Substituting Values After Differentiation

Why the Order of Steps Matters

In a related-rates problem, quantities change over time but are linked by a constraint. The constraint may be a geometric relation or another equation that holds throughout the motion. The goal is often to find one rate at a particular instant, using the values and other rates known at that instant.

The crucial order is: first differentiate the general constraint with respect to time; then substitute the values and known rates for the specified instant. Before differentiation, the quantities that change must remain variables. Substituting their current values too early can turn the constraint into a numerical equality and discard the dependence that relates their rates.

Changing Quantities and Instantaneous Values

Let \(x(t)\) and \(y(t)\) be quantities that vary with time \(t\). Their rates of change are \(x'(t)\) and \(y'(t)\). An equation such as \(F(x(t),y(t))=C\) describes a constraint: it is assumed to hold as the quantities change, not merely at one selected instant.

At a particular time \(t_0\), we may know the values \(x(t_0)\) and \(y(t_0)\), as well as one rate such as \(x'(t_0)\). Those are data for evaluating the differentiated constraint. They do not replace the time-dependent quantities before differentiation.

For example, if \(x(t)^2+y(t)^2=25\) holds over an interval, the quantities are not fixed at \(x=3\) and \(y=4\). Those values might describe just one instant. Differentiating the general constraint gives an equation involving \(x(t)\), \(y(t)\), \(x'(t)\), and \(y'(t)\). Only after obtaining that equation should we use \(x=3\) and \(y=4\) to evaluate it at the instant in question.

Why Differentiating First Works

Suppose \(F\) is differentiable and a constraint \(F(x(t),y(t))=C\) holds for the times under consideration, where \(C\) is constant. By the chain rule, the derivative of the left side is the partial derivative with respect to \(x\) multiplied by \(x'(t)\), plus the partial derivative with respect to \(y\) multiplied by \(y'(t)\). The derivative of the constant on the right is zero.

Here \(F_x\) and \(F_y\) denote the partial derivatives of \(F\) with respect to its first and second variables. This equation is a rate relation: it records how the changing quantities must vary together in order for the original constraint to remain true.

The result depends on the constraint holding over time and on the differentiability needed to apply the chain rule. If an equation is known only at one isolated instant, differentiating it as though it held over an interval is not justified. In a related-rates setup, the constraint represents a continuing relationship, and the specified instant supplies the values at which its derivative is evaluated.

A Reliable Substitution Procedure

A useful way to keep the roles of the general variables and instant-specific data clear is to separate the work into stages.

The order is not a cosmetic preference. The differentiated equation contains information about rates. A numerical version of the original constraint usually contains only values, so it cannot by itself determine how those values are changing.

Worked Examples

A Common Mistake: Substituting Too Early

Consider the constraint \(z(t)^3+x(t)^2=17\) from the last example. One might substitute the instant’s values \(z=2\) and \(x=3\) into the original equation first, obtaining \(8+9=17\). That is a true numerical statement, but it no longer displays how either quantity changes over time.

Differentiating that numerical equality gives \(0=0\). This is not a rate equation and cannot determine \(z'\). The values have replaced the changing quantities before the chain rule could act on them.

The correction is to differentiate the original time-dependent constraint before substituting:

At the specified instant, this becomes \(12z'+24=0\), which does contain the information needed to solve for \(z'\). The numerical constraint verifies that the instant’s values are compatible with the relationship; the differentiated constraint determines how the rates are related.

Using the Method Carefully

Keep three kinds of information distinct while working: the constraint that holds over time, the values of the quantities at the instant, and the rates known or requested at that instant. The constraint is differentiated; the instant-specific values and known rates are substituted afterward.

A fixed constant in the constraint can be used as written throughout. For example, differentiating \(p(t)q(t)=36\) gives a zero on the right because \(36\) is constant. By contrast, a quantity such as \(p(t)\) is not constant just because its value happens to be \(4\) at one instant. Its derivative \(p'(t)\) may still be nonzero.

The final substitution into the rate equation is also a useful check. After solving for the requested rate, place it back into the evaluated equation and verify that both sides agree. This confirms the algebra in the rate calculation, although it does not replace checking that the original model and given instant-specific values are appropriate.