Thread: 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 1200π m^2/min

they did not show any working !

also

differencite this

F(x) = √ (x-1)/(x+1)

an simplify!

cheers !

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

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



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

A=area
r=radius
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:
= sq m/min

thanks buddy , any luck with the second question?

Originally Posted by arsenal12345

thanks buddy , any luck with the second question?

depends ... which is the function you posed?




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