Question

A cube of edge length s is inscribed in a sphere of radius r. write the surface areasurface area upper ff of a cubecube as a function of the radius r. ​(b) find the instantaneous rate of change of the surface areasurface area upper ff with respect to radius r of the sphere sphere. ​(c) evaluate the rate of change of upper ff at r equals 2r=2 and r equals 8r=8. ​(d) if r is measured in millimeters millimeters and upper ff is measured innbsp square nbsp square millimeters​, what units

Answer 1

Given that the length of the cube's edges is $s$, then the length of the diagonal along any face of the cube is, by the Pythagorean theorem,

$\sqrt{s^2+s^2}=s\sqrt2$

Then the diagonal along the opposite vertices of the cube (i.e. the line segment through the center of the cube joining its opposite corners) is, again by the Pythagorean theorem,

$\sqrt{(s\sqrt2)^2+s^2}=s\sqrt3$

The radius of the sphere, $r$, corresponds to half the length of the cube's diagonal; that is,

$r=\dfrac{\sqrt3}2s$

The surface area of the cube in terms of its edge length $s$ is

$A=6s^2$

which means the area, rewritten in terms of the sphere's radius $r$, is

$A=6\left(\dfrac2{\sqrt3}r\right)^2=8r^2$

Hence the rate of change of the cube's surface area with respect to a change in the sphere's radius is

$\dfrac{\mathrm dA}{\mathrm dr}=16r$

When $r=2$, we have

$\dfrac{\mathrm dA}{\mathrm dr}\bigg|_{r=2}=16(2)=32$

while when $r=8$, we get

$\dfrac{\mathrm dA}{\mathrm dr}\bigg|_{r=8}=16(8)=128$

I'm not sure what part (d) reads, so I'll leave that to you...