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

Click For Summary
SUMMARY

The discussion focuses on calculating the rate of change of the volume of a right circular cone with respect to height and radius. The volume formula is given as \( V = \frac{1}{3} \pi r^2 h \). For a constant radius, the rate of change of volume with respect to height is expressed as \( \frac{dh}{dt} = \frac{3}{\pi r} \frac{dV}{dt} \). Conversely, when height is constant, the rate of change of volume with respect to radius is \( \frac{dV}{dr} = \frac{2\pi}{3} rh \).

PREREQUISITES
  • Understanding of calculus, specifically differentiation.
  • Familiarity with the volume formula for a right circular cone.
  • Knowledge of the relationship between variables in multivariable calculus.
  • Basic algebra for manipulating equations and solving for variables.
NEXT STEPS
  • Study the application of partial derivatives in multivariable calculus.
  • Learn about the geometric interpretation of volume changes in cones.
  • Explore the concept of related rates in calculus.
  • Investigate the implications of constant variables in differential equations.
USEFUL FOR

Students of calculus, mathematics educators, and professionals in engineering or physics who require a solid understanding of volume change dynamics in geometric shapes.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
(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.
 
Physics news on Phys.org
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:

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

$$\frac{dV}{dr}=\frac{2\pi}{3}rh$$
 
this problem was posted on Linkedin
#1k
1k+
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K