*Relating area derivatives to perimeters, volume derivatives to surface areas.*
We begin by investigating how the area of a square (or circle) varies as its side length (or radius), \( x \), changes.

The area of a square with variable side length \( x \gt 0 \) is \( A_s(x) = x^2 \); for a slightly different side length \( x + \Delta x \), the area is \( A_s(x + \Delta x) = (x + \Delta x)^2 \).

Similarly, the area of a circle with variable radius \( x \gt 0 \) is \( A_c(x) = \pi x^2 \); for a slightly different side length \( x + \Delta x \), the area is \( A_c(x + \Delta x) = \pi(x + \Delta x)^2 \).

Note that, for the square, \( \dfrac{dA_s}{dx} = 2x \), which is half the perimeter of the original square, while for the circle, \( \dfrac{dA_c}{dx} = 2\pi x \), which is identically equal to the perimeter of the original circle.

The geometry behind this is revealed in the rightmost shape in each pair of shapes shown.

In the case of the square, we see that increasing the side length of the square by \( \Delta x \) adds area \( \Delta A_s \), equal to the area of the grey strips shown on two sides of the square, plus the area of the small black square in the corner, i.e. \( \Delta A_s = 2x \Delta x + (\Delta x)^2 \). The area of the small black square is \( \Delta x ^2\), and we see why only the \( 2x \) term survives when we first find the derivative:
\[ \dfrac{dA_s}{dx} = \lim_{\Delta x \to 0} ~ \dfrac{\Delta A_s}{\Delta x} = \lim_{\Delta x \to 0} ~ \dfrac{2x \Delta x + (\Delta x)^2}{\Delta x} = \lim_{\Delta x
\to 0} ~ (2x + \Delta x) = 2x \]

The circle's variation is similar, but less obvious geometrically. In this case, \( \Delta A_c \) equals the area of the grey band of width \( \Delta x \) around the circle, which is \[ \Delta A_c = \pi (x + \Delta x)^2 - \pi x^2 = 2\pi x \Delta x + \pi (\Delta x)^2 \]

The first term is the area of a strip of length \( 2\pi x \) (the inner circumference of the grey band), and width \( \Delta x \). The second term represents the additional area due to the longer outer circumference \( 2\pi ( x + \Delta x) \) of the grey band, which approaches \( 2\pi x \) as \( \Delta x \to 0 \).

Thus, the derivative is \( \displaystyle \dfrac{dA}{dx} = \lim_{\Delta x \to 0} ~ \dfrac{dA}{dx} = \lim_{\Delta x \to 0} ~ \dfrac{2 \pi x \Delta x + \pi (\Delta x)^2}{\Delta x} = 2\pi x \).

- Discover how this idea generalizes by relating volume derivatives to surface areas.
- Show that for a cube with variable side length \( x \gt 0 \), the derivative of the volume, \( V \), is identically equal to half the surface area, \( S \). Make a sketch similar to that for the square to illustrate why this result holds.
- Show that for a sphere of variable radius \( x \gt 0 \), the derivative of the volume
,\( V \), is identically to the surface area. Explain the geometry briefly.

- Show that, for a pyramid with square base of variable side length \( x \gt 0 \)) and height \( kx \), there is only one value of the constant \( k \) for which the derivative of the volume \( V \) is identically equal to half the surface area \( S \).
- For a cylinder of variable radius \( x \gt 0 \) and height \( kx \), find a value of \( k \) which makes the derivative of the volume \( V \) equal to half the surface area \( S \). Then, find a second value of \( k \) such that \( \dfrac{dV}{dx} = S \).
- Show that, for a ‘pill capsule’ consisting of hemispherical caps of variable radius \( x \gt 0 \) on a cylinder of length \( kx \), there is no value of \( k \gt 0 \) for which the volume derivative \( \dfrac{dV}{dx} \) is identically equal to half the surface area, \( S \).
- For each of the solids in parts b), c), and d), determine all possible values of the constant \( a \) for which there is a positive constant \( k \) such that \( \dfrac{dV}{dx} = aS \).