Related Rates in Changing Solids

Changing Volume and Changing Dimensions

A filling tank, draining funnel, or expanding balloon changes over time. Its volume and dimensions are related by a geometric formula, so a known volume rate can determine how quickly a dimension is changing at a particular instant.

The key step is to treat the dimension as a function of time and differentiate the volume formula with respect to time. This is an application of the chain rule. The resulting rate is instantaneous: it describes what is happening at the specified moment, not necessarily at every moment.

Core method: Express the volume in terms of the changing dimension, differentiate with respect to time, then substitute the measurements and rate given for the instant in question.

Volume Formulas and Time Rates

For a cylinder of radius \(r\) and height \(h\), a cone of base radius \(r\) and height \(h\), and a sphere of radius \(r\), the volume formulas are:

V_{\text{cylinder}}=\pi r^2h,\qquad V_{\text{cone}}=\frac{1}{3}\pi r^2h,\qquad V_{\text{sphere}}=\frac{4}{3}\pi r^3.

When a dimension changes with time \(t\), its time rate is written, for example, as \(dr/dt\) or \(dh/dt\). The volume rate is \(dV/dt\). If the volume can be written as \(V=F(x)\), where \(x=x(t)\), the chain rule gives:

\frac{dV}{dt}=F'(x)\frac{dx}{dt}.

This form applies directly when one changing dimension determines the volume. A cylinder with fixed radius and changing liquid height is one such case. For a cone filled with liquid, both the liquid radius and height change, but the cone’s fixed shape supplies a relation between them. That relation lets us rewrite the volume using just one changing dimension.

Rate signs: A positive time derivative indicates an instantaneous increase in that quantity; a negative time derivative indicates an instantaneous decrease. A differentiable quantity can be increasing and still have derivative zero at an instant, so a zero derivative does not by itself mean the quantity is constant.

Why Differentiating the Volume Works

The volume is not an independent quantity in these problems: at every time, it is determined by the dimensions through the appropriate formula. If one changing dimension \(x(t)\) determines the volume, then \(V(t)=F(x(t))\). By the chain rule, the instantaneous volume rate is the sensitivity of volume to \(x\), namely \(F'(x)\), multiplied by the instantaneous rate at which \(x\) changes.

This multiplication explains the structure of a related-rates equation. The factor \(F'(x)\) measures how much the volume changes per unit change in the dimension at the current size; \(dx/dt\) measures how quickly that dimension changes per unit time. Their product has units of volume per time, as \(dV/dt\) should.

For example, if \(V=\pi r^2h\) and \(r\) is fixed while \(h\) changes, then \(V\) is a function of \(h(t)\) alone. Differentiating gives \(dV/dt=\pi r^2\,dh/dt\). For a cone whose liquid surface stays geometrically similar to the full cone, the radius is not fixed; instead, it is proportional to the liquid height. Substituting that proportionality first gives a volume formula in one variable, which can then be differentiated.

Worked Example: Filling a Cylindrical Tank

Worked Example: Finding the Rising Water Level

A vertical cylindrical tank has a fixed radius of \(3\) m. Water enters at \(dV/dt=6/25\) m\(^3\)/min. Find the rate at which the water height \(h\) is rising.

Since the cylinder’s radius is fixed, its water volume is \(V=\pi r^2h\). Substitute \(r=3\):

V=\pi(3)^2h=9\pi h.

Differentiate with respect to time. The radius and \(9\pi\) are constant:

\frac{dV}{dt}=9\pi\frac{dh}{dt}.

Substitute the given volume rate and solve:

\frac{6}{25}=9\pi\frac{dh}{dt}, \qquad \frac{dh}{dt}=\frac{6}{225\pi} =\frac{2}{75\pi}\ \text{m/min}.

The height rate is positive, consistent with the tank filling. Check by substitution into the differentiated volume formula:

9\pi\left(\frac{2}{75\pi}\right) =\frac{18}{75} =\frac{6}{25}\ \text{m}^3/\text{min}.

This recovers the specified volume rate.

Cones: Use the Fixed Shape

For liquid in a conical container, the liquid’s surface radius \(r\) changes with its depth \(h\). If the container is a fixed cone and the liquid forms a smaller cone with the same proportions, similar triangles give \(r/h=R/H\), where \(R\) and \(H\) are the container’s fixed radius and height. This proportion is the extra relationship needed to express the liquid volume in terms of one changing dimension.

For a fixed cone shape: If the full cone has radius \(R\) and height \(H\), then liquid depth \(h\) and liquid-surface radius \(r\) satisfy \(r=(R/H)h\). This relation assumes the liquid surface is horizontal and the liquid occupies a smaller cone similar to the container.

Worked Example: Filling a Conical Container

Worked Example: Relating Cone Depth to Filling Rate

A conical container is \(9\) m high and has a top radius of \(3\) m. It is being filled so that \(dV/dt=\pi\) m\(^3\)/min. Find \(dh/dt\) when the liquid depth is \(3\) m.

Similar triangles give \(r/h=3/9=1/3\), so \(r=h/3\). Substitute this into the cone volume formula:

V=\frac{1}{3}\pi r^2h =\frac{1}{3}\pi\left(\frac{h}{3}\right)^2h =\frac{\pi h^3}{27}.

Differentiate with respect to time, using the chain rule:

\frac{dV}{dt} =\frac{\pi}{27}\cdot 3h^2\frac{dh}{dt} =\frac{\pi h^2}{9}\frac{dh}{dt}.

At \(h=3\) m, the coefficient of \(dh/dt\) is \(\pi(3)^2/9=\pi\). Thus:

\pi=\pi\frac{dh}{dt}, \qquad \frac{dh}{dt}=1\ \text{m/min}.

Check by substituting \(h=3\) and \(dh/dt=1\) into the differentiated volume relation:

\frac{\pi(3)^2}{9}(1)=\pi\ \text{m}^3/\text{min}.

The result matches the given filling rate. The positive depth rate is consistent with the liquid level rising.

Spheres: Expansion and Shrinkage

A sphere’s volume depends only on its radius, so its volume rate is obtained by differentiating with respect to time. The radius need not be fixed; treating it as \(r(t)\) gives the relation between the instantaneous volume rate and radius rate. A negative volume rate, as in a shrinking or melting sphere, produces a negative radius rate.

Worked Example: A Shrinking Sphere

Worked Example: Finding a Melting Sphere’s Radius Rate

A spherical ice ball has radius \(5\) cm at a particular instant. Its volume is decreasing at \(20\pi\) cm\(^3\)/min. Find \(dr/dt\) at that instant.

Because the volume is decreasing, the signed volume rate is \(dV/dt=-20\pi\) cm\(^3\)/min. Begin with the sphere formula and differentiate with respect to time:

V=\frac{4}{3}\pi r^3, \qquad \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

Substitute \(r=5\) cm and the signed volume rate:

-20\pi=4\pi(5)^2\frac{dr}{dt} =100\pi\frac{dr}{dt}.

Divide both sides by \(100\pi\):

\frac{dr}{dt}=-\frac{1}{5}\ \text{cm/min}.

Check the result in the differentiated volume formula:

4\pi(5)^2\left(-\frac{1}{5}\right) =100\pi\left(-\frac{1}{5}\right) =-20\pi\ \text{cm}^3/\text{min}.

The computed rate reproduces the given volume decrease, and its negative sign indicates that the radius is instantaneously shrinking.

A Common Mistake: Treating a Cone’s Radius as Fixed

In the cone example, a tempting but incorrect step is to differentiate \(V=\frac{1}{3}\pi r^2h\) as though \(r\) were constant, writing \(dV/dt=\frac{1}{3}\pi r^2\,dh/dt\). That calculation omits the changing radius. In the filled cone, \(r=h/3\), so the correct procedure is to substitute the relation before differentiating:

V=\frac{\pi h^3}{27} \quad\Longrightarrow\quad \frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}.

At \(h=3\) m, this coefficient is \(\pi\), whereas the incorrect fixed-radius expression would give \(\frac{1}{3}\pi(1)^2=\pi/3\) as the coefficient of \(dh/dt\). It misses the volume change caused by the widening liquid surface. The mistake would predict \(dh/dt=3\) m/min for a volume rate of \(\pi\) m\(^3\)/min, but substituting that rate into the correct relation gives \(3\pi\) m\(^3\)/min, not \(\pi\) m\(^3\)/min.

Choosing and Checking the Relation

Before differentiating, decide which dimensions are fixed and what geometric constraint links any dimensions that change together. In the examples, the cylinder’s radius is fixed, the cone’s radius is tied to its depth by similar triangles, and a sphere’s volume depends only on its radius. If a problem gives no relation between multiple changing dimensions, the volume formula alone may not determine each rate separately.

After differentiating, substitute the measurements and rates for the specified instant. Keep rates signed: use a negative \(dV/dt\) for decreasing volume and a negative dimension rate for a dimension that is instantaneously decreasing. Finally, substitute the result back into the differentiated equation. This check confirms the arithmetic and whether the equation reproduces the stated rate.

Related rates are useful because a volume rate is often easier to measure or specify than a dimension rate. The chain rule translates between them, provided the solid’s geometry and the changing dimensions are modeled correctly. The method does not say that a dimension changes at a constant rate; it gives its instantaneous rate at the moment being analyzed.

Check Your Understanding

For each question, identify the volume relation and the dimension that changes before calculating.

A cylinder has fixed radius \(2\) m. Write \(dV/dt\) in terms of \(dh/dt\).
A conical container has height \(6\) m and top radius \(2\) m. What relation connects the liquid radius \(r\) and depth \(h\) while the liquid forms a similar cone?
For a sphere, derive the relation between \(dV/dt\) and \(dr/dt\).
If a sphere’s volume is decreasing at an instant, what sign should its volume rate have? What does the sign of its radius rate indicate?
Why is it incorrect to hold the liquid-surface radius fixed when differentiating the volume of a filling cone?