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Related Rates Models · Tutorial 298 of 999

Choosing a Constraint for Multi-Step Problems

Identify the constraints linking the quantities in a changing system, differentiate them in a useful order, and check the resulting rate with the original relationships.

Intermediate 9 min read

What You'll Learn

  • Distinguish the constraint describing the geometry from the equation defining the requested quantity.
  • Choose and combine equations that link a known motion rate to an unknown rate.
  • Differentiate time-dependent constraints using the chain rule.
  • Find a volume rate from a changing dimension subject to a geometric restriction.
  • Find an angular rate from the motion of an object along a path.
  • Check signs, units, and results against the original constraints.

Why One Equation May Not Be Enough

A related-rates problem may describe a physical motion but ask about a different quantity. A slider may move horizontally while the question asks how fast a box’s volume is changing. A person may walk along a path while the question asks how fast an observer’s angle of sight is changing. In these problems, the rate you are given is not necessarily the rate you need.

The key is to choose equations that connect the quantities in stages. One equation describes a geometric or physical constraint; another defines the requested quantity. Differentiating those relationships links the known rate to the unknown one.

Definition: A constraint is an equation that must remain true as a system changes. In a multi-step related-rates problem, use constraints to connect the changing quantity whose rate is known to the quantity whose rate is requested.

For example, a rectangle might have a fixed perimeter, while its area and the volume of a box built from it change. The perimeter equation links the two changing side lengths; the volume equation links those side lengths to the requested volume. The equations work together, but they play different roles.

A Reliable Order for Choosing Equations

Start by naming the quantities that change as functions of time. Then identify what the question asks for and what rate is given. Next, write the equations that connect those quantities. Usually, it is helpful to differentiate the constraint first to find an intermediate rate, then use the equation for the target quantity.

1
Identify the target and the given rate.
Write down the requested derivative and the rate supplied in the problem, including their units.
2
List the changing quantities.
Assign a variable to each quantity that changes with time. Treat fixed measurements as constants.
3
Write the constraints.
Choose relationships that connect the changing quantities, and separately write the equation for the requested quantity.
4
Differentiate and connect the rates.
Differentiate with respect to time, then substitute the given rate and the measurements at the specified instant.
5
Interpret and check.
Include units, interpret the sign, and check that the result is consistent with the original relationships.

In general, do not substitute the instant’s measurements into a relationship before differentiating if doing so would turn changing quantities into constants. First differentiate the relationship in its variable form; then use the values that apply at the instant.

Why the Constraints Link the Rates

Suppose two changing quantities \(x(t)\) and \(y(t)\) obey a constraint \(F(x,y)=0\). Because the constraint remains true as time passes, differentiating both sides with respect to time gives

$$ \frac{d}{dt}F(x(t),y(t)) = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} =0. $$

This is the chain rule: the change in \(F\) comes from the changes in both of its time-dependent inputs. If the coefficient of \(dy/dt\) is nonzero, the equation can be solved for \(dy/dt\). Once that intermediate rate is known, a second relationship can connect it to the target rate.

For instance, if a target quantity \(Q\) is given by \(Q=G(x,y)\), then

$$ \frac{dQ}{dt} = \frac{\partial G}{\partial x}\frac{dx}{dt} + \frac{\partial G}{\partial y}\frac{dy}{dt}. $$

Substituting the rate found from the constraint gives \(dQ/dt\) in terms of the available information. The method works because each equation supplies a relationship that remains true throughout the motion, not merely at one instant.

Important condition: Differentiate equations that describe the changing system over the relevant interval. A measurement that is fixed may be treated as a constant; a measurement that changes must remain a variable until after differentiation.

Worked Example: A Changing Box with Fixed Perimeter

Worked Example: A Changing Box with Fixed Perimeter

A rectangular base has a fixed perimeter of \(40\) feet. Its length is increasing at \(1\) foot per minute. At the instant when the length is \(12\) feet, the box’s fixed height is \(3\) feet. How fast is the box’s volume changing?

Step 1: Name the quantities and identify the target. Let \(l\) and \(w\) be the base’s length and width, in feet. The given rate is \(dl/dt=1\) ft/min, and the requested rate is \(dV/dt\).

Step 2: Use the perimeter constraint to find the width rate. The perimeter remains \(40\) feet, so

$$ 2l+2w=40. $$

Differentiate with respect to time:

$$ 2\frac{dl}{dt}+2\frac{dw}{dt}=0. $$

Divide by \(2\) and substitute \(dl/dt=1\) ft/min:

$$ \frac{dw}{dt}=-\frac{dl}{dt}=-1\text{ ft/min}. $$

At the instant in question, the constraint gives \(2(12)+2w=40\), so \(24+2w=40\), \(2w=16\), and \(w=8\) feet. The width is decreasing while the length increases.

Step 3: Use the volume equation. The height \(h=3\) feet is constant, so \(V=lwh\). Differentiate:

$$ \frac{dV}{dt} = h\left(w\frac{dl}{dt}+l\frac{dw}{dt}\right). $$

Substitute \(h=3\), \(w=8\), \(dl/dt=1\), \(l=12\), and \(dw/dt=-1\):

$$ \frac{dV}{dt} = 3\left(8(1)+12(-1)\right) = 3(8-12) = -12\text{ ft}^3/\text{min}. $$

Check: Eliminate \(w\) using \(w=20-l\). Then \(V=3l(20-l)=60l-3l^2\), so \(dV/dt=(60-6l)(dl/dt)\). At \(l=12\) and \(dl/dt=1\), this is \((60-72)(1)=-12\text{ ft}^3/\text{min}\), matching the result. The negative sign means the volume is decreasing at that instant.

Worked Example: A Box with a Fixed Diagonal

Worked Example: A Box with a Fixed Diagonal

A rectangular base has a fixed diagonal of \(10\) centimeters. Its sides are \(x\) and \(y\), and the box has a fixed height of \(2\) centimeters. At the instant when \(x=6\) cm and \(y=8\) cm, \(x\) is increasing at \(1\) cm/s. Find the rate of change of the box’s volume.

Step 1: Use the diagonal constraint. By the Pythagorean theorem, \(x^2+y^2=100\). Differentiate with respect to time:

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

Divide by \(2y\) and substitute the given values:

$$ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{6}{8}(1) = -\frac{3}{4}\text{ cm/s}. $$

Step 2: Differentiate the volume equation. Since the height is fixed, \(V=2xy\). The product rule gives

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

Substitute the instantaneous values and both side rates:

$$ \frac{dV}{dt} = 2\left(8(1)+6\left(-\frac{3}{4}\right)\right) = 2\left(8-\frac{18}{4}\right) = 2\left(\frac{7}{2}\right) = 7\text{ cm}^3/\text{s}. $$

Check: From the diagonal constraint, \(y=\sqrt{100-x^2}\), so \(V=2x\sqrt{100-x^2}\). Differentiating with respect to time gives \(dV/dt=2(y-x^2/y)(dx/dt)\). At \(x=6\), \(y=8\), and \(dx/dt=1\), this is \(2(8-36/8)=2(7/2)=7\text{ cm}^3/\text{s}\), as before. The volume is increasing even though \(y\) is decreasing.

Worked Example: From Walking Speed to Angular Rate

Worked Example: From Walking Speed to Angular Rate

An observer stands \(5\) meters from a straight path. A person walks along the path, moving away from the point nearest the observer at \(4\) meters per second. At the instant the person is \(12\) meters from that nearest point, how fast is the angle of sight changing? Measure \(\theta\) from the line perpendicular to the path.

Step 1: Choose the geometric constraint. The observer, the nearest point on the path, and the person form a right triangle. Its fixed perpendicular side is \(5\) meters, and its changing path distance is \(x\). Thus, \(\tan\theta=x/5\), or \(\theta=\arctan(x/5)\).

Step 2: Differentiate to connect the motion to the angle. The chain rule gives

$$ \frac{d\theta}{dt} = \frac{1}{1+(x/5)^2}\cdot\frac{1}{5}\frac{dx}{dt} = \frac{5}{25+x^2}\frac{dx}{dt}. $$

Step 3: Substitute the instant’s values. With \(x=12\) m and \(dx/dt=4\) m/s,

$$ \frac{d\theta}{dt} = \frac{5}{25+144}(4) = \frac{20}{169}\text{ rad/s}. $$

Check: At this instant, \(\tan\theta=12/5\), so \(\sec^2\theta=1+(12/5)^2=169/25\). Differentiating \(\tan\theta=x/5\) gives \(\sec^2\theta(d\theta/dt)=(1/5)(dx/dt)=4/5\). Therefore \(d\theta/dt=(4/5)(25/169)=20/169\text{ rad/s}\), matching the result. The angle increases as the person moves farther along the path.

A Common Mistake: Ignoring a Changing Dimension

In the fixed-perimeter box example, a tempting but incorrect calculation is \(dV/dt=wh(dl/dt)=8(3)(1)=24\text{ ft}^3/\text{min}\). That calculation treats the width as constant while the length changes. But the perimeter constraint requires \(dw/dt=-1\text{ ft/min}\); the width is not fixed.

The product rule includes the effect of both changing sides:

$$ \frac{dV}{dt} = h\left(w\frac{dl}{dt}+l\frac{dw}{dt}\right) = 3\left(8(1)+12(-1)\right) = -12\text{ ft}^3/\text{min}. $$

The incorrect answer misses the constraint that links the dimensions. A useful check is to ask: Does every changing quantity in the target equation have its rate accounted for?

Choosing Constraints That Do Useful Work

A problem may contain several facts, but not every fact is needed to find the requested rate. Select equations that form a usable chain from the known motion to the target. Keep fixed quantities distinct from changing ones, and check that each intermediate rate has a clear role.

  • For a target volume, look for a geometric constraint linking its changing dimensions and a separate formula for volume.
  • For a target angle, look for a right-triangle or other geometric relationship connecting the angle to a changing distance.
  • Differentiate before substituting instantaneous values when the variables are changing.
  • Track units through each step; for example, a volume rate has units of cubic units per time, while an angular rate is measured in radians per time.
  • Interpret a negative sign as a decrease in the chosen positive direction, not automatically as an error.

Check Your Understanding

For each question, identify the constraint needed to connect the given motion to the requested rate.

  1. A rectangle has fixed perimeter, and its length is changing. What equation relates the width rate to the length rate?
  2. A rectangle’s diagonal is fixed while one side increases. What geometric equation should be differentiated to find the other side’s rate?
  3. A person moves along a straight path at a known speed, and an observer’s distance from the path is fixed. What relationship connects the path distance to the observer’s angle?
  4. Why should you usually differentiate a relationship before substituting the measurements at a particular instant?
  5. In a volume formula containing two changing side lengths and a fixed height, what rate terms must appear when differentiating the volume?

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