Water leaking onto a floor forms a circular pool. The area of the pool increases at a rate of 25π cm²/min. How fast is the radius of the pool increasing when the radius is 6 cm?

Question

Christopher Allers · Accepted Answer

#dr/dt = (25)/(12)#cm/min

so from the question, we know that #dA/dt=25pi#, this means that the area of the circular puddle is increasing constantly at this rate. so in order to find how fast the radius is increasing, we must first determine a relationship between the two values. So, for a circle that is the Area formula #Area = pi*r^2#. The next part involves related rates and the chain rule. We know that #dr/dt =dA/dt * dr/dA# (the rate of radius change expressed as two other rates). So, #A=pi*r^2 # #dA/dr = 2pi*r# #dr/dA = 1/(2pi*r)# using chain rule now. #dr/dt = dA/dt *dr/dA# #dr/dt =25pi * 1/(2pi*r)# Sub in 6cm for radius #dr/dt = (25pi)/(2*6pi)# #dr/dt = (25)/(12)#cm/min

Charles Anctil · Answer

undefined To find the rate at which the radius is increasing, use the formula for the rate of change of area with respect to time: $ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $. Plug in the given values: $25\pi = 2\pi (6) \frac{dr}{dt} $. Solve for $ \frac{dr}{dt} $: $ \frac{dr}{dt} = \frac{25\pi}{12} $ cm/min.