Geometric Related-Rates Problem-Solving Workshop

Turning a Geometric Situation into Rates

In a related-rates problem, several quantities change with time and geometry links them together. The central task is not to differentiate a formula chosen at random. It is to describe the situation with variables, write a relationship that is true throughout the motion, and differentiate that relationship with respect to time.

This method applies to a point moving under a geometric constraint, to dimensions of a changing triangle, and to quantities such as the depth and volume of liquid in a container. The answer is a rate at a particular instant, so the values at that instant must be used after the relationship has been differentiated.

Definition: A related rate is a time derivative of a quantity that changes as part of a system of quantities connected by a relationship. If \(x=x(t)\), its rate is \(dx/dt\). A geometric constraint connects the changing variables, allowing one rate to be found from others.

A Reliable Problem-Solving Process

Before calculating, distinguish between quantities that change and values that are merely given at one instant. Use a variable for each changing quantity, and include units in the setup. Then express the geometry as an equation valid during the motion.

Define the variables.
Name each changing length, area, volume, or position, and specify its unit.

Write the geometric constraint.
Use the relevant relation, such as the Pythagorean theorem, a similarity ratio, or a volume formula.

Differentiate with respect to time.
Apply the chain rule to every quantity that depends on time.

Substitute the instant’s data and solve.
Only now insert the given lengths and rates, then interpret the sign and units.

Check the result.
Confirm it satisfies the differentiated constraint and has the correct units and direction.

Why Differentiating the Constraint Works

Suppose changing quantities \(x(t)\) and \(y(t)\) satisfy a geometric equation \(F(x(t),y(t))=0\) throughout an interval of time. Since the left side is identically zero as time passes, its derivative is zero as well. Applying the chain rule gives the rate equation.

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

This equation explains why the rates cannot be chosen independently: they must preserve the constraint. If the equation can be solved for the unknown rate at the instant, the known rates and dimensions determine it. Substitution should happen after differentiation because the dimensions may vary before and after that instant.

Key rule: Differentiate the relationship between the changing quantities with respect to time, using the chain rule. Then evaluate the resulting rate equation at the specified instant.

Worked Example: A Ladder and Its Moving Endpoints

Worked Example: A Ladder Sliding Along a Wall

A \(10\)-meter ladder rests with one end on the ground and the other against a vertical wall. At a particular instant, the foot is \(6\) meters from the wall and is moving away from it at \(2/3\) meter per second. How fast is the top moving?

Let \(x(t)\) be the foot’s horizontal distance from the wall and \(y(t)\) the top’s height, both measured in meters. The ladder’s length is fixed, so the Pythagorean theorem gives \(x^2+y^2=100\). At the instant of interest, \(x=6\), and the positive height is \(y=\sqrt{100-36}=8\). Also, \(dx/dt=2/3\) meter per second.

Differentiate the constraint with respect to time:

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

Substitute \(x=6\), \(y=8\), and \(dx/dt=2/3\):

2(6)\left(\frac{2}{3}\right)+2(8)\frac{dy}{dt}=0, \qquad 8+16\frac{dy}{dt}=0, \qquad \frac{dy}{dt}=-\frac{1}{2}.

The top is moving downward at \(1/2\) meter per second. The negative sign indicates that its height is decreasing, not that the speed is negative.

Check: At this instant, the derivative of the fixed-length constraint is \(2(6)(2/3)+2(8)(-1/2)=8-8=0\), as required. Both terms have units of square meters per second.

Worked Example: A Triangle with Changing Legs

Worked Example: Area and Hypotenuse Rates

A right triangle has legs \(x\) and \(y\), measured in meters. At an instant, \(x=6\) meters and \(y=8\) meters. The first leg is increasing at \(0.5\) meter per second while the second is decreasing at \(0.2\) meter per second. Find the rates of change of the triangle’s area and hypotenuse at that instant.

Let \(A\) denote the area in square meters and \(z\) the hypotenuse in meters. The defining geometric relationships are \(A=xy/2\) and \(z^2=x^2+y^2\). Here, \(dx/dt=0.5\) meter per second and \(dy/dt=-0.2\) meter per second.

Differentiate the area relationship:

\frac{dA}{dt} = \frac{1}{2}\left(y\frac{dx}{dt}+x\frac{dy}{dt}\right).

Substituting the instant’s values gives

\frac{dA}{dt} = \frac{1}{2}\left(8(0.5)+6(-0.2)\right) = \frac{1}{2}(4-1.2) = 1.4\ \text{m}^2/\text{s}.

Now \(z=\sqrt{6^2+8^2}=10\) meters. Differentiate the Pythagorean relationship and solve for \(dz/dt\):

2z\frac{dz}{dt} = 2x\frac{dx}{dt}+2y\frac{dy}{dt}, \qquad \frac{dz}{dt} = \frac{x(dx/dt)+y(dy/dt)}{z}.

Therefore,

\frac{dz}{dt} = \frac{6(0.5)+8(-0.2)}{10} = \frac{3-1.6}{10} = 0.14\ \text{m}/\text{s}.

The area is increasing at \(1.4\) square meters per second, and the hypotenuse is increasing at \(0.14\) meter per second. The decreasing leg does not force either result to be negative; the combined contributions determine each rate.

Check: Substituting the hypotenuse rate into the differentiated constraint gives \(2(10)(0.14)=2(6)(0.5)+2(8)(-0.2)=2.8\) square meters per second. The area formula gives \(dA/dt\) in square meters per second, as expected.

Worked Example: Filling a Conical Container

Worked Example: Liquid Depth in a Cone

A conical container is \(8\) meters high and has a top radius of \(6\) meters. Water is poured in at \(3\pi\) cubic meters per minute. How fast is the water depth increasing when the water is \(4\) meters deep?

Let \(h(t)\) be the water depth in meters and \(r(t)\) the radius of the water’s circular surface in meters. Similar triangles give \(r/h=6/8=3/4\), so \(r=3h/4\). The water’s volume is the volume of a cone, \(V=\pi r^2h/3\). Use the similarity relation to express volume using just \(h\):

V = \frac{\pi}{3}\left(\frac{3h}{4}\right)^2h = \frac{3\pi}{16}h^3.

Differentiate with respect to time:

\frac{dV}{dt} = \frac{9\pi}{16}h^2\frac{dh}{dt}.

At \(h=4\) meters, with \(dV/dt=3\pi\) cubic meters per minute,

3\pi = \frac{9\pi}{16}(4)^2\frac{dh}{dt} = 9\pi\frac{dh}{dt}, \qquad \frac{dh}{dt} = \frac{1}{3}\ \text{m}/\text{min}.

The depth is increasing at \(1/3\) meter per minute. The positive sign matches the fact that water is being added.

Check: At this instant, \(r=3\) meters and \(dr/dt=(3/4)(dh/dt)=1/4\) meter per minute. Differentiating the original volume formula gives

\frac{dV}{dt} = \frac{\pi}{3}\left(2r\frac{dr}{dt}h+r^2\frac{dh}{dt}\right) = \frac{\pi}{3}\left(2(3)\left(\frac{1}{4}\right)(4)+9\left(\frac{1}{3}\right)\right) = 3\pi\ \text{m}^3/\text{min}.

This agrees with the stated filling rate. The units also agree: volume per unit time.

Common Pitfall: Substituting Before Differentiating

In the ladder example, the relation is \(x^2+y^2=100\). A tempting but incorrect approach is to replace \(x\) and \(y\) with their instant values first, obtaining \(6^2+8^2=100\), and then to differentiate that numerical equality. This gives \(0=0\), which contains no information about either rate.

The correction is to preserve the time-dependent quantities until after differentiation:

x(t)^2+y(t)^2=100 \quad\Longrightarrow\quad 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.

Only then should the instant’s values \(x=6\) and \(y=8\) be substituted. Another frequent error is to omit a chain-rule factor: \(d(x^2)/dt\) is \(2x(dx/dt)\), not merely \(2x\). Writing units beside rates helps catch related mistakes: the unknown \(dy/dt\) in the ladder problem must have units of meters per second.

Interpreting the Result

A related rate describes instantaneous change, not the total change over a time interval. Its sign gives direction: a positive rate means the chosen quantity increases, and a negative rate means it decreases. Its units are the units of the quantity divided by time, such as meters per second or cubic meters per minute.

Before accepting a result, check that the chosen constraint describes the geometry throughout the motion, that every changing quantity was differentiated, and that the answer satisfies the resulting rate equation. When the geometry provides multiple relationships, as in a changing triangle, each can offer an independent check.

Check Your Understanding

For each question, define the changing quantities and identify the constraint or formula you would differentiate.

A \(13\)-meter ladder has its foot \(5\) meters from a wall and moving away at \(1\) meter per second. What additional geometric value is needed before finding the top’s rate?
For a right triangle with changing legs \(x\) and \(y\), write a time-derivative formula for its area.
Why should the instant’s numerical side lengths usually be substituted after differentiating?
If a container’s volume is measured in cubic meters and time in minutes, what units should \(dV/dt\) have?
In the conical-container example, why can the radius not be treated as constant while the water depth changes?