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Related Rates Problems · Tutorial 644 of 1495

Isolating the Requested Rate in a Multi-Variable Equation

Use algebra carefully after differentiating a multi-variable constraint to solve for the particular rate the problem asks for.

Intermediate 9 min read

What You'll Learn

  • Identify the coefficient of a requested rate in a differentiated constraint
  • Rearrange a rate equation without losing terms or signs
  • Substitute instantaneous values only after differentiating
  • Check a solved rate by substitution and by units
  • Recognize when a zero coefficient prevents direct isolation

Why isolating the requested rate matters

A related-rates equation can contain several unknown-looking rates even when a problem asks for just one. Differentiating the constraint gives a relationship among those rates; it does not automatically tell you which one to report. The remaining task is algebraic: collect the terms, solve for the requested rate, and then evaluate using the data for the specified instant.

This step deserves care. A missed term or sign can change the result, and dividing by a coefficient that is zero can produce an invalid calculation. The goal here is to make the isolation step systematic and checkable.

Key idea: Differentiate the constraint while the quantities are changing. Then rearrange the resulting equation to isolate the requested rate. Substitute the instant’s data only after the rate equation has been formed.

A general form for the rate equation

Suppose changing quantities \(x\), \(y\), and \(z\) satisfy a constraint \(F(x,y,z)=C\), where \(C\) is constant. Treat each quantity as a function of time. The chain rule gives

$$ \frac{d}{dt}F(x,y,z) = F_x(x,y,z)\frac{dx}{dt} + F_y(x,y,z)\frac{dy}{dt} + F_z(x,y,z)\frac{dz}{dt} =0. $$

Here \(F_x\), \(F_y\), and \(F_z\) are the partial derivatives of \(F\) with respect to its indicated variables. If the requested rate is \(dy/dt\), collect its term and move the other rate terms to the opposite side:

$$ F_y(x,y,z)\frac{dy}{dt} = -F_x(x,y,z)\frac{dx}{dt} -F_z(x,y,z)\frac{dz}{dt}. $$

If \(F_y(x,y,z)\ne 0\) at the instant under consideration, division gives

$$ \frac{dy}{dt} = -\frac{ F_x(x,y,z)\frac{dx}{dt} + F_z(x,y,z)\frac{dz}{dt} }{ F_y(x,y,z) }. $$

This expression is evaluated at the specified instant using the values of the quantities and the rates that are given. The condition \(F_y\ne 0\) matters: if the coefficient is zero, this equation cannot be divided by it to obtain \(dy/dt\). The equation may then impose a condition on the other rates, or it may fail to determine \(dy/dt\) at that instant.

Theorem: For a differentiable constraint \(F(x(t),y(t),z(t))=C\), if \(F_y\ne 0\) at the instant of interest, the differentiated constraint determines \(dy/dt\) by the formula above. This follows from the chain rule and algebraic rearrangement.

A reliable isolation procedure

Before calculating, identify exactly which rate is requested. Then keep every rate in the differentiated equation until you have collected its terms. This prevents an easy-to-miss contribution, especially when the original constraint contains a product or a nonlinear expression.

1
Name the target rate.
Mark the derivative the problem asks you to find, such as \(dy/dt\).
2
Differentiate the full constraint.
Apply the chain rule and product rule wherever required; keep the rates as symbols.
3
Collect and isolate.
Move all terms not containing the target rate to the other side. Divide only if its coefficient is nonzero.
4
Evaluate and check.
Substitute the instant’s data, preserve signs and units, and put the result back into the differentiated equation.

Worked examples

Worked Example: Solving for one rate in a symmetric constraint

Suppose two lengths \(x\) and \(y\), measured in centimeters, satisfy \(x^2+xy+y^2=C\), where \(C\) is constant. At an instant when \(x=2\) cm, \(y=1\) cm, and \(dx/dt=3\) cm/min, find \(dy/dt\).

Differentiate each term with respect to time. The product \(xy\) contributes both \(y\,dx/dt\) and \(x\,dy/dt\):

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

Collect the terms containing \(dy/dt\), and collect the other terms on the right:

$$ (x+2y)\frac{dy}{dt} = -(2x+y)\frac{dx}{dt}. $$

At the given instant, \(x+2y=2+2(1)=4\), which is nonzero, and \(2x+y=2(2)+1=5\). Therefore,

$$ \frac{dy}{dt} = -\frac{5(3)}{4} = -\frac{15}{4}\ \text{cm/min}. $$

Check by substituting into the full differentiated equation. Its four terms are \(2(2)(3)=12\), \(1(3)=3\), \(2(-15/4)=-15/2\), and \(2(1)(-15/4)=-15/2\). Their sum is \(12+3-15/2-15/2=0\), as required. The negative sign means \(y\) is decreasing at this instant.

Worked Example: Isolating a rate when three rates appear

Suppose lengths \(x,y,z\), measured in centimeters, satisfy \(x^2+yz=17\ \text{cm}^2\). At an instant, \(x=3\) cm, \(y=4\) cm, \(z=2\) cm, \(dx/dt=1\) cm/min, and \(dz/dt=-2\) cm/min. Find \(dy/dt\).

The stated values satisfy the constraint because \(3^2+(4)(2)=9+8=17\). Differentiate the constraint:

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

Move the known-rate terms to the right, then divide by \(z\), which is \(2\) cm and therefore nonzero:

$$ z\frac{dy}{dt} = -2x\frac{dx}{dt} -y\frac{dz}{dt}, $$ $$ \frac{dy}{dt} = -\frac{2x\frac{dx}{dt}+y\frac{dz}{dt}}{z} = -\frac{2(3)(1)+(4)(-2)}{2} = -\frac{6-8}{2} = 1\ \text{cm/min}. $$

Check in the differentiated equation: \(2(3)(1)+2(1)+4(-2)=6+2-8=0\). All three terms have units \(\text{cm}^2/\text{min}\). Thus the result satisfies both the equation and its units.

Worked Example: Isolating a rate with a dimensionless factor

Let \(x\) and \(y\) be lengths in centimeters and \(z\) a dimensionless quantity. Suppose \((x+y)z=18\) cm. At an instant, \(x=1\) cm, \(y=2\) cm, \(z=6\), \(dy/dt=2\) cm/min, and \(dz/dt=-1\ \text{min}^{-1}\). Find \(dx/dt\).

The given values satisfy the constraint: \((1+2)(6)=18\). Differentiate the product \((x+y)z\):

$$ z\left(\frac{dx}{dt}+\frac{dy}{dt}\right) + (x+y)\frac{dz}{dt} =0. $$

Collect the target rate and isolate it:

$$ z\frac{dx}{dt} = -z\frac{dy}{dt} -(x+y)\frac{dz}{dt}, $$ $$ \frac{dx}{dt} = -\frac{dy}{dt} -\frac{x+y}{z}\frac{dz}{dt}. $$

Substitute the data:

$$ \frac{dx}{dt} = -2 -\frac{1+2}{6}(-1) = -2+\frac{1}{2} = -\frac{3}{2}\ \text{cm/min}. $$

Check in the differentiated equation: \(6(-3/2+2)+(1+2)(-1)=6(1/2)-3=3-3=0\). Both terms have units \(\text{cm/min}\). The negative result indicates that \(x\) is decreasing.

Why the rearrangement works

After differentiation, the equation is a sum of contributions from all the changing quantities. In the general constraint, the contribution associated with \(y\) is \(F_y\,dy/dt\). The remaining terms account for the changes in the other variables. Moving those terms to the other side leaves the \(y\)-contribution equal to their negative sum. If its coefficient \(F_y\) is nonzero, ordinary division gives one value for \(dy/dt\).

This argument also explains why substituting values too early is risky. The differentiated equation shows which rate has which coefficient before the numbers obscure the structure. Once the target rate is isolated, evaluating at the specified instant is direct substitution into a formula.

A common mistake: losing a product-rule term

In the first example, the term \(xy\) differentiates to \(y\,dx/dt+x\,dy/dt\). Omitting its first contribution gives the incorrect equation

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

Using the example’s values in this incorrect equation gives \(12+4\,dy/dt=0\), so the mistaken calculation yields \(dy/dt=-3\ \text{cm/min}\). But substitution into the correct differentiated equation gives \(15+4(-3)=3\ne 0\). The proposed rate fails the check.

The correction is to differentiate the product completely before collecting terms:

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

Including both terms gives \(15+4\,dy/dt=0\), hence \(dy/dt=-15/4\ \text{cm/min}\), which satisfies the full equation.

When direct isolation is not possible

The coefficient of the requested rate must be checked before dividing. For example, if a differentiated constraint has the form \(a(t)\,dy/dt+b(t)=0\), then at an instant when \(a(t)\ne 0\), it gives \(dy/dt=-b(t)/a(t)\). If \(a(t)=0\), division is invalid. When \(b(t)\ne 0\), the equation is inconsistent with the assumed data; when \(b(t)=0\), it reduces to \(0=0\) and does not determine \(dy/dt\) at that instant. In either case, do not report a rate by dividing by zero.

A final check should be more than a glance at the arithmetic. Substitute the proposed rate back into the differentiated equation and verify that the two sides agree. Also check that the units of every term in that equation match. These checks often expose a missing contribution, a sign error, or an incorrect coefficient.

Check Your Understanding

Answer these questions using the isolation procedure.

  1. If a differentiated constraint is \(3\,dx/dt+2\,dy/dt-5\,dz/dt=0\), isolate \(dy/dt\) in terms of the other two rates.
  2. In a product term \(xz\), what two terms appear when differentiating with respect to time?
  3. Why should the instant’s numerical data generally be substituted after the constraint is differentiated?
  4. What condition is required to divide by the coefficient of the requested rate?
  5. If a proposed rate is \(dy/dt=4\) and substitution makes the differentiated equation read \(7=0\), what does that tell you?

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