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

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Discussion Overview

The discussion revolves around the rates of change of the volume of a cone with respect to its height and radius, specifically under conditions where either the radius or height is held constant. Participants explore the mathematical relationships involved in these changes.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants derive the formula for the rate of change of volume with respect to height when the radius is constant, stating that $$\frac{dh}{dt}=\frac{3}{\pi r}\frac{dV}{dt}$$.
  • Others suggest that time variables may not be necessary and propose differentiating directly with respect to height or radius.
  • One participant expresses confusion over the complexity of the answers, expecting simpler results for $$\frac{dV}{dh}$$ and $$\frac{dV}{dr}$$.
  • Some participants clarify that the expressions for $$\frac{dV}{dh}$$ and $$\frac{dV}{dr}$$ can be stated as $$\frac{dV}{dh}=\frac{\pi}{3}r^2$$ and $$\frac{dV}{dr}=\frac{2\pi}{3}rh$$, respectively.

Areas of Agreement / Disagreement

Participants express differing views on whether to include time variables in their calculations. There is no consensus on the best approach to take, as some prefer a direct differentiation method while others follow a time-dependent approach.

Contextual Notes

Participants note the need to clarify assumptions regarding the constancy of height or radius and the implications of differentiating with respect to time versus directly with respect to the variables.

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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