Sand falls from a conveyor belt at the rate of 10 [itex] \frac{ft^{3}}{min} [/itex] onto a conical pile. The radius of the base of the pile is always equal to half the pile's height. How fast is the height growing when the pile is 5 ft high?(adsbygoogle = window.adsbygoogle || []).push({});

So [tex] r = \frac{1}{2} h [/tex]. That means when [tex] h = 5 [/tex] , [tex] r = 2.5 [/tex]. We want to find [tex] \frac{dh}{dt} [/tex]. I know the volume of a cone is: [tex] \frac{1}{3}\pi r^{2}h [/tex].

[tex] \frac{dV}{dt} = \frac{1}{3}\pi r^{2} h \frac{dh}{dt} [/tex]. So would I just do:

[tex] 10 = \frac{1}{3}\pi (2.5)^{2}(5) \frac{dh}{dt} [/tex] and solve for [tex] \frac{dh}{dt} [/tex]?

Thanks

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Related Rates

**Physics Forums | Science Articles, Homework Help, Discussion**