Why a Cube’s Size Matters
A cube’s volume depends on the length of its side. If that side changes over time, then the volume changes too. The two rates are related, but they are not equal: the same increase in side length produces a larger volume increase for a larger cube.
This tutorial develops that relationship by differentiating the cube-volume formula with respect to time. The result lets us find either rate when the other is known, provided we also know the cube’s side length at the instant of interest.
Volume and Side Length
Let \(s(t)\) be the side length of a cube at time \(t\), and let \(V(t)\) be its volume. For a cube, volume is the cube of its side length:
Here, \(s(t)\) must be nonnegative because it represents a length. In a rate problem, we usually consider an instant when \(s(t)>0\). We assume the side length is differentiable with respect to time near that instant, so its instantaneous rate of change exists.
The units matter. If \(s\) is measured in centimeters and \(t\) in minutes, then \(\frac{ds}{dt}\) is in centimeters per minute, while \(\frac{dV}{dt}\) is in cubic centimeters per minute.
Differentiate the Volume Relation
Because \(s\) depends on time, differentiating \(V=s^3\) with respect to \(t\) requires the chain rule. The side-length rate is part of the derivative; it cannot be omitted.
The first formula is useful when the side-length rate is known and the volume rate is required. The rearranged formula is useful when the volume rate is known and the side-length rate is required. In either case, the side length must be evaluated at the same instant as the given rate.
Why the Rate Formula Works
The derivative can be justified directly using the product rule. Write the volume as a product of three identical factors, \(V=s\cdot s\cdot s\). Since each factor depends on time, the product rule gives one differentiated factor at a time:
\(V(t)=s(t)s(t)s(t)\).
\(\frac{dV}{dt}=\frac{ds}{dt}ss+s\frac{ds}{dt}s+ss\frac{ds}{dt}\).
Each term is \(s^2\frac{ds}{dt}\), so \(\frac{dV}{dt}=3s^2\frac{ds}{dt}\).
This derivation shows why the factor \(3s^2\) appears: a small change in side length affects each of the cube’s three equal length factors. It also makes clear why the formula relates instantaneous rates, rather than finite changes over a time interval.
Finding Side-Length Growth from Volume Growth
Worked Example: A Cube with Increasing Volume
A cube has side length \(4\) cm at an instant. At that instant its volume is increasing at \(96\) cm³/min. Find the rate at which its side length is increasing.
Use the rate relationship and solve for \(\frac{ds}{dt}\):
Substitute \(s=4\) cm and \(\frac{dV}{dt}=96\) cm³/min:
The side length is increasing at \(2\) cm/min. Check by substituting this rate into the original relationship:
This matches the given volume rate, so the result is consistent.
The side length is essential to this calculation. A volume rate alone does not determine a unique side-length rate: the same volume change corresponds to different side-length changes at different cube sizes.
Finding Volume Growth from Side-Length Growth
Worked Example: A Cube Whose Side Is Growing
A cube has side length \(3\) m, and its side length is increasing at \(0.5\) m/s. Find its volume rate at that instant.
Use \(\frac{dV}{dt}=3s^2\frac{ds}{dt}\). Substitute \(s=3\) m and \(\frac{ds}{dt}=0.5\) m/s:
Thus, the volume is increasing at \(13.5\) m³/s. Check by substituting the calculated volume rate back into the relationship:
The calculation reproduces the stated result. The units also agree: \(\text{m}^2\cdot\text{m}/\text{s}=\text{m}^3/\text{s}\).
This example illustrates the size effect. The coefficient multiplying the side-length rate is \(3s^2\). At a larger side length, that coefficient is larger, so the same side-length rate produces a greater volume rate.
Decreasing Volume and Negative Rates
A rate is signed. If a cube is shrinking, its side-length rate is negative. Its volume rate is also negative when its side length is positive, because \(3s^2\) is then positive. Keeping the sign helps distinguish growth from shrinkage.
Worked Example: A Shrinking Cube
A cube has volume \(125\) cm³ at an instant, and its volume is decreasing at \(30\) cm³/min. Find the side-length rate at that instant.
First find the positive side length. Since \(V=s^3\),
A decreasing volume has rate \(\frac{dV}{dt}=-30\) cm³/min. Use the rearranged rate formula:
The side length is decreasing at \(0.4\) cm/min. Check using the original relationship:
This agrees with the given volume rate, including its negative sign.
A Common Mistake: Dropping the Side-Length Rate
A frequent error is to differentiate \(V=s^3\) and write \(\frac{dV}{dt}=3s^2\). That expression is missing the factor \(\frac{ds}{dt}\). The power rule gives the derivative with respect to \(s\); the chain rule is also needed because \(s\) changes with time.
For example, if \(s=3\) m and \(\frac{ds}{dt}=0.5\) m/s, the incorrect calculation would give \(3(3)^2=27\) and might label that as the volume rate. But \(27\) has units of square meters, not cubic meters per second, so it cannot be the requested rate.
The correction is to include the side-length rate:
Checking units is a useful safeguard: the factor \(3\) is dimensionless, \(s^2\) has units of area, and \(\frac{ds}{dt}\) has units of length per time. Their product has units of volume per time.
Interpreting the Conversion
For a positive side length, \(\frac{dV}{dt}\) and \(\frac{ds}{dt}\) always have the same sign, because the multiplier \(3s^2\) is positive. An expanding cube has both rates positive; a shrinking cube has both rates negative. This conclusion follows directly from the rate formula.
The reverse calculation requires care near \(s=0\). Dividing by \(3s^2\) is valid only when \(s>0\). At \(s=0\), that denominator is zero, so the rearranged formula cannot be used to infer a side-length rate from a volume rate. The original differentiated relation remains the correct formula, but it gives \(\frac{dV}{dt}=0\) at that instant whenever \(\frac{ds}{dt}\) is finite. Rate problems that ask for the side-length rate from volume data therefore need a positive side length at the instant being considered.
The relationship is useful whenever a cube’s dimensions are changing and one type of rate is measured while the other is sought. The key steps are to use the current side length, preserve the sign of the given rate, and include \(\frac{ds}{dt}\) when differentiating with respect to time.
Check Your Understanding
Use \(V=s^3\) and \(\frac{dV}{dt}=3s^2\frac{ds}{dt}\) to answer these questions.
- A cube has side length \(2\) cm and its side grows at \(1\) cm/min. What is its volume rate?
- A cube has side length \(6\) m and its volume grows at \(324\) m³/s. What is its side-length rate?
- If a cube is shrinking while its side length is positive, what are the signs of its volume rate and side-length rate?
- Why can the volume rate not be converted to a side-length rate using \(\frac{1}{3s^2}\) when \(s=0\)?
- What units result when \(s\) is in meters and \(\frac{ds}{dt}\) is in meters per second in the formula for \(\frac{dV}{dt}\)?
