Why Differentiate a Constraint with Respect to Time?
In a related-rates situation, quantities may change while remaining linked by a geometric or physical relationship. For example, the area of a circle depends on its radius, and the legs of a right triangle are linked when its hypotenuse is fixed. The relationship holds as time passes, so its quantities should be treated as functions of time.
If a quantity is called \(x\), write it as \(x(t)\) when it changes with time. Its instantaneous rate is \(x'(t)=\frac{dx}{dt}\). A constraint such as \(x^2+y^2=169\) then means that \(x(t)^2+y(t)^2=169\) throughout the interval being considered.
The result is an equation connecting instantaneous rates. Once that equation has been found, values at a particular instant can be used to calculate an unknown rate.
The Chain Rule for Time-Dependent Quantities
If \(x=x(t)\) is differentiable, then differentiating a power of \(x\) requires the chain rule. In particular,
The factor \(\frac{dx}{dt}\) is essential: the inside quantity \(x(t)\) itself changes with time. More generally, for a differentiable function \(F\),
For a relation involving two time-dependent quantities, differentiate each term with respect to \(t\). For example, if \(x(t)^2+y(t)^2=169\), then
This is implicit differentiation with respect to time: the equation relates \(x\) and \(y\), but neither needs to be solved explicitly as a function of the other before differentiating.
The Product Rule for a Constraint
When two changing quantities are multiplied, use the product rule. If \(x=x(t)\) and \(y=y(t)\) are differentiable, then
Therefore, if their product is constant, \(x(t)y(t)=C\), differentiating gives
Both terms appear because either factor can change. The product is not differentiated by multiplying the two rates, nor by differentiating just one factor.
Why the Differentiation Works
The method follows directly from treating the constraint as an equation of functions. Here is the argument for a right-triangle constraint with constant hypotenuse.
If the legs are \(x(t)\) and \(y(t)\), and the fixed hypotenuse is \(c\), the Pythagorean theorem gives \(x(t)^2+y(t)^2=c^2\).
Because the identity holds as \(t\) varies, its derivative is \(2x(t)x'(t)+2y(t)y'(t)=0\). The chain rule supplies \(x'(t)\) and \(y'(t)\).
If \(y(t)\ne 0\), rearranging gives \(y'(t)=-\frac{x(t)}{y(t)}x'(t)\). The equation follows by differentiating the original identity, not by assuming one leg is a function already solved from the other.
This derivation assumes the legs are differentiable functions of time. The division in the final formula also requires \(y(t)\ne 0\). Keeping these conditions in view helps distinguish a valid rate equation from an expression that cannot be used at a particular point.
Worked Example: Rate of Change of a Circle’s Area
Worked Example: Rate of Change of a Circle’s Area
The radius \(r(t)\) of a circle is \(4\) meters at an instant and is increasing at \(\frac{3}{2}\) meters per minute. Find the instantaneous rate of change of its area at that instant.
Step 1: Write the area as a time-dependent quantity. The circle-area formula is \(A(t)=\pi r(t)^2\).
Step 2: Differentiate with respect to time. By the chain rule,
Step 3: Substitute the values for this instant.
The area is increasing at \(12\pi\) square meters per minute at this instant. To check the rate formula, differentiating \(A(t)=\pi r(t)^2\) again by the chain rule gives \(2\pi r(t)r'(t)\), the same expression used above.
Worked Example: A Fixed Right-Triangle Constraint
Worked Example: Finding a Leg’s Rate of Change
A right triangle has a fixed hypotenuse of \(13\) centimeters. At an instant, one leg has length \(x=5\) centimeters and is increasing at \(3\) centimeters per second. The other leg has length \(y=12\) centimeters. Find \(\frac{dy}{dt}\) at that instant.
Step 1: Write the constraint. The Pythagorean theorem gives \(x(t)^2+y(t)^2=13^2=169\).
Step 2: Differentiate the whole equation with respect to time.
Step 3: Substitute the instantaneous values and solve.
Thus the other leg’s instantaneous rate is \(-\frac{5}{4}\) centimeters per second. Check by substituting into the differentiated constraint: \(2(5)(3)+2(12)(-\frac{5}{4})=30-30=0\). The original constraint also holds at the stated instant, since \(5^2+12^2=25+144=169\).
Worked Example: Differentiating a Product Constraint
Worked Example: Two Quantities with a Fixed Product
Two positive quantities \(x(t)\) and \(y(t)\) satisfy \(x(t)y(t)=48\). At an instant, \(x=6\), \(y=8\), and \(\frac{dx}{dt}=\frac{3}{2}\). Find \(\frac{dy}{dt}\) at that instant.
Step 1: Differentiate using the product rule.
The derivative of the constant \(48\) is \(0\), so \(y\frac{dx}{dt}+x\frac{dy}{dt}=0\).
Step 2: Substitute the given values and solve.
Therefore, \(\frac{dy}{dt}=-2\) in the units of \(y\) per unit of time. Check the differentiated constraint: \(8(\frac{3}{2})+6(-2)=12-12=0\). The stated values also satisfy the original constraint because \(6\cdot 8=48\).
A Common Mistake: Omitting the Rate Factor
When differentiating \(x(t)^2\), it is incorrect to write only \(2x\). That expression is the derivative of \(x^2\) with respect to \(x\), not with respect to time when \(x\) changes with time. The chain rule requires the additional factor \(\frac{dx}{dt}\).
For the triangle in the earlier example, omitting that factor would produce the incorrect equation \(2x+2y\frac{dy}{dt}=0\). Substituting \(x=5\) and \(y=12\) would then give
This does not satisfy the correctly differentiated constraint with \(\frac{dx}{dt}=3\): \(2(5)(3)+2(12)(-\frac{5}{12})=30-10=20\ne 0\). The correction is to differentiate \(x(t)^2\) as \(2x(t)\frac{dx}{dt}\), giving the verified rate \(-\frac{5}{4}\).
Differentiate Before Using Instantaneous Values
A constraint relates quantities throughout their motion, whereas numerical values describe only one instant. Differentiate the relationship while the quantities are still variables; then substitute the values and known rates for the instant in question. Substituting values into the original constraint first may simply produce a numerical identity, which no longer contains the changing quantities needed to find their rates.
The essential sequence is: represent changing quantities as functions of time, differentiate the full constraint using the appropriate rules, and then evaluate the resulting rate equation at the specified instant. The chain rule handles composed quantities such as \(r(t)^2\); the product rule handles products such as \(x(t)y(t)\); and implicit differentiation applies these rules directly to equations linking quantities.
Check Your Understanding
For each question, identify the constraint and the differentiation rule that applies before calculating.
- If \(u(t)^2+v(t)^2=100\), what equation results from differentiating with respect to \(t\)?
- If \(A(t)=\pi r(t)^2\), write \(\frac{dA}{dt}\) in terms of \(r\) and \(\frac{dr}{dt}\).
- If \(p(t)q(t)=36\), what equation relates \(\frac{dp}{dt}\) and \(\frac{dq}{dt}\)?
- For a right triangle with fixed hypotenuse, why does differentiating the squared leg lengths produce rate factors?
- Why should you differentiate a constraint before substituting values for a particular instant?
