Derivative of Area: Why does dA/dr relate to circ. or dV/dr relate to surface area?

apple2357

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So, i have come across the fact that when you find dA/dr for the area of circle you get the Circumference?
And the dV/dr ( V=Volume of a sphere) you get Surface Area.

1) I don't understand why dA/dr relates to circumference or dV/dr relates to surface area?
2) Why doesn't this work for a square,say when A= x^2 to give perimeter of 4x ?
 
So, i have come across the fact that when you find dA/dr for the area of circle you get the Circumference?
And the dV/dr ( V=Volume of a sphere) you get Surface Area.

1) I don't understand why dA/dr relates to circumference or dV/dr relates to surface area?
2) Why doesn't this work for a square,say when A= x^2 to give perimeter of 4x ?
Quick answer is that circles and squares have different properties.

That is why (dA/dr = C) circle gives you the maximum enclosed area for a given perimeter (circumference). A similar rule applies for spheres (dV/dr = S).
 
So, i have come across the fact that when you find dA/dr for the area of circle you get the Circumference?
And the dV/dr ( V=Volume of a sphere) you get Surface Area.

1) I don't understand why dA/dr relates to circumference or dV/dr relates to surface area?
2) Why doesn't this work for a square,say when A= x^2 to give perimeter of 4x ?

When you increase the radius by dr, you add a ring around the circle that increases the area by dA = C dr. (Unwrapped, it is approximately a rectangle with length C and thickness dr.)

For a sphere, increasing the radius by dr, you add a uniform "coat of paint" to the volume, increasing it by dV = S dr. (Volume is "base area" times thickness, if we ignore the curvature.)

But for a square, adding dx to each side adds a strip of that width to only two sides, not all four. The change in area is not P dr, but P/2 dr.
 
For the area of the circle, it is given as a function of the radius. The area of the square is given as a function of its side lengths. Radii and side lengths are inherently different things, so taking the derivative with respect to them should certainly show different behavior.

You can reformulate the area of a square, too, to have the same behavior. Writing it as a function of something other than its side length. Observe:

Take a square of side length x, so its area is \(\displaystyle A(x)=x^2\). Let's create a new variable \(\displaystyle s=x/2\) (half the side of the square... let's pretend it is a kind of "radius" for the square. It is, in fact, the radius of an inscribed circle). Now the area of the original square is given by \(\displaystyle A(s)=4s^2\), and its Perimiter is \(\displaystyle P(s)=8s\). Note that \(\displaystyle dA/ds = P(s)\).

For the circle, let \(\displaystyle x=\sqrt{\pi} r\), and let's call this a "side" of the circle. Then the area of the circle is \(\displaystyle A(x) = x^2\), and \(\displaystyle C(x) =2\sqrt{\pi}x\). Then \(\displaystyle dA/dx \neq C(x)\).

Hopefully this shows you that it depends on how the variables are constructed.
 
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Thanks for your thoughts. Been playing around and thinking about a square of length 2x ( rather than x) , so the area A = 4x^2.
Now if i find dA/dx i get 8x which is exactly the perimeter- which is exactly what the doan above says.
Thanks
 
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