Why Check Units and Signs?
A related-rate calculation can use the correct differentiation rule and still produce an answer that is wrong or misinterpreted. A unit check can reveal that a length has mistakenly been treated as a rate, or that quantities with incompatible units have been added. A sign check can reveal that a decreasing quantity has been reported as increasing.
The key is to interpret each derivative as a signed rate at a particular instant. Its units describe what quantity is changing per unit time, and its sign describes the direction of change. Neither is an optional label: both are part of the answer.
For example, if a radius is measured in centimeters and time in seconds, \(\frac{dr}{dt}\) is measured in centimeters per second. A statement that the radius is decreasing at \(0.4\) centimeters per second means \(\frac{dr}{dt}=-0.4\ \text{cm}/\text{s}\), not \(+0.4\ \text{cm}/\text{s}\). The positive number \(0.4\) is the magnitude of the rate of decrease.
Three Checks to Make
At a stated instant, check a related-rate calculation in three connected ways:
- Units: Does every derivative have units of the quantity being differentiated per unit time? Do all terms being added have the same units?
- Sign: Does the sign match the described direction of change? Is the rate a signed derivative, or has the problem asked for a positive magnitude?
- Instant: Were the given values and rates evaluated at the same stated instant, and do they satisfy the original geometric or physical constraint?
A unit check does not prove a calculation is correct: two different quantities can have the same units. A sign check also depends on the physical interpretation and the chosen variables. But either check can expose a mistake before an answer is accepted.
Worked Example: Area Changing as a Circle Shrinks
Worked Example: Area Changing as a Circle Shrinks
At an instant, a circular region has radius \(r=5\ \text{cm}\), and its radius is decreasing at \(0.4\ \text{cm}/\text{s}\). Find and interpret the rate of change of its area at that instant.
First express the stated direction as a signed rate: \(\frac{dr}{dt}=-0.4\ \text{cm}/\text{s}=-\frac{2}{5}\ \text{cm}/\text{s}\). The area is \(A=\pi r^2\). Differentiating with respect to time gives \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\).
Now substitute the radius and radius rate at the same instant: \[ \frac{dA}{dt} =2\pi(5\ \text{cm})\left(-\frac{2}{5}\ \text{cm}/\text{s}\right) =-4\pi\ \text{cm}^2/\text{s}. \] The units agree: radius times radius per time gives area per time. The negative sign means the area is decreasing. Thus the area is decreasing at \(4\pi\ \text{cm}^2/\text{s}\), while its signed rate of change is \(-4\pi\ \text{cm}^2/\text{s}\).
Check: Substituting the signed radius rate into the differentiated area formula gives \(2\pi(5)(-2/5)=-4\pi\), as required. Since \(A=\pi r^2\) increases with positive \(r\), a decreasing radius is consistent with a decreasing area.
Why Differentiating a Constraint Gives Rates
Related rates arise when changing quantities are tied by a constraint. The chain rule explains why differentiating that constraint produces an equation among their rates. Suppose differentiable functions \(x(t)\) and \(y(t)\) satisfy \(F(x(t),y(t))=C\), where \(C\) is constant and \(F\) is differentiable. The chain rule gives the following relation wherever the functions and derivatives are defined:
The last equality holds because the derivative of the constant \(C\) is zero. Each term in the sum must have units matching the rate of change of \(F\): if \(F\) has units \(U\), those terms have units \(U/T\). This is useful as a check, but it does not replace checking whether the variables and rates are physically appropriate.
Write the relationship among the changing quantities, including fixed quantities where relevant.
Apply the chain rule to each time-dependent quantity. The resulting terms should represent rates with compatible units.
Use values and rates that refer to the same instant, and keep their signs.
Check units, substitute the computed rate into the differentiated relation, and explain what its sign means.
Worked Example: A Ladder Sliding Against a Wall
Worked Example: A Ladder Sliding Against a Wall
A \(10\)-foot ladder rests against a vertical wall. Let \(x\) be the distance from the wall to the foot of the ladder and \(y\) the height of its top above the ground. At an instant, \(x=6\ \text{ft}\), \(y=8\ \text{ft}\), and the foot moves away from the wall at \(\frac{dx}{dt}=3\ \text{ft}/\text{s}\). Find and interpret the rate at which the top moves.
The ladder length is constant, so its constraint is \(x^2+y^2=(10\ \text{ft})^2\). Differentiate with respect to time: \(\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.\) Both terms have units \(\text{ft}^2/\text{s}\): a length multiplied by a length-per-time rate.
Substitute the data and solve: \[ 2(6\ \text{ft})(3\ \text{ft}/\text{s}) + 2(8\ \text{ft})\frac{dy}{dt} =0, \] \[ 36\ \text{ft}^2/\text{s} + 16\ \text{ft}\frac{dy}{dt} =0, \] \[ 16\ \text{ft}\frac{dy}{dt}=-36\ \text{ft}^2/\text{s}, \qquad \frac{dy}{dt}=-\frac{36}{16}\ \text{ft}/\text{s} =-\frac{9}{4}\ \text{ft}/\text{s}. \] Dividing \(\text{ft}^2/\text{s}\) by \(\text{ft}\) gives \(\text{ft}/\text{s}\), as required.
The negative sign means that the top is moving down. Its rate of decrease in height is \(9/4\ \text{ft}/\text{s}\). Check: Substitution into the differentiated constraint gives \(36\ \text{ft}^2/\text{s}+16\ \text{ft}(-9/4\ \text{ft}/\text{s})=36-36=0\ \text{ft}^2/\text{s}\). The constraint is satisfied, and the negative vertical rate agrees with the foot moving farther from the wall.
When a Sign Describes Direction, Not Speed
A derivative is not automatically a speed. The derivative records both magnitude and direction relative to the chosen positive direction for the variable. In the ladder example, \(y\) is height measured upward, so a negative \(\frac{dy}{dt}\) means the height decreases. If instead a question asks for the speed of the top moving vertically, it asks for a nonnegative magnitude: \(\left|\frac{dy}{dt}\right|=9/4\ \text{ft}/\text{s}\). The units remain the same, but the speed does not carry the direction encoded by the derivative’s sign.
Likewise, the phrase “rate of decrease” is often used for a positive magnitude. Read the wording carefully. If a quantity is decreasing, its signed derivative is negative; its rate of decrease, when requested as a magnitude, is the positive opposite of that derivative. Keeping these meanings separate prevents an answer from being numerically plausible but semantically wrong.
Worked Example: Checking a Time-Dependent Radius
Worked Example: Checking a Time-Dependent Radius
Near \(t=2\ \text{s}\), suppose a sphere’s radius is modeled by \[ r(t)=4\ \text{cm}-(1\ \text{cm}/\text{s}^3)(t-2\ \text{s})^3. \] Find the sphere’s volume rate of change at \(t=3\ \text{s}\), and check the units and sign.
The model is dimensionally consistent: \((t-2\ \text{s})\) has units of seconds, so \((1\ \text{cm}/\text{s}^3)(t-2\ \text{s})^3\) has units of centimeters. It can therefore be subtracted from \(4\ \text{cm}\). At \(t=3\ \text{s}\), \[ r(3\ \text{s}) =4\ \text{cm}-(1\ \text{cm}/\text{s}^3)(1\ \text{s})^3 =3\ \text{cm}. \]
Differentiate the model: \[ \frac{dr}{dt} =-3(1\ \text{cm}/\text{s}^3)(t-2\ \text{s})^2, \] where the derivative of \(t-2\ \text{s}\) with respect to \(t\) is \(1\). At \(t=3\ \text{s}\), \[ \frac{dr}{dt} =-3(1\ \text{cm}/\text{s}^3)(1\ \text{s})^2 =-3\ \text{cm}/\text{s}. \] The radius is decreasing at that instant.
The sphere’s volume is \(V=\frac{4}{3}\pi r^3\). Differentiate with respect to time: \(\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). Now substitute the values at \(t=3\ \text{s}\): \[ \frac{dV}{dt} =4\pi(3\ \text{cm})^2(-3\ \text{cm}/\text{s}) =-108\pi\ \text{cm}^3/\text{s}. \] The units are \(\text{cm}^2\) times \(\text{cm}/\text{s}\), giving \(\text{cm}^3/\text{s}\). The negative sign means the volume is decreasing.
Check: With \(r=3\ \text{cm}\) and \(\frac{dr}{dt}=-3\ \text{cm}/\text{s}\), the differentiated volume formula gives \(4\pi(9)(-3)=-108\pi\ \text{cm}^3/\text{s}\). Because \(r>0\), a decreasing radius is consistent with a decreasing volume.
A Common Mistake: Losing the Sign or the Units
Suppose the circle in the first example is shrinking at \(0.4\ \text{cm}/\text{s}\). A common incorrect calculation is to insert the positive magnitude as if it were the signed derivative: \[ \frac{dA}{dt} =2\pi(5\ \text{cm})(+0.4\ \text{cm}/\text{s}) =+4\pi\ \text{cm}^2/\text{s}. \] The units are plausible, but the sign says the area is increasing. That contradicts the stated shrinking radius. The correction is to use \(\frac{dr}{dt}=-0.4\ \text{cm}/\text{s}\), giving \(\frac{dA}{dt}=-4\pi\ \text{cm}^2/\text{s}\).
Units can expose a different kind of mistake. The expression \(4\ \text{cm}-(t-2\ \text{s})^3\) is not a valid radius model when \(t\) is measured in seconds: it subtracts a quantity measured in seconds cubed from a length. The correction in the preceding example includes a coefficient \(1\ \text{cm}/\text{s}^3\), so the cubic term is measured in centimeters. Differentiating that corrected model also gives a rate in \(\text{cm}/\text{s}\). A formula should be checked both before and after differentiation.
Check Your Understanding
For each question, keep the signed derivative distinct from any positive magnitude such as speed or rate of decrease.
- A radius measured in centimeters is decreasing at \(0.2\ \text{cm}/\text{s}\). What is the sign and unit of \(\frac{dr}{dt}\)?
- If \(A=\pi r^2\), what units should \(\frac{dA}{dt}\) have when \(r\) is measured in meters and time in seconds?
- A ladder constraint is \(x^2+y^2=L^2\), with fixed \(L\). After differentiating with respect to time, what units should each term have if \(x\) and \(y\) are lengths?
- A calculation gives \(\frac{dy}{dt}=-2\ \text{m}/\text{s}\), where \(y\) is height measured upward. State the direction of motion and the vertical speed.
- Why is \(4\ \text{cm}-(t-2\ \text{s})^3\) dimensionally inconsistent as a radius model, and what units must a coefficient multiplying the cubic term have?