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Differentiating Constraints with Respect to Time

Learn to differentiate time-dependent constraints correctly while retaining the rate of change of every variable.

Intermediate 9 min read

What You'll Learn

  • Differentiate an equation that holds between changing quantities over time
  • Apply the Chain Rule to powers of time-dependent variables
  • Apply the product rule when a constraint contains products
  • Keep each variable’s rate symbol in the differentiated equation
  • Recognize and correct the common error of omitting a time derivative

From a Constraint to a Rate Equation

A constraint is an equation that links quantities. If those quantities change over time, the equation links functions of time rather than fixed numbers. Differentiating the equation with respect to time turns the constraint into a relationship between their rates of change.

The essential point is that a changing quantity must be treated as a function of time. For example, if \(x\) depends on time, then \(x^2\) is a composite function: the square of \(x(t)\). The Chain Rule therefore contributes both a power factor and the rate \(dx/dt\).

In this tutorial, \(x\) and \(y\) are shorthand for differentiable functions \(x(t)\) and \(y(t)\). The notation \(dx/dt\) and \(dy/dt\) denotes their rates of change with respect to time. We differentiate the whole constraint with respect to \(t\), applying the usual differentiation rules together with the Chain Rule.

Key idea: If an equation links quantities that vary with time, differentiate both sides with respect to \(t\). A term involving a changing variable must include that variable’s rate of change.

The Rules Needed for Time Differentiation

Suppose \(x=x(t)\) is differentiable. For a positive integer \(n\), the Chain Rule gives

$$ \frac{d}{dt}\bigl(x^n\bigr) = n x^{n-1}\frac{dx}{dt}. $$

The factor \(n x^{n-1}\) comes from differentiating the power with respect to \(x\); the factor \(dx/dt\) comes from the dependence of \(x\) on time. In particular, \(d(x^2)/dt=2x\,dx/dt\), not just \(2x\).

If both \(x\) and \(y\) depend on time, the product rule gives

$$ \frac{d}{dt}(xy) = x\frac{dy}{dt} + y\frac{dx}{dt}. $$

Both terms are needed: either factor can change, so the product’s rate includes the contribution from the change in each factor. A constant \(C\), by contrast, has derivative \(dC/dt=0\).

Rule: Differentiate each term of the constraint with respect to time. Use the Chain Rule for powers of time-dependent variables, the product rule for products of such variables, and zero for a constant.

Why Differentiating the Constraint Works

A constraint between changing quantities is an identity over the time interval in which it holds. If two differentiable functions are equal throughout an interval, their derivatives are equal at each interior time in that interval. Thus, differentiating both sides preserves the equality and produces a necessary relationship between the rates.

The following derivation explains how the Chain Rule enters. Let \(x=x(t)\) be differentiable and let \(n\) be a positive integer. Define \(g(u)=u^n\), so that \(g'(u)=nu^{n-1}\). Since \(x^n=g(x(t))\), the Chain Rule gives

$$ \frac{d}{dt}\bigl(x(t)^n\bigr) = g'\bigl(x(t)\bigr)\frac{dx}{dt} = n x(t)^{n-1}\frac{dx}{dt}. $$

This derivation shows why the rate \(dx/dt\) cannot be omitted: the power is applied to \(x(t)\), not to a fixed number. For a product, the product rule likewise treats both factors as functions of time.

1
Interpret the variables as functions.
Read \(x\) and \(y\) as \(x(t)\) and \(y(t)\), even when the argument \(t\) is not written every time.
2
Differentiate each side with respect to time.
Use the Chain Rule, product rule, sum rule, and the derivative of a constant as appropriate.
3
Retain the rate symbols.
Write \(dx/dt\), \(dy/dt\), or the corresponding rate for every changing quantity that was differentiated.

Worked Examples

Worked Example: A Constraint with Two Powers

Suppose differentiable quantities \(x(t)\) and \(y(t)\) satisfy \(x^3+y^2=20\) throughout an interval. Differentiate the constraint with respect to time.

Differentiate the left side term by term. For the first term, apply the Chain Rule to \(x^3\):

$$ \frac{d}{dt}(x^3) = 3x^{3-1}\frac{dx}{dt} = 3x^2\frac{dx}{dt}. $$

For the second term, apply it to \(y^2\):

$$ \frac{d}{dt}(y^2) = 2y^{2-1}\frac{dy}{dt} = 2y\frac{dy}{dt}. $$

The right side is constant, so its derivative is zero. Therefore the differentiated constraint is

$$ 3x^2\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. $$

Check: Differentiating the first term gives \(3x^2\,dx/dt\), differentiating the second gives \(2y\,dy/dt\), and differentiating \(20\) gives \(0\). These are exactly the terms in the resulting equation.

Worked Example: A Product and a Power

Suppose \(x(t)\) and \(y(t)\) satisfy \(xy+y^2=18\). Differentiate every term with respect to time.

For the product \(xy\), use the product rule. The derivative of the first factor is \(dx/dt\), and the derivative of the second is \(dy/dt\):

$$ \frac{d}{dt}(xy) = \frac{dx}{dt}\,y + x\frac{dy}{dt} = y\frac{dx}{dt} + x\frac{dy}{dt}. $$

For the square \(y^2\), use the Chain Rule, and differentiate the constant on the right:

$$ \frac{d}{dt}(y^2) = 2y\frac{dy}{dt}, \qquad \frac{d}{dt}(18)=0. $$

Combining these derivatives gives

$$ y\frac{dx}{dt} + x\frac{dy}{dt} + 2y\frac{dy}{dt} = 0. $$

Check: The product contributes two terms, \(y\,dx/dt\) and \(x\,dy/dt\). The square contributes \(2y\,dy/dt\), and the constant contributes zero. Each original term has been differentiated once, with both product contributions present.

Worked Example: A Product of a Power and a Variable

Suppose \(x(t)\) and \(y(t)\) satisfy \(x^2y+y^3=30\). The first term is a product, so differentiate it using the product rule. Its first factor is \(x^2\), whose time derivative is \(2x\,dx/dt\); its second factor is \(y\), whose derivative is \(dy/dt\). Hence

$$ \begin{aligned} \frac{d}{dt}(x^2y) &= \frac{d}{dt}(x^2)\,y + x^2\frac{dy}{dt}\\ &= \left(2x\frac{dx}{dt}\right)y + x^2\frac{dy}{dt}\\ &= 2xy\frac{dx}{dt} + x^2\frac{dy}{dt}. \end{aligned} $$

For the other variable power, the Chain Rule gives

$$ \frac{d}{dt}(y^3) = 3y^2\frac{dy}{dt}. $$

Since the derivative of \(30\) is zero, the differentiated constraint is

$$ 2xy\frac{dx}{dt} + x^2\frac{dy}{dt} + 3y^2\frac{dy}{dt} = 0. $$

Check: Applying the product rule to \(x^2y\) gives \(2xy\,dx/dt+x^2\,dy/dt\). Applying the Chain Rule to \(y^3\) gives \(3y^2\,dy/dt\), and the constant gives zero. The displayed equation contains each of these contributions.

A Common Error: Leaving Out the Rate

A frequent mistake is to differentiate a power as if its base were a fixed number. For example, this calculation is wrong when \(x=x(t)\):

$$ \frac{d}{dt}(x^2)=2x. $$

The expression \(2x\) is the derivative of \(x^2\) with respect to \(x\), not with respect to time. Since \(x\) itself changes with time, the Chain Rule is required. The correct calculation is

$$ \frac{d}{dt}\bigl(x(t)^2\bigr) = 2x(t)\frac{dx}{dt}. $$

The same issue occurs in a product. Writing \(d(xy)/dt=xy'\) or keeping only one product-rule term misses the other factor’s contribution. The full derivative is \(y\,dx/dt+x\,dy/dt\). A useful check is to ask whether every changing factor has had its rate accounted for.

Reading the Result as a Rate Relationship

After differentiation, the resulting equation is a constraint on the rates and the current variable values. It does not claim that each rate is separately zero. For instance, an equation with a right side of zero can hold because different terms balance one another.

It is also important to distinguish the original constraint from its differentiated form. The original equation describes which values of the quantities are linked. The differentiated equation describes how their rates must be related while that constraint continues to hold. Both statements come from the same time-dependent relationship, but they convey different information.

Takeaway: Differentiate a time-dependent constraint as an identity in \(t\). Apply the Chain Rule to each power, the product rule to each product, and preserve a rate symbol for every changing quantity.

Check Your Understanding

For each question, treat every variable shown as a differentiable function of time.

  1. Differentiate \(a^4+b^2=9\) with respect to time.
  2. What two terms does the product rule produce when differentiating \(pq\) with respect to time?
  3. Differentiate \(r^2s+s^3=12\) with respect to time.
  4. Explain why \(d(u^5)/dt\) includes \(du/dt\) when \(u\) depends on time.
  5. Identify the error in the claim \(d(v^3)/dt=3v^2\) for a time-dependent \(v\), and write the correct derivative.

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