Related Rates in Four Different Settings
A related-rates problem asks how quickly one quantity changes when another quantity changes. The quantities may describe positions, angles, lengths of rope, or the dimensions of a volume. The common mathematical structure is that the quantities are connected by a constraint, and that constraint remains true as time passes.
This workshop brings together four settings: two objects moving on perpendicular roads, a rotating beam, a pulley with a fixed rope length, and water filling a cylindrical tank. In every case, the essential sequence is the same: define the changing quantities, write the constraint, differentiate with respect to time, then substitute the values for the instant of interest.
Why Differentiating the Constraint Works
Suppose a geometric relationship can be written as \(F(x,y)=0\), where \(x\) and \(y\) both vary with time. The constraint says that the pair \((x(t),y(t))\) continues to lie on the same relationship. Differentiating both sides with respect to time gives
The chain rule accounts for the fact that \(x\) and \(y\) change as time changes. For example, if \(x^2+y^2=z^2\), then
Each derivative appears because its corresponding quantity depends on time. This is the reason substitution comes later: differentiating first preserves the dependence of the quantities on time. The calculations below apply that idea in several geometries.
Worked Example: Two Cars on Perpendicular Roads
Worked Example: Two Cars on Perpendicular Roads
One car travels east, away from an intersection, at \(40\) km/h. A second car travels north, away from the same intersection, at \(30\) km/h. At a particular instant, the eastbound car is \(3\) km from the intersection and the northbound car is \(4\) km from it. How quickly is the distance between the cars changing at that instant?
Let \(x\) and \(y\) be the cars’ respective distances from the intersection, and let \(z\) be the distance between them. The roads are perpendicular, so the Pythagorean relationship is
Both cars are moving away from the intersection, so \(dx/dt=40\) km/h and \(dy/dt=30\) km/h. At the instant in question, \(x=3\) km and \(y=4\) km. First find the separation:
Differentiate the constraint with respect to time:
Solve for the requested rate:
Now substitute the instant’s values and rates:
The distance between the cars is increasing at \(48\) km/h. Check by putting the result into the differentiated constraint: the left side is \(2(3)(40)+2(4)(30)=240+240=480\), and the right side is \(2(5)(48)=480\). The two sides agree, and the positive sign matches the fact that both cars are moving away from the intersection.
Worked Example: A Rotating Beam
Worked Example: A Rotating Beam
A beam rotates about a fixed point on the ground. A vertical wall is \(12\) m from that point. The beam’s angle \(\theta\), measured from the ground, is increasing at \(\pi/6\) radians per second. How quickly is the point where the beam meets the wall moving vertically when \(\theta=\pi/4\)?
Let \(h\) be the height where the beam meets the wall. The right triangle has horizontal side \(12\) m and angle \(\theta\), so the tangent relationship gives
Differentiate with respect to time. The chain rule gives a factor of \(d\theta/dt\):
At \(\theta=\pi/4\), \(\sec^2(\pi/4)=2\), and the angular rate is \(\pi/6\) radians per second. Therefore,
The point on the wall is moving upward at \(4\pi\) m/s at that instant. To check, substitute the result into the differentiated constraint: \(12\sec^2(\pi/4)(\pi/6)=12(2)(\pi/6)=4\pi\) m/s, as calculated. This relationship uses radians for the angular rate; the derivative of \(\tan\theta\) is applied with \(\theta\) measured in radians.
Worked Example: A Pulley with a Fixed Rope
Worked Example: A Pulley with a Fixed Rope
A rope passes over a pulley \(5\) m above level ground. One end is attached to a cart moving along the ground, while the other end hangs vertically and supports a load. At an instant, the cart is \(12\) m horizontally from the point directly below the pulley, and the hanging part of the rope is \(7\) m long. The cart is moving toward that point at \(3\) m/s. How quickly is the hanging length changing?
Let \(x\) be the cart’s horizontal distance from the point below the pulley, and let \(y\) be the vertical hanging length. The slanted rope segment has length \(\sqrt{x^2+5^2}\). Since the total rope length \(L\) is fixed, the constraint is
The cart moves toward the point below the pulley, so \(dx/dt=-3\) m/s. Differentiate the fixed-length constraint:
The given position is consistent with the rope geometry: the slanted segment is \(\sqrt{12^2+25}=\sqrt{169}=13\) m, and \(13+7=20\) m. Substitute \(x=12\) m and \(dx/dt=-3\) m/s into the differentiated constraint:
Thus,
The hanging length is increasing at \(36/13\) m/s, so the load is moving downward at that speed. Check by substituting into the differentiated constraint: \((12/13)(-3)+36/13=-36/13+36/13=0\), as required for a fixed rope length. The positive sign for \(dy/dt\) is also sensible: as the cart moves closer to the pulley, the slanted segment shortens, so the hanging segment must lengthen.
Worked Example: Filling a Cylindrical Tank
Worked Example: Filling a Cylindrical Tank
Water flows into a vertical cylindrical tank of radius \(2\) m at a constant rate of \(3\) m\(^3\)/min. How quickly is the water depth rising when the depth is \(1.5\) m? Assume the tank is not overflowing.
Let \(h\) be the water depth and \(V\) the volume of water. For a cylinder with fixed radius \(r=2\) m, the volume constraint is
Differentiate with respect to time. The radius is constant, while both \(V\) and \(h\) vary:
The inflow rate is \(dV/dt=3\) m\(^3\)/min, so
The water depth is rising at \(3/(4\pi)\) m/min. Check by substituting into the differentiated volume equation: \(4\pi(3/(4\pi))=3\) m\(^3\)/min, which matches the given inflow. The depth \(1.5\) m does not appear in the final rate because a cylinder with fixed vertical sides has the same horizontal cross-sectional area at every depth.
A Common Mistake: Forgetting the Rates
A frequent error is to differentiate a geometric constraint as if the changing lengths were constants. For the perpendicular-road problem, an incorrect step would be
This drops the rates \(dx/dt\) and \(dy/dt\). The chain rule requires
The incorrect line would give \(2(3)+2(4)=2(5)(dz/dt)\), or \(dz/dt=7/5\) km/h. That value fails the actual differentiated constraint: its right side would be \(2(5)(7/5)=14\), while the correct left side is \(480\). The rates are essential parts of the chain rule, not optional factors.
A Reliable Workflow
These problems use different physical stories, but a consistent workflow helps keep the reasoning organized.
State what each variable measures and its units. Choose positive directions so that the signs of the rates have a clear meaning.
Translate the fixed geometry or physical condition into an equation, such as a right-triangle relationship, rope-length equation, or volume formula.
Use the chain rule for every variable that changes. Keep any fixed quantities, such as a radius or rope length, constant.
Insert positions, angles, and known rates only after differentiation, then solve for the requested rate.
Substitute it into the differentiated equation, verify units, and ask whether its sign and magnitude agree with the physical situation.
The derivative equation is the central check: it must balance exactly after the rate is found. A sign check provides a second kind of verification. For example, two cars moving away from an intersection should become farther apart, while a cart moving toward a pulley can cause the hanging rope segment to grow.
Check Your Understanding
For each question, identify the constraint before deciding how to differentiate.
- Two objects move on perpendicular paths. What distance relationship connects their distances from the intersection to their separation?
- For a point where a rotating beam meets a wall, what relationship connects the beam’s angle to the point’s height?
- In the pulley example, why is the cart’s horizontal rate negative when it moves toward the point below the pulley?
- For a vertical cylindrical tank of fixed radius, which quantity in the volume formula changes with time?
- Why must the rates of changing variables be included when differentiating a constraint?
