular film on the surface of the water. If the radius of the circle increases at the rate of 3 meters per min, how fast is the area of the circle increasing when the radius is 200 meters ?

π x r^2= area of circle

therefore
π x200^2=125663.71

125663.71x3 = 376991.12

according to the book the answear is wrong as they have 1200
π m^2/min

they did not show any working !

also

differencite this

F(x) = √ (x-1)/(x+1)

an simplify!

cheers !

2. ## Re: related rates

Originally Posted by arsenal12345
oil from a leaking oil tanker radiates outward in the form of a circular film on the surface of the water. If the radius of the circle increases at the rate of 3 meters per min, how fast is the area of the circle increasing when the radius is 200 meters ?

π x r^2= area of circle

therefore π x200^2=125663.71

125663.71x3 = 376991.12 ... this is incorrect

according to the book the answear is wrong as they have 1200π m^2/min
$\displaystyle A = \pi r^2$

take the derivative of the above equation w/r to time ...

$\displaystyle \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$

sub in your givens ... you'll get the answer the text provides.

3. ## Re: related rates

A=area
t=time

A(t)=Pi*r(t)^2
dA/dt=2*Pi*r*dr/dt

at r=200, dr/dt=3 we obtain rate of increase of area as:
$\displaystyle 2\pi (200)3$ = $\displaystyle 1200 \pi$ sq m/min

4. ## Re: related rates

thanks buddy , any luck with the second question?

5. ## Re: related rates

$\displaystyle dF/dx=\frac{1}{\sqrt{x-1} (x+1)^{3/2}}$

6. ## Re: related rates

Originally Posted by arsenal12345
thanks buddy , any luck with the second question?
depends ... which is the function you posed?

$\displaystyle \frac{\sqrt{x-1}}{x+1}$

or

$\displaystyle \sqrt{\frac{x-1}{x+1}}$