### Video Transcript

In this video, weβre talking about
surface area to volume ratios. Weβll study these ratios for common
three-dimensional shapes, including cubes and spheres. And weβll also come to see how
these ratios change as shape size changes.

We can get started by reminding
ourselves of what the surface area and volume are for cubes and spheres. Given a cube of side length π, we
know that the area of one of the cubeβs faces is π times π, or π squared. And then, there are six faces over
all, four sides and one on top and one on bottom, which tells us that the total
surface area of a cube of side length π, and weβll call this π sub π, is equal to
six times π squared. And then, as we think about the
volume of this cube, weβll call that π sub π, we know that thatβs equal to the
area of one of the cubeβs faces, π times π, multiplied by the cubeβs depth, which
is also π. So, that the cubeβs volume is the
side length multiplied by itself three times, π cubed.

As a next step, we can compare the
cubeβs volume to its surface area by taking this ratio. This is equal to π cubed divided
by six π squared, which means that two factors of π cancel from numerator and
denominator, leaving us with a volume to surface area ratio for a cube of the side
length π divided by six. In other words, this ratio is
proportional to the side length. Note that this is for the ratio of
volume to service area, but we could equally consider the inverse, surface area to
volume, and still the relation between surface area and volume would remain the
same. We just need to keep an eye on
which quantity is in the numerator and which is in the denominator.

Knowing this about a cube, we can
then consider similar properties for a sphere. Given a sphere of radius π, the
surface area of that sphere, we can call it π sub π , is four times π times that
radius squared. And then its volume, what weβll
call π sub π , is equal to four-thirds times π times π cubed. And then, once again, we can solve
for this ratio, the volume of our shape to its surface area. When we calculate this fraction, we
find that not only do two factors of our dimension of interest, in this case the
sphereβs radius, cancel out, but so do the factors of four and π in numerator and
denominator. This leaves us with π divided by
three so that for a sphere, the volume to the surface area of the sphere is
proportional to its radius π.

So far, weβve been considering
shapes whose dimensions of interest, either the side length π or the radius π, are
fixed. But itβs interesting to see for a
shape whose size changes, how this change can affect the ratio of volume to surface
area. For example, say that we have three
spheres. The radius of the first is π
, that
of the second is twice that, and that of the third is three times π
. We can label these sphere one, two,
and three. And recall from earlier that the
ratio of each spheres volume to its surface area is equal to its radius divided by
three. This means that for sphere one,
that ratio is π
divided by three, while for sphere two, itβs two π
divided by
three, and for sphere three, itβs three π
divided by three or simply π
.

We see then that as the size of the
sphere changes, so does this ratio. Itβs increasing as π
gets larger
and larger. And it turns out that the same
trend occurs if we consider a cube instead of a sphere. But considering the shape of a
sphere, in particular these three we have here, we can see that as the radius
increases, sphere volume grows at a faster rate than surface area, while as the
radius decreases, sphere surface area actually changes more than volume.

So, as we consider these ratios of
volume to surface area, thereβs some subtlety involved in how this ratio changes as
the size of a given shape changes. Nonetheless, we can always say that
in the case of a cube, this ratio is proportional to the cubeβs side length, and in
the case of a sphere, itβs proportional to the radius. Though weβve limited our discussion
so far to cubes and spheres, the trends that weβre discovering apply in general to
three-dimensional shapes. Even if we had some very irregular
three-dimensional object, if the overall size of that object changes, say, by being
scaled up or scale down, then the shapeβs volume will grow at a faster rate than its
surface area as that size increases, while when it decreases to a relatively small
size, its surface area will change at a faster rate.

Knowing all this about surface area
and volume for three-dimensional objects, letβs get some practice now through an
example exercise.

The volume π of a sphere of radius
π is given by the formula π is equal to four-thirds times π times π cubed. The surface area π΄ of the same
sphere is given by π΄ equals four times π times π squared. Which of the following equations
correctly gives the ratio of the volume of the sphere to the surface area of the
sphere, π divided by π΄? (A) π divided by π΄ is equal to
one-third π cubed. (B) π divided by π΄ is equal to
one-third times π times π. (C) π divided by π΄ is equal to
one-fourth times π times π squared. (D) π divided by π΄ is equal to
one-third times π. (E) π divided by π΄ is equal to
one-third times π squared.

OKay, so in this exercise, weβre
considering a sphere that has a radius π. Weβre told the equations for the
volume as well as the surface area of the sphere in terms of its radius. And we want to solve for the ratio
of its volume to that surface area. We can do this by writing out the
expression for these two terms and then dividing the volume equation by the area
one. When we do this, weβre effectively
creating one equation out of two fractions. And notice that the fraction on the
left is exactly the ratio we want to solve for.

So then, looking on the right-hand
side of this equality, letβs see what will cancel out. Thereβs a factor of four in both
numerator and denominator, also a factor of π. And moreover, we can see that two
factors of the radius π cancel from top and bottom. When we remove all the cancelled
factors, weβre left with π, the radius of the sphere, divided by three. Looking across our five answer
options, we see that option (D) expresses the same ratio. The volume of a sphere to its
surface area is equal to one-third times its radius.

Letβs look now at a second example
exercise.

The graph shows how the volume π,
the surface area π΄, and the volume to surface area ratio, π divided by π΄, vary
with the radius of a sphere. Which line represents π? Which line represents π΄? Which line represents π divided by
π΄?

Looking at this graph, we see that
both the vertical and the horizontal axes are unlabeled, but that we have these
three curves, the red one, the blue one, and the purple one. Weβre told that one of these three
curves represents how the volume of a sphere varies with its radius. Another one represents how its
surface area changes with its radius. And then the third line shows how
the ratio of volume to surface area changes with sphere radius. We want to figure out which line is
which.

We notice that all three lines
start at the origin. And then, for some smaller values
of our sphere radius, the purple line is above the red and the blue lines. Eventually though, the blue line
crosses the purple line and the red one does as well. And then, as our radius continues
to increase, the red line crosses the blue line and begins to grow fastest of all
three.

At this point, we can recall the
mathematical equations for the volume of a sphere and its surface area. A sphere of radius π has a volume
four-thirds times π times π cubed, which means that its volume is proportional to
its radius cubed. And then, the surface area of a
sphere is equal to four times π times its radius squared, meaning that this
quantity is proportional to π squared. Once we know the formulas for π
and π΄, we can take their ratio and find that itβs equal to π divided by three,
which tells us that this fraction is proportional to π. Weβll want to focus on these
proportionalities as we figure out which line on our graph corresponds to which
quantity.

The line that corresponds to π,
the answer to the first part of our question, will be a line that follows a cubic
progression. Likewise, the line that represents
π΄ will follow a quadratic progression. And the one that represents π over
π΄ will be linear. Knowing this, we can look at our
graph and note that the red line, even though it starts out below the blue and the
purple line, eventually exceeds them both as the sphere radius increases. So then, by the time it leaves this
graph we see, the red line has the steepest slope or gradiant. In other words, itβs increasing at
the fastest rate. Of the three functional forms weβre
looking for, cubic, quadratic, and linear, we know that itβs the cubic form that
increases in this manner. So, weβll say that the line on our
graph representing π is the red line because as the sphere radius gets larger and
larger, this line increases at the greatest rate.

Next, we want to identify which
line represents the surface area of the sphere. And as we said, this line will be
represented by a quadratic function. Of the lines remaining on our
graph, we see that one, the blue line, has a curve to it as a quadratic function
would, while the purple line looks to be straight. So then, weβll say that itβs the
blue line that represents the sphereβs surface area as its radius changes. And this leaves us with just one
last line, which we expect will represent π divided by π΄. As we saw, we expect this line to
be linear, that is, to have a constant slope or gradient. And indeed, we see that the purple
line fits this description. And so, summing up, the red line
represents the volume π, the blue line represents the surface area π΄, and itβs the
purple line that represents π divided by π΄.

Letβs look now at one more example
exercise.

Consider a cube that initially has
side length π₯. If π₯ increases by a factor of two,
by what factor does the volume of the cube change? By what factor does the surface
area of the cube change?

Alright, so hereβs our cube that
weβre told initially has a side length π₯. And then, weβre to imagine that π₯
increases by a factor of two. This would lead to a cube that
looks like this, with side length two π₯. When this change happens, we first
want to know, by what factor does the volume of the cube change? Now, if we have a cube where π is
the length of each side, then we can recall that the volume of that cube is equal to
π cubed. This means that if we call the
volume of the cube with side length π₯ βπ sub π₯β, then by our equation for the
volume of a cube, that volume is equal to π₯ cubed, while the volume of the cube
with side length two π₯, we can call it π sub two π₯, is equal to two π₯ quantity
cubed or eight times π₯ cubed.

To answer our question, to say by
what factor the volume of the cube changes, weβll want to divide our final cube
volume by our initial volume. When we do this, we see that itβs
equal to eight times π₯ cubed divided by π₯ cubed. So, the factors of π₯ cubed cancel,
telling us that when we double the side length of our cube from π₯ to two π₯, its
volume has increased by a factor of eight. This is the answer to the first
part of our question.

In part two, we want to say by what
factor does the surface area of the cube change. If we again consider a cube of side
length capital πΏ, then the surface area of that cube is equal to six times πΏ
squared. This is because each of the cubeβs
faces has an area of πΏ squared and there are six faces in total. So, if we call π΄ sub π₯ the
surface area of our cube when itβs side length is π₯, then thatβs equal to six times
π₯ squared. And then, if π΄ sub two π₯ is the
cubeβs surface area after its side length has doubled, then thatβs equal to six
times two π₯ quantity squared or six times four times π₯ squared, which equals 24π₯
squared.

Once more, weβll want to take the
ratio of the final quantity to the initial quantity. When we divide 24π₯ squared by six
π₯ squared, the π₯ squareds cancel out. And then, 24 divided by six is
four. So, this is our answer to how the
surface area of the cube changes with this change in side length. All together then, when we double
the side length of a cube, its volume increases by a factor of eight and its surface
area grows by a factor of four.

Letβs summarize now what weβve
learned about surface area to volume ratios. In this lesson, after recalling
that for a cube with side length π, the surface area of that cube is six π
squared, and its volume is π cubed, we saw that the ratio of a cubeβs volume to its
surface area is its side length divided by six, in other words, that this ratio is
proportional to the side length. Related to this, we saw that for a
sphere of radius π, its surface area being four ππ squared and its volume being
four-thirds ππ cubed means that the ratio of a sphereβs volume to its surface area
is equal to its radius divided by three, that is, that this fraction is proportional
to the radius.

We saw that as the size of a shape
changes, say, by increasing in radius, the ratio of its volume to surface area
changes as well. As the shapeβs characteristic
dimension increases, so does this ratio, while as the shape size decreases, the
ratio follows suit. And lastly, we saw that these
trends describe not just cubes and spheres, but three-dimensional shapes in
general.