MHB *Volume of a cone change of rate of volume with respect to h and r

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The discussion focuses on finding the rate of change of the volume of a cone with respect to its height and radius. For a constant radius, the rate of change of volume with respect to height is expressed as dh/dt = (3/πr)dV/dt, leading to dV/dh = (πr/3). Conversely, when height is constant, the rate of change of volume with respect to radius is dV/dr = (2π/3)rh. Participants clarify that differentiation should be done with respect to the stated variable without introducing time as a factor. The conversation emphasizes deriving these relationships through calculus principles applied to the volume formula of a cone.
karush
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(a) Find the rate of change of the volume with respect to the height if the radius is constant

vol of right circular cone is $$V=\frac{1}{3} \pi r^2 h$$

from this $$h=\frac{3V}{\pi r^2}$$

$$\frac{dh}{dt}=\frac{3}{\pi r}\frac{dV}{dt}$$

$$\frac{\pi r}{3}\frac{dh}{dt}=\frac{dV}{dt}$$

not sure about this we don't have t or rate of change of height:confused:

this next question is the same except height is constant

(b) Find the rate of change of the volume with respect to the radius if the height is constant.
 
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Re: Volume of a cone change of rate of volumn in respect to h and r

karush said:
(a) Find the rate of change of the volume with respect to the height if the radius is constant

vol of right circular cone is $$V=\frac{1}{3} \pi r^2 h$$

from this $$h=\frac{3V}{\pi r^2}$$

$$\frac{dh}{dt}=\frac{3}{\pi r}\frac{dV}{dt}$$

$$\frac{\pi r}{3}\frac{dh}{dt}=\frac{dV}{dt}$$

not sure about this we don't have t or rate of change of height:confused:

this next question is the same except height is constant

(b) Find the rate of change of the volume with respect to the radius if the height is constant.
They are looking for expressions for dh/dt and dr/dt in terms of variables. So solve your equation in a) for dh/dt.

-Dan
 
Re: Volume of a cone change of rate of volumn in respect to h and r

so this is the ans for (a)

$$\frac{dh}{dt}=\frac{3}{\pi r}\frac{dV}{dt}$$
 
Re: Volume of a cone change of rate of volumn in respect to h and r

The way I interpret these problems, there is no need to introduce a variable for time. You simply need to differentiate with respect to the stated variable.
 
Re: Volume of a cone change of rate of volumn in respect to h and r

so just took out dt...

$$dh=\frac{3}{\pi r}dV$$
 
Re: Volume of a cone change of rate of volumn in respect to h and r

karush said:
(a) Find the rate of change of the volume with respect to the height if the radius is constant

(b) Find the rate of change of the volume with respect to the radius if the height is constant.

a) You are being asked to find $$\frac{dV}{dh}$$.

b) You are being asked to find $$\frac{dV}{dr}$$.
 
Re: Volume of a cone change of rate of volumn in respect to h and r

$$ \displaystyle
dh=\frac{3}{\pi r}dV
\text { then }
\frac{dV}{dh}
=\frac{\pi\text{ r}}{3}
$$
$$\text{ and }$$
$$r=\sqrt{\frac{3V}{\pi h}}$$
$$\text { so }$$
$$dr=\frac{\sqrt{3}}{2h\sqrt{\frac{\pi v}{h}}}dV$$
$$\text{ and }$$
$$\frac{dV}{dr}=\frac{2h\sqrt{\frac{\pi v}{h}}}{\sqrt{3}}$$

I was expecting something more simple for answer?
 
Re: Volume of a cone change of rate of volumn in respect to h and r

What I meant to do is as follows:

Given:

$$V=\frac{\pi}{3}r^2h$$

then:

$$\frac{dV}{dh}=\frac{\pi}{3}r^2$$

$$\frac{dV}{dr}=\frac{2\pi}{3}rh$$
 
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