57
$\begingroup$

The derivative of the volume of a sphere, with respect to its radius, gives its surface area. I understand that this is because given an infinitesimal increase in radius, the change in volume will only occur at the surface. Similarly, the derivative of a circle's area with respect to its radius gives its circumference for similar reasons.

A similar logic can be applied to other simple geometric shapes and the pattern holds. For a cylinder, it becomes more interesting. Its volume is $\pi r^2h$, with the height $h$ and radius $r$ being independent.

Incremental increases in a cylinders height will "add a circle" to the bottom of the cylinder, so it makes sense to me that the derivative of volume with respect to height is the are of that circle $\pi r^2$. Incremental changes to its radius, will add a "tube" around the cylinder, whose area is $2\pi rh$, the derivative of volume with respect to radius.

I thought I had found a nice heuristic reason for why the derivatives of volumes give surface areas, but it breaks down for the cone. The volume is $\frac{\pi r^2 h}{3}$. Assuming the angle $\theta$ of the cone remains constant, $h$ and $r$ are dependent ($r=h\tan\theta$).

Now, the derivative of the cone's volume with respect to height gives the area of the base of the cone, this intuitively makes sense, but the derivative with respect to radius doesn't seem to match any geometric quantity.

$V=\dfrac{\pi r^3}{3\tan\theta}~$ and $~\dfrac{\text{d}V}{\text{d}r}=\dfrac{\pi r^2}{\tan\theta}$

Similarly, I'd hoped to find some way to drive the surface area of the cone without the base $\pi rl$ where $l$ is the distance from the point of the cone to the edge of the circle at its base. However, the derivative with respect to $l$ also doesn't seem to have any meaningful geometric significance.

What's going on here? Is it an issue of not defining the shape rigorously enough? Do I need to parametrise the volume rigorously in some way?

$\endgroup$
5
  • 3
    $\begingroup$ Ultimately what is true is that the derivative of the volume is the surface area if the shapes are parametrized by moving the boundary at unit speed in the normal direction where the normal direction makes sense. Thinking instead about area and perimeter, for a square you need to change the side length at rate $2$, and then the rate of change of the area is $4s$ which is the perimeter. For an equilateral triangle you need to change the side length at rate $\sqrt{3}$ and then the rate of change of the area is $3s$ which is the perimeter. $\endgroup$
    – Ian
    Commented Mar 13, 2023 at 18:58
  • 1
    $\begingroup$ This makes sense if you think about the volume of the new region, which will have cross-sectional area above each surface given for short time by the area of the original surface, and then a thickness that changes at unit speed. $\endgroup$
    – Ian
    Commented Mar 13, 2023 at 19:00
  • $\begingroup$ Sorry, I guess for the equilateral triangle the correct rate of change of the side length is $2\sqrt{3}$. Still you get the point, you need to move the sides at unit speed in their normal direction, move the vertices with velocity given by the sum of the velocities of the adjacent sides, and then the side lengths change at a rate based on that.) $\endgroup$
    – Ian
    Commented Mar 14, 2023 at 13:46
  • $\begingroup$ @Ian "the derivative of the volume is the surface are if [...]" I guess this has something to do with differential geometry. Where can I read more on this? $\endgroup$ Commented Mar 19, 2023 at 12:45
  • 1
    $\begingroup$ @blundered_bishop About this in particular I think you want the co-area formula. $\endgroup$
    – Ian
    Commented Mar 19, 2023 at 14:43

2 Answers 2

51
$\begingroup$

If you continuously enlarge a solid cone by adding material to the base, then every inch of added height corresponds to an inch-thick layer added to the circular base, so the rate of change of volume is equal to the area of the base times the rate of change of the height. That is, $dV/dh=A_\text{base}=\pi r^2$.

However, if you enlarge the cone by adding material to the "cap", then adding an inch of height (or radius) does not correspond to exactly an inch-thick layer added to the cap. The added thickness is actually $\Delta t=\Delta r \cos\theta=\Delta h \sin\theta$, which you can see by drawing a couple of triangles (with a segment of length $t$ perpendicular to the surface of the cone and with an endpoint at the center of the base). The rate of change of volume is the area of the cap times the rate at which we add thickness to the cap: $dV=A_\text{cap}dt$, so we have $A_\text{cap}=\frac{dV}{dt}=\frac1{\cos\theta} \frac{dV}{dr}=\frac{\pi rh}{\cos\theta}=\pi rl$, where $l$ is the slant height.

$\endgroup$
3
  • 1
    $\begingroup$ I see what you mean, and thank you for your explanation. You have also nicely explained why despite that $dV/dh$ can represent the increment to both the either the cap or the base of the cone, it makes sense that they describe the same quatinty. $\endgroup$
    – Numeral
    Commented Mar 12, 2023 at 13:49
  • $\begingroup$ I'm not sure what the image you describe should look like, can you add it? $\endgroup$
    – ljaniec
    Commented Mar 16, 2023 at 22:19
  • $\begingroup$ "every inch" or ... "every centimeter" ... :) $\endgroup$
    – Jean Marie
    Commented Mar 17, 2023 at 9:25
3
$\begingroup$

My intuition tells me that you should be able to find such a formula by choosing the right parameterization. A couple of motivating examples:

First example: a disc

Consider a disk of radius r. The area in terms of the radius is $\pi r^2$ and in terms of the diameter $\pi (\frac{d}{2})^2$. If we take the derivatives of these expressions with respect to each parameter we find:

  • $2 \pi r$ for the radius, which is the circumference, But
  • $\frac{1}{2} \pi d$ for the diameter, which is not the circumference.

The parameterization affects how "fast" the area grows, and varying the diameter makes the area of the disk "grow too slowly" for the change in area to be the circle that surrounds it.

Second example: an equilateral triangle

Consider an equilateral triangle of side $s$. The area is $A(s)=\frac{\sqrt{3}}{4}s^2$ and the perimeter is $P(s)=3s$. The derivative of the area with respect to $s$ is $\frac{dA}{ds} = \frac{\sqrt{3}}{2}s$, which is not the perimeter.

So let's introduce a new parameter $q$: $s = kq$, for $k$ a constant. We have $A(q)=\frac{\sqrt{3}}{4}k^2q^2$ and $P(q)=3kq$. If we set $\frac{dA}{dq} = P(q)$ we find $k=\frac{\sqrt{3}}{2}$. This gives a parameter of $q = \frac{2}{\sqrt{3}}s$, which happens to be the diameter of the circumscribed circle of our triangle!

So if we parametrize the area of an equilateral triangle by the diameter of its circumscribed circle, then the rate of change of the area is the circumference.

I suspect a similar approach could give you a parametrization with the same property for the cone. As a first step, you could try doing the same for a cone whose base is equal to its side length and see what you get. Who knows, maybe the parameter you'll find will have an interpretation in terms of the radius of the circumscribed sphere.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .