Home > Calculus > Calculating Volume using Integrals

How fast is the volume changing with respect to time when the radius is changing with respect to time when the radius is changing at a rate of dr/dt=1.5 feet per second and r=2 feet?

Answer 1

#18.84955592153876\ \text{cubic ft. per sec}#

The volume of cone at instant of time #t# is #V=\frac{\pi}{3}r^3#

Now, differentiating volume V w.r.t. time #t#, the rate of change of volume #(\frac{dV}{dt})# of cone is #\frac{dV}{dt}=\frac{d}{dt}(\frac{\pi}{3}r^3)# #=\frac{\pi}{3}(3r^2)\frac{dr}{dt}# #=\pi r^2\frac{dr}{dt}# #=\pi (2)^2(1.5)\quad (\text{setting} \ r=2, \frac{dr}{dt}=1.5 )# #=18.84955592153876\ \text{cubic ft. per sec#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

To find how fast the volume is changing with respect to time when the radius is changing at a rate of ( \frac{dr}{dt} = 1.5 ) feet per second and ( r = 2 ) feet, you can use the formula for the volume of a sphere, ( V = \frac{4}{3} \pi r^3 ), and differentiate it with respect to time using the chain rule. Then substitute the given values for ( r ) and ( \frac{dr}{dt} ) into the expression to find the rate of change of volume with respect to time.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.