Related Rates with Multiple Changing Inputs

When Several Quantities Change

In a related-rates problem, the quantities in a model vary as time passes. Often, more than one input changes, so the rate of an output depends on several rates at once. A useful example is the ideal gas law: pressure, volume, amount of gas, and temperature may all vary during a process.

The key is to regard each changing quantity as a function of time, differentiate the relationship with respect to time, and only then substitute the instantaneous values and rates. This keeps the contribution from each changing input visible and makes it possible to isolate a rate that is not given.

In what follows, a quantity written without an explicit \(t\), such as \(P\), denotes its value at the instant being considered; \(P'\) denotes its instantaneous rate of change. A fixed constant has derivative zero. In particular, the gas constant \(R\) in the ideal gas law is treated as fixed.

The General Rate Equation

Suppose quantities \(x_1,\ldots,x_k\) depend on time and satisfy a differentiable relation \(F(x_1,\ldots,x_k)=0\). The chain rule gives the rate equation along their changing values:

Each changing input contributes its partial derivative multiplied by its own rate. If the relation is instead written as \(Q=f(x_1,\ldots,x_k)\), then \(Q\) may also change, and its rate is the corresponding sum of these contributions. These are applications of the multivariable chain rule; the task in a related-rates calculation is to apply it to the model and solve the resulting equation.

Worked Example: Solving for the Temperature Rate

Why Differentiating Captures All the Changes

The rate equation is not an additional physical assumption. It follows because the original relationship remains true as time changes. For the gas law, define \(G(t)=P(t)V(t)-n(t)RT(t)\). If the gas-law relation holds throughout the process, then \(G(t)=0\) throughout, so \(G'(t)=0\).

Differentiate each term. The product rule applies to \(P(t)V(t)\) and \(n(t)T(t)\); the constant \(R\) remains a factor. Thus:

Rearranging gives exactly the rate equation used above. Each term appears because the corresponding factor can change; treating one changing factor as fixed would omit its contribution. The same reasoning works for any differentiable constraint: differentiate the constraint along the time-dependent quantities, and its derivative must remain zero.

Worked Example: Solving for the Pressure Rate

Worked Example: A Constraint with Three Changing Inputs

Isolating a Rate and Knowing When It Is Possible

After differentiation and substitution, a related-rates problem usually becomes a linear equation in the unknown rate. In the gas-law examples, the coefficient of \(T'\) is \(Rn\), and the coefficient of \(P'\) is \(V\). If that coefficient is nonzero, ordinary algebra isolates the unknown. For example, solving for \(T'\) gives:

The condition \(Rn\ne 0\) matters: division by zero is not defined. If the coefficient of the unknown rate is zero, the differentiated equation may impose a condition on the known rates without determining the unknown. It may also be inconsistent with those rates. One must inspect the resulting equation rather than divide automatically.

In other models, the coefficient depends on the instantaneous values. For the constraint \(xy^2+z^2=13\), the coefficient of \(z'\) is \(2z\). When \(z=3\), it is nonzero and the rate is determined. If instead \(z=0\), this equation at that instant would have no \(z'\) term; it would not determine \(z'\) from that differentiated constraint alone.

A Common Mistake: Substituting Before Differentiating

A tempting but incorrect approach is to put all instantaneous values into the model first and then differentiate the resulting numerical equality. In the first gas-law example, substituting the values into \(PV=nRT\) gives \(6=6\), a true numerical statement. But this equation contains no time-varying quantities left to differentiate, so it gives no information about \(T'\).

The values describe one instant; they do not replace the functions that change over time. The correct order is to differentiate while the quantities are still functions, then evaluate the resulting rate equation at the instant. For example, differentiating first produces \(P'V+PV'=R(n'T+nT')\), which contains the rates needed to solve the problem.

Why Multiple-Input Rates Matter

Many models connect quantities that change together. In the ideal gas law, pressure can respond to changing volume, amount of gas, and temperature. In a geometric constraint, the rate of one coordinate can depend on the rates of several others. A single-variable calculation that overlooks any changing input generally produces an incomplete rate equation.

Writing each changing quantity as a function of time makes the dependence explicit. Differentiating the whole relationship accounts for every contribution, while the given instantaneous data turn the resulting expression into an equation that can be solved. The final check confirms that the proposed rate satisfies the differentiated model at that instant.