When to Substitute Instantaneous Values

Why the Order of Steps Matters

A related-rates problem describes quantities that change together over time. A geometric or physical relationship connects those quantities, and differentiation with respect to time connects their rates. The order matters: if you replace a changing quantity with its value at one instant too early, you may turn it into a constant and erase the very rate you are trying to find.

The reliable approach is to keep changing quantities as functions of time, differentiate the relationship, and only then substitute the values and rates that apply at the requested instant. This tutorial focuses on diagnosing the error and seeing why that order works.

Changing Quantities and Instantaneous Values

Suppose \(x\) represents a distance that changes over time. To emphasize that dependence, write \(x=x(t)\). Its instantaneous value \(x(t_0)=6\) tells us the distance at the single time \(t_0\). It does not mean that the distance equals \(6\) at every time.

For example, if \(x(t)=6+3(t-t_0)\), then \(x(t_0)=6\), but \(\frac{dx}{dt}=3\). Replacing \(x(t)\) by \(6\) before differentiating instead gives the constant function \(x(t)=6\), whose derivative is \(0\). The replacement changes the function, not merely its notation.

The usual related-rates process begins with a relationship that holds throughout the motion or process. If \(x(t)\) and \(y(t)\) satisfy that relationship, differentiating it with respect to \(t\) produces an equation involving \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), and possibly the quantities themselves. The values at a particular time are then used in that differentiated equation.

Why Differentiation Must Come First

The reason can be stated precisely. Suppose a constraint \(F(x(t),y(t))=0\) holds as the quantities change, and \(F\) is differentiable. The chain rule gives the rate equation

Here \(F_x\) and \(F_y\) are the partial derivatives of \(F\) with respect to its first and second variables. This equation describes how the changing quantities’ rates must relate at each time where the constraint holds and the derivatives exist.

At the specified time \(t_0\), we can then evaluate this equation using \(x(t_0)\), \(y(t_0)\), and any given rates. If we instead replace \(x(t)\) by \(x(t_0)\) in the original equation before differentiating, the resulting expression no longer describes the changing quantity \(x(t)\). Differentiating that frozen value gives zero, not the original rate.

Worked Example: A Sliding Ladder

The Ladder Error: Freezing Both Dimensions

A tempting but incorrect calculation is to substitute \(x=6\) and \(y=8\) into the original constraint first:

This is true, but it is only a numerical statement about the ladder at that instant. It contains no changing quantities left to differentiate. Differentiating the numerical equation gives \(0=0\), which cannot determine either rate.

A closely related error is to solve for the height, substitute \(x=6\), and then differentiate the resulting numerical value:

Now \(8\) is a fixed number, so differentiating it gives zero. But the ladder’s height is not fixed at \(8\) feet throughout its motion. The substitution has discarded the relationship that describes how the height changes.

Worked Example: Volume of a Changing Sphere

The Sphere Error: Substituting the Radius First

If the radius \(r=3\) is substituted into the volume formula before differentiating, the result is \(V=\frac43\pi(3)^3=36\pi\) cubic centimeters. That correctly gives the sphere’s volume at the specified instant. It does not describe the sphere’s volume at other times. Differentiating the fixed number \(36\pi\) gives zero, which incorrectly suggests that the volume is not changing.

The radius value is needed, but it is needed after differentiation. The derivative equation still contains \(r\), so substituting \(r=3\) there calculates the volume’s rate at the correct instant without freezing the radius during the differentiation.

When Early Substitution Is Safe

Not every substitution before differentiation is an error. A value can be substituted early if it is a genuine constant for the whole process being modeled, rather than a changing quantity’s value at one instant.

The distinction is not whether a number is known. It is whether the quantity is constant over time. A fixed ladder length, a fixed rectangle width, and fixed numerical coefficients may be substituted as constants. A ladder’s distance from the wall at one instant or a sphere’s radius at one instant may not be treated as constant if that dimension is changing.

A Practical Diagnostic

Before substituting a number, ask what the number represents. If it is the value of a changing quantity “at the instant,” keep the corresponding variable in the equation until after differentiating. If it is a fixed parameter that does not vary during the modeled process, it can be treated as a constant.

One useful check is to inspect the equation after substitution. If the substitution has turned every changing quantity into a number, then differentiating that numerical equation cannot reveal the original rates. Return to the time-dependent relationship and differentiate before evaluating.

This order is essential because a related-rates equation describes change, not just a configuration at one moment. Differentiation captures the way quantities vary together; evaluation supplies the measurements for the particular moment of interest.