Rates of Change in Sphere Volume and Surface Area

How a Changing Radius Changes a Sphere

A sphere’s volume and surface area both depend on its radius. If the radius changes over time, both measurements change too. Related-rates calculations describe how quickly those changes happen at a particular instant.

Let $r=r(t)$ be the radius of a sphere at time $t$. Assume that $r(t)>0$ and is differentiable at the time under consideration. Its volume $V$ and surface area $S$ are

V=\frac{4}{3}\pi r^3, \qquad S=4\pi r^2.

These formulas express volume and area in terms of the radius. Because the radius depends on time, volume and surface area also depend on time. Differentiating each formula with respect to time gives equations for their instantaneous rates.

Key idea: For a sphere whose radius varies with time, $$ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}, \qquad \frac{dS}{dt}=8\pi r\frac{dr}{dt}. $$ The rates depend on both the sphere’s radius at that instant and the rate at which the radius is changing.

Deriving the Rate Equations

The chain rule applies because $V$ and $S$ are functions of $r$, and $r$ is a function of $t$. In particular, differentiating a power of $r(t)$ with respect to $t$ introduces the factor $dr/dt$.

Differentiate the volume formula.
Starting from $V=\frac{4}{3}\pi r^3$, treat $r$ as a function of time. The constant multiple rule and chain rule give $\frac{dV}{dt}=\frac{4}{3}\pi\cdot 3r^2\frac{dr}{dt}$.

Simplify the volume rate.
Since $\frac{4}{3}\cdot 3=4$, this becomes $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$.

Differentiate the surface-area formula.
Starting from $S=4\pi r^2$, the chain rule gives $\frac{dS}{dt}=4\pi\cdot 2r\frac{dr}{dt}$.

Simplify the surface-area rate.
Therefore, $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$.

This derivation shows why the radius rate appears in both equations: a change in time changes the radius, and that change is passed through the volume or area formula by the chain rule. The powers of the radius differ because volume is proportional to $r^3$, while surface area is proportional to $r^2$.

To use the equations, identify the radius and radius rate at the instant, then substitute those values. The units follow from the quantities involved: if $r$ is measured in centimeters and $t$ in minutes, then $dr/dt$ is measured in centimeters per minute, $dV/dt$ in cubic centimeters per minute, and $dS/dt$ in square centimeters per minute.

Interpreting the Signs

The radius $r$ is positive for a nondegenerate sphere, and $\pi$ is positive. Therefore, the signs of both $dV/dt$ and $dS/dt$ match the sign of $dr/dt$. If the radius is increasing, the volume and surface area are increasing. If the radius is decreasing, both are decreasing.

A negative rate does not mean that volume or surface area is negative. It means the quantity is getting smaller as time increases. For example, $dV/dt=-20\text{ cm}^3/\text{s}$ means that the volume is decreasing at $20\text{ cm}^3$ per second at that instant.

Worked Example: An Expanding Sphere

A spherical balloon has radius $3\text{ cm}$ at an instant, and its radius is increasing at $2\text{ cm}/\text{min}$. Find the instantaneous rates of change of its volume and surface area.

First record the values and units:

r=3\text{ cm}, \qquad \frac{dr}{dt}=2\text{ cm}/\text{min}.

Use the volume-rate equation:

\frac{dV}{dt} =4\pi r^2\frac{dr}{dt} =4\pi(3\text{ cm})^2(2\text{ cm}/\text{min}) =4\pi(9\text{ cm}^2)(2\text{ cm}/\text{min}) =72\pi\text{ cm}^3/\text{min}.

Use the surface-area-rate equation:

\frac{dS}{dt} =8\pi r\frac{dr}{dt} =8\pi(3\text{ cm})(2\text{ cm}/\text{min}) =48\pi\text{ cm}^2/\text{min}.

Both rates are positive, as expected for an expanding sphere. Check by substituting the values back into the rate equations: $4\pi(3)^2(2)=72\pi$ and $8\pi(3)(2)=48$. The units are respectively cubic centimeters per minute and square centimeters per minute.

When the Radius Is Decreasing

The same formulas work when the sphere shrinks. In that case, $dr/dt$ is negative, so the calculated volume and surface-area rates are negative as well. Keep the negative sign in the rate equation; it records the direction of change.

Worked Example: A Shrinking Sphere

A sphere has radius $5\text{ cm}$, and the radius is decreasing at $0.4\text{ cm}/\text{s}$. Find the rates of change of volume and surface area.

Because the radius is decreasing, its rate is negative:

r=5\text{ cm}, \qquad \frac{dr}{dt}=-0.4\text{ cm}/\text{s}.

For volume, substitute into the rate equation:

\frac{dV}{dt} =4\pi(5\text{ cm})^2(-0.4\text{ cm}/\text{s}) =4\pi(25\text{ cm}^2)(-0.4\text{ cm}/\text{s}) =-40\pi\text{ cm}^3/\text{s}.

For surface area:

\frac{dS}{dt} =8\pi(5\text{ cm})(-0.4\text{ cm}/\text{s}) =-16\pi\text{ cm}^2/\text{s}.

Both rates are negative, confirming that both measurements are decreasing. Substitution checks the results: $4\pi(25)(-0.4)=-40\pi$, and $8\pi(5)(-0.4)=-16\pi$. The units are $\text{cm}^3/\text{s}$ and $\text{cm}^2/\text{s}$, respectively.

Finding the Radius Rate from a Volume Rate

Sometimes the radius rate is not given. If the radius and volume rate are known, solve the volume-rate equation for $dr/dt$. For $r>0$, the coefficient $4\pi r^2$ is nonzero, so division by it is valid:

\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}.

Once the radius rate is found, substitute it into the surface-area-rate equation if the surface-area rate is also needed. This is still an instantaneous calculation: use the radius and volume rate at the same instant.

Worked Example: Using a Volume Rate to Find a Surface-Area Rate

At an instant, a sphere has radius $4\text{ cm}$, and its volume is increasing at $96\pi\text{ cm}^3/\text{min}$. Find the radius rate and the surface-area rate.

Begin with the volume-rate equation and isolate the radius rate:

\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^2} = \frac{96\pi\text{ cm}^3/\text{min}} {4\pi(4\text{ cm})^2} = \frac{96\pi\text{ cm}^3/\text{min}} {64\pi\text{ cm}^2} = \frac{3}{2}\text{ cm}/\text{min}.

Now use this radius rate to find the surface-area rate:

\frac{dS}{dt} =8\pi r\frac{dr}{dt} =8\pi(4\text{ cm})\left(\frac{3}{2}\text{ cm}/\text{min}\right) =48\pi\text{ cm}^2/\text{min}.

Check the inferred radius rate in the volume equation:

4\pi(4\text{ cm})^2\left(\frac{3}{2}\text{ cm}/\text{min}\right) =4\pi(16)\left(\frac{3}{2}\right)\text{ cm}^3/\text{min} =96\pi\text{ cm}^3/\text{min}.

This matches the given volume rate. The positive radius rate is consistent with the stated increase in volume.

A Direct Relationship Between the Two Rates

When $r>0$, the surface-area rate equation can also be compared with the volume-rate equation. Multiplying $dS/dt=8\pi r\,dr/dt$ by $r/2$ gives

\frac{r}{2}\frac{dS}{dt} = \frac{r}{2}\left(8\pi r\frac{dr}{dt}\right) = 4\pi r^2\frac{dr}{dt} = \frac{dV}{dt}.

Thus, at a positive radius, the rates satisfy

\frac{dV}{dt}=\frac{r}{2}\frac{dS}{dt}.

This relationship follows from the two rate formulas; it does not say that volume equals radius times surface area. It connects their instantaneous rates at the same radius and time. The units also agree: multiplying an area rate by a length gives a volume rate.

Common Mistake: Forgetting the Radius Rate

A frequent error is to differentiate with respect to the radius and then treat the result as if it were already a rate with respect to time. For example, starting from $V=\frac{4}{3}\pi r^3$, a mistaken calculation might be

\frac{dV}{dt}=4\pi r^2.

This expression is missing $dr/dt$. The derivative $4\pi r^2$ is the rate of change of volume with respect to radius, not with respect to time. Applying the chain rule gives the correct calculation:

\frac{dV}{dt} = \frac{dV}{dr}\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}.

The units expose the mistake. The expression $4\pi r^2$ has units of area, while $dV/dt$ must have units of volume per time. Multiplying by $dr/dt$, which has units of length per time, gives the required units of volume per time.

Why These Rates Matter

Volume and surface area describe different geometric measurements, so their rates have different units and different dependence on the radius. For the same radius rate, the volume rate includes $r^2$, while the surface-area rate includes $r$. As a result, the numerical rates respond differently to the size of the sphere.

For a related-rates calculation, keep the quantities distinct: $r$ is a length, $dr/dt$ is a length rate, $V$ is a volume, and $S$ is an area. Differentiate first, then use the values corresponding to the instant of interest. Finally, check the sign and units. These checks often reveal a missing chain-rule factor or an incorrectly assigned rate.

Check Your Understanding

Use the sphere rate equations to answer the following questions. Include units and retain signs in your answers.

A sphere has radius $2\text{ m}$ and its radius increases at $0.5\text{ m}/\text{s}$. What are $dV/dt$ and $dS/dt$?
A sphere’s radius is decreasing. What signs must $dV/dt$ and $dS/dt$ have when its radius is positive?
At an instant, $r=3\text{ cm}$ and $dV/dt=-36\pi\text{ cm}^3/\text{min}$. Find $dr/dt$.
Why does differentiating $V=\frac{4}{3}\pi r^3$ with respect to time produce a factor of $dr/dt$?
For positive radius, express $dV/dt$ in terms of $r$ and $dS/dt$.

Rates of Change in Sphere Volume and Surface Area

What You'll Learn

How a Changing Radius Changes a Sphere

Deriving the Rate Equations