How a Cylinder’s Volume Changes
A cylinder’s volume depends on two dimensions: its radius and its height. If either dimension changes, the volume may change as well. When both change, their effects contribute to the volume rate together.
The key is to describe each changing dimension as a function of time before differentiating. Then the volume formula becomes a relationship between time-dependent quantities, and differentiation gives the instantaneous rate of change.
The General Rate Formula
Suppose \(r\) and \(h\) are differentiable functions of time. Differentiating \(V=\pi r^2h\) with respect to \(t\) requires both the product rule and the chain rule. The result is
The first term accounts for the effect of a changing radius, while the second accounts for the effect of a changing height. Each term uses the current dimensions and the corresponding rate at the instant in question.
If one dimension is fixed, its rate is zero. For example, if the height is fixed, then \(\frac{dh}{dt}=0\), and the formula reduces to \(\frac{dV}{dt}=2\pi rh\frac{dr}{dt}\). If the radius is fixed, then \(\frac{dr}{dt}=0\), giving \(\frac{dV}{dt}=\pi r^2\frac{dh}{dt}\).
Why the Formula Works
To justify the result, regard \(r\) and \(h\) as functions of \(t\). First apply the product rule to \(r(t)^2h(t)\); then apply the chain rule to \(r(t)^2\):
The two terms are necessary because the volume is a product involving both dimensions. Differentiating only the radius factor would omit the change in volume caused by a changing height; differentiating only the height factor would omit the effect of a changing radius.
Worked Example: Radius Changes While Height Is Fixed
Worked Example: Radius Changes While Height Is Fixed
A cylinder has a fixed height of \(10\) cm. At an instant when its radius is \(3\) cm, the radius is increasing at \(0.2\) cm/s. Find the rate of change of its volume.
Step 1: Identify the dimensions and rates. At the instant, \(r=3\) cm, \(h=10\) cm, and \(\frac{dr}{dt}=0.2\) cm/s. Since the height is fixed, \(\frac{dh}{dt}=0\) cm/s.
Step 2: Apply the general formula.
Check. With \(t=0\) at this instant, the radius is \(r(t)=3+0.2t\) and the height remains \(10\). Thus \(V(t)=10\pi(3+0.2t)^2\), so \(\frac{dV}{dt}=20\pi(3+0.2t)(0.2)\). At \(t=0\), this is \(20\pi(3)(0.2)=12\pi\ \text{cm}^3/\text{s}\), as required. The positive rate indicates that the volume is increasing.
Worked Example: Height Changes While Radius Is Fixed
Worked Example: Height Changes While Radius Is Fixed
A cylinder has a fixed radius of \(4\) m. At a particular instant its height is \(8\) m and is decreasing at \(0.5\) m/min. Find the volume rate at that instant.
Step 1: Record the signed rates. Here \(r=4\) m, \(h=8\) m, \(\frac{dr}{dt}=0\) m/min because the radius is fixed, and \(\frac{dh}{dt}=-0.5\) m/min because the height is decreasing.
Step 2: Substitute into the rate formula.
Check. With \(t=0\) at the instant, \(h(t)=8-0.5t\) while \(r(t)=4\). Therefore \(V(t)=16\pi(8-0.5t)\), and \(\frac{dV}{dt}=16\pi(-0.5)=-8\pi\ \text{m}^3/\text{min}\). The negative sign confirms that volume is decreasing.
When Both Radius and Height Change
If both dimensions vary, their effects can reinforce or oppose one another. An increasing radius contributes a positive term, while a decreasing height contributes a negative term. The net volume rate is their sum, so the volume may increase, decrease, or momentarily stay constant.
Worked Example: Radius and Height Both Increase
At an instant, a cylinder has radius \(2\) cm and height \(5\) cm. Its radius is increasing at \(0.3\) cm/s, and its height is increasing at \(0.4\) cm/s. Find the volume rate.
Step 1: Set out the instantaneous values. We have \(r=2\) cm, \(h=5\) cm, \(\frac{dr}{dt}=0.3\) cm/s, and \(\frac{dh}{dt}=0.4\) cm/s.
Step 2: Calculate each contribution separately.
Step 3: Add the contributions and state the units.
Check. Take \(t=0\) at this instant and write \(r(t)=2+0.3t\), \(h(t)=5+0.4t\). Differentiating \(V(t)=\pi(2+0.3t)^2(5+0.4t)\) by the product and chain rules gives \(\frac{dV}{dt}=\pi\bigl(2(2+0.3t)(0.3)(5+0.4t)+(2+0.3t)^2(0.4)\bigr)\). At \(t=0\), this is \(\pi(2(2)(0.3)(5)+(2)^2(0.4))=\pi(6+1.6)=7.6\pi\ \text{cm}^3/\text{s}\), matching the result. Since both dimensions are increasing here, the positive volume rate is consistent with the changing dimensions.
Signs and Units
Rate signs describe direction, not magnitude. A positive \(\frac{dr}{dt}\) means the radius is increasing; a negative \(\frac{dh}{dt}\) means the height is decreasing. Keep these signs when substituting. The formula then determines the net direction of volume change.
Units provide a useful check. If \(r\) and \(h\) are measured in cm and the dimension rates in cm/s, then each term has units \(\text{cm}^3/\text{s}\). For example, the first term combines \(r\), \(h\), and \(\frac{dr}{dt}\), giving \(\text{cm}\cdot\text{cm}\cdot\text{cm}/\text{s}\). The second combines \(r^2\) and \(\frac{dh}{dt}\), giving the same units. Both must be volume per time.
Common Mistake: Forgetting One Changing Dimension
Suppose \(r=2\) cm and \(h=5\) cm, with \(\frac{dr}{dt}=0.3\) cm/s and \(\frac{dh}{dt}=0.4\) cm/s. A common incorrect calculation is to use only the radius term and report \(\frac{dV}{dt}=2\pi(2)(5)(0.3)=6\pi\ \text{cm}^3/\text{s}\). That calculation silently treats the height as fixed, even though the problem says it is changing.
The correction is to include the height contribution, \(\pi r^2\frac{dh}{dt}=\pi(2)^2(0.4)=1.6\pi\). Thus the correct total is \(6\pi+1.6\pi=7.6\pi\ \text{cm}^3/\text{s}\). The first term alone is appropriate only when \(\frac{dh}{dt}=0\).
A Reliable Calculation Sequence
For cylinder-volume rate problems, organize the calculation so that dimensions, rates, and fixed quantities are not confused:
Use \(V=\pi r^2h\), with \(r\) and \(h\) understood as time-dependent if they change.
Use both terms in \(\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}\).
Use negative rates for decreasing dimensions and zero for fixed dimensions.
Keep units through the calculation, interpret the sign, and verify the result using the differentiated volume expression when useful.
The central distinction is whether a dimension is fixed or changing. The formula handles either situation, but only if each rate is represented correctly. Writing the general formula first makes it harder to omit a contribution or apply a changing rate to a fixed dimension.
Check Your Understanding
Answer these questions using the general cylinder-volume rate formula.
- Write the formula for \(\frac{dV}{dt}\) when both \(r\) and \(h\) depend on time.
- A cylinder’s height is fixed. What is \(\frac{dh}{dt}\), and which term in the general formula disappears?
- If the radius is fixed while the height decreases, what sign should \(\frac{dV}{dt}\) have?
- When radius and height are both changing, why must their contributions be added rather than choosing just one?
- If lengths are measured in meters and time in seconds, what units should \(\frac{dV}{dt}\) have?
