# Thread: How do I solve the rate of volume change of a cone?

1. ## How do I solve the rate of volume change of a cone?

The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 5 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 30 inches?

? cubic inches per second

If someone could help solve this so I can use it to study for my exam that'd be spectacular!

2. Originally Posted by pseizure2000
The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 5 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 30 inches?

? cubic inches per second

If someone could help solve this so I can use it to study for my exam that'd be spectacular!
first, we must know the formula for the volume of a right-circular cone.

$\displaystyle V = \frac 13 \pi r^2 h$

now we differentiate implicitly with respect to $\displaystyle t$

$\displaystyle \Rightarrow \frac {dV}{dt} = \frac 23 \pi rh~\frac {dr}{dt} + \frac 13 \pi r^2~\frac {dh}{dt}$

now, we were given all the variables besides $\displaystyle \frac {dV}{dt}$, which is what we must find. so just plug them in. (Note, when something is decreasing, we treat its rate as negative)

3. Hello, pseizure2000!

The radius of a right circular cone is increasing at a rate of 3 inches per second
and its height is decreasing at a rate of 5 inches per second.
At what rate is the volume of the cone changing
when the radius is 40 inches and the height is 30 inches?

The volume of a cone is: .$\displaystyle V \;=\;\frac{\pi}{3}r^2h$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \;=\;\frac{\pi}{3}r^2\left(\frac{dh}{dt}\right) + \frac{\pi}{3}2rh\left(\frac{dr}{dt}\right)$

We are given: .$\displaystyle \frac{dr}{dt} = 3,\;\frac{dh}{dt} = -5,\;r = 40,\;h = 30$

Plug them in: .$\displaystyle \frac{dV}{dt}\;=\;\frac{\pi}{3}(40^2)(-5) + \frac{2\pi}{3}(40)(30)(3) \;=\;-\frac{8000\pi}{3} + \frac{7200\pi}{3}$

Therefore: .$\displaystyle \frac{dV}{dt} \;=\;-\frac{800\pi}{3}$ in³/sec

4. thank you! very well done and easy to understand!

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### rate of change of volume of a cone

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