Solving Related Rates Problems
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.1: “Related Rates” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Which equation relates the quantities
Two cars leave an intersection at the same time. Car A travels north at 60 km/h; Car B travels east at 80 km/h.
Before computing: sketch the situation. Label the distance of Car A from the intersection as $y$, Car B’s distance as $x$, and the distance between the cars as $z$.
- What equation relates $x$, $y$, and $z$?
- Predict: after 1 hour, what is the rate at which the distance between the cars is growing?
The first question sets up the problem. The second requires differentiating that equation.
Naming the quantities
Every related-rates problem involves multiple quantities changing simultaneously. The approach:
- Name the quantities: what is changing?
- Find the equation that ties them together: it is usually geometry, volume, or a physical law.
- Differentiate with respect to time: the chain rule connects the rates.
- Read off the answer: identify units, check the sign.
The hardest step is usually step 2. Identifying the equation that ties the quantities together takes the most practice.
Prerequisite Check
Quick Reference
Five-step strategy (use on every problem):
- Draw and label: picture the situation; name every changing quantity as a function of time.
- Identify the equation: write a single equation relating the relevant quantities.
- Differentiate: take $d/dt$ of both sides using the chain rule.
- Substitute: plug in known values at the moment in question.
- Solve and check: find the unknown rate; check units and sign.
Key Concepts
1. The Strategy Applied: Two Cars
Example 1. Car A goes north at 60 km/h, Car B goes east at 80 km/h. How fast are they moving apart after 1 hour?
Step 1: Let $x(t)$ = Car B’s distance east (km), $y(t)$ = Car A’s distance north (km), $z(t)$ = distance between cars (km).
Step 2: Pythagorean theorem: $z^2 = x^2 + y^2$.
Step 3: Differentiate: \[ 2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}. \]
Step 4: After 1 hour: $x = 80$, $y = 60$, $dx/dt = 80$, $dy/dt = 60$, $z = \sqrt{6400 + 3600} = 100$ km. \[ 2(100)\frac{dz}{dt} = 2(80)(80) + 2(60)(60) = 12800 + 7200 = 20000. \] \[ \frac{dz}{dt} = \frac{20000}{200} = 100 \text{ km/h}. \]
Check: Does this make sense? The two cars travel at 60 and 80 km/h; their separation grows at 100 km/h (the hypotenuse, confirming the 60-80-100 right triangle scaling). Prediction confirmed.
2. Choosing the Right Equation
The hardest skill: identifying which equation to differentiate.
| Situation | Equation to use |
|---|---|
| Object on a screen/wall, object moving | Similar triangles |
| Two quantities with fixed-length constraint | Pythagorean theorem |
| Changing area, volume | Area/volume formula |
| Angle and side lengths | Trig ratio ($\tan\theta = opp/adj$) |
| Temperature, pressure | Physical law (Boyle, Newton cooling) |
When the problem involves an angle, reach for a trig ratio first. When it involves a triangle, reach for the Pythagorean theorem or similar triangles. When it involves a 3D shape, reach for the volume formula.
Example 2. A street light at height 5 m illuminates a 1.8 m tall person walking away from the base at 1.5 m/s. How fast is the shadow lengthening?
Step 1: Let $x$ = person’s distance from pole (m), $s$ = shadow length (m). Both change with $t$.
Step 2: Similar triangles (light, top of person, tip of shadow): \[ \frac{5}{x + s} = \frac{1.8}{s}. \] Cross-multiply: $5s = 1.8(x + s)$, so $3.2s = 1.8x$, giving $s = \frac{1.8}{3.2}x = \frac{9}{16}x$.
Step 3: Differentiate: \[ \frac{ds}{dt} = \frac{9}{16}\frac{dx}{dt}. \]
Step 4: $dx/dt = 1.5$ m/s.
Step 5: $ds/dt = (9/16)(1.5) = 13.5/16 \approx 0.844$ m/s.
The shadow grows at a constant rate, independent of position. (Because $s \propto x$, the relationship is linear.)
3. Multiple Representations
Graphical: For the shadow problem, the graph of shadow length $s$ vs. time $t$ is a straight line (since $s = (9/16)x = (9/16)(1.5t) = 0.844t$). The slope of this line is $ds/dt = 0.844$ m/s.
Tabular:
| $t$ (s) | $x = 1.5t$ (m) | $s = 0.844t$ (m) | $ds/dt$ (m/s) |
|---|---|---|---|
| 0 | 0 | 0 | 0.844 |
| 5 | 7.5 | 4.22 | 0.844 |
| 10 | 15 | 8.44 | 0.844 |
The constant rate of shadow growth is confirmed by both the formula and the table.
4. Checking Units and Sign
Units: The unit of $dz/dt$ is km/h (km per hour) because $z$ is in km and $t$ is in hours. If your answer has wrong units, an algebra error has occurred.
Sign: A positive rate means the quantity is increasing; negative means decreasing. If you compute a negative $dz/dt$ for the two-cars problem, something is wrong, because the distance between the cars is increasing. If you get a negative $ds/dt$ for a shadow that is lengthening, the sign is wrong.
5. Why the Five-Step Strategy Is in That Order
Students often want to substitute the specific numerical values first, then differentiate. This fails because differentiation operates on a variable equation (relating general functions of $t$); substituting makes it a constant equation whose derivative is zero.
The strategy forces you to keep the equation general (steps 1-3) and substitute only at the moment of interest (step 4). The specific moment exists in a universe where the equation holds at every moment; substituting pins down “which moment.”
Named Misconception: composition-is-not-chaining
In related-rates problems, the chain rule connects $dz/dt$ to $dx/dt$ and $dy/dt$. A common confusion: treating these as if you can simply chain the rates by multiplying them together, e.g., writing $dz/dx \cdot dy/dt$ without a proper chain-rule setup.
The correct chain rule when $z^2 = x^2 + y^2$ gives $2z\,dz/dt = 2x\,dx/dt + 2y\,dy/dt$. The rates add (after weighting by the geometry); they do not multiply in a chain.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Substituting before differentiating | Plugging $x = 80, y = 60$ into $z^2 = x^2 + y^2$ first | Differentiate $z^2 = x^2 + y^2$ in general; substitute after |
| Wrong equation | Using area formula when the problem involves similar triangles | Identify the constraint that ties the quantities; draw a picture |
| Unit mismatch | Mixing km and m in the same equation | Convert all lengths to the same unit before setting up |
Common Misconceptions
the rates in a related-rates problem can be combined by chaining them in a product without deriving the full rate equation from the constraint.
This is the composition-is-not-chaining error. In the two-cars problem, $\frac{dz}{dt}$ is not simply $\frac{dz}{dx} \cdot \frac{dx}{dt}$; because $z$ depends on both $x$ and $y$, the correct equation is $2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$, which comes from differentiating the constraint $z^2 = x^2 + y^2$. The rates add (weighted by geometry) rather than multiply in a chain.
it does not matter whether numerical values are substituted into the constraint equation before or after differentiating.
This is the rate-as-fixed-number error. Substituting $x = 80$ and $y = 60$ into $z^2 = x^2 + y^2$ before differentiating yields $z^2 = 10000$, a constant whose derivative is zero. The constraint equation describes a relationship that holds at every moment in time and must be differentiated in that general form; only after obtaining the rate equation should specific moment values be substituted.
Leveled Practice
Level 1 -- Setting Up the Equation
Problem 1. A 10 m ladder leans against a wall. Its base slides out at 1 m/s. Identify $x$, $y$, and the equation relating them. Do not compute the rate yet.
Show answer
$x$ = distance of base from wall (m), $y$ = height of top on wall (m). Equation: $x^2 + y^2 = 100$.
Problem 2. Water flows into a rectangular pool (5 m by 8 m) at $3$ m$^3$/min. Let $h$ = depth (m). Write the equation relating $V$ and $h$ and differentiate with respect to $t$.
Show answer
$V = 40h$. Differentiating: $dV/dt = 40\,dh/dt$. With $dV/dt = 3$: $dh/dt = 3/40 = 0.075$ m/min.
Level 2 -- Full Problems
Problem 3. A person stands 5 m from the base of a 12 m tall building and walks away at 2 m/s. A spotlight shines from the top of the building. How fast is the person’s shadow (cast on the ground beyond them) moving away from the building when the person is 8 m from the building?
Show answer
Let $d$ = person’s distance from building, $s$ = shadow tip’s distance from building.
Similar triangles: $\frac{12}{s} = \frac{12 - 1.8}{s - d}$. The problem does not mention the person’s height; let’s say the person is 1.8 m tall.
Similar triangles (building top to shadow tip, and person’s head to shadow tip): $\frac{12}{s} = \frac{1.8}{s - d}$.
$12(s - d) = 1.8s$: $12s - 12d = 1.8s$, so $10.2s = 12d$, giving $s = (12/10.2)d = (20/17)d$.
$ds/dt = (20/17)\,dd/dt = (20/17)(2) = 40/17 \approx 2.35$ m/s.
Shadow tip moves away from building at about 2.35 m/s. (The rate is constant, independent of $d$.)
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) Two cars approach an intersection from perpendicular roads, both at 30 km/h. How fast is the distance between them decreasing when each is 0.5 km from the intersection?
(b) (Mid) In the two-cars-leaving problem (Example 1), the cars travel at speeds $v_A$ (north) and $v_B$ (east). Find a formula for $dz/dt$ at time $t$ in terms of $v_A$, $v_B$, and $t$.
(c) (Ceiling) The shadow problem (Example 2) gave $ds/dt = (9/16)\,dx/dt$ regardless of position. Interpret this: what geometric property of similar triangles explains why the shadow rate is a constant multiple of the walking rate, independent of position?
Show answer
(a) Let $x(t) = y(t) = 0.5 - 30t$ km (both approaching). $z^2 = x^2 + y^2 = 2x^2$.
$2z\,dz/dt = 4x\,dx/dt$.
At $x = y = 0.5$: $z = 0.5\sqrt{2}$. $dx/dt = -30$ (approaching). $2(0.5\sqrt{2})\,dz/dt = 4(0.5)(-30)$. $\sqrt{2}\,dz/dt = -60$. $dz/dt = -60/\sqrt{2} = -30\sqrt{2} \approx -42.4$ km/h. The distance decreases at about 42.4 km/h.
(b) $x = v_B t$, $y = v_A t$, $z = \sqrt{v_B^2 + v_A^2}\,t$.
$2z\,dz/dt = 2x\,dx/dt + 2y\,dy/dt = 2v_B^2 t + 2v_A^2 t$.
$dz/dt = (v_A^2 + v_B^2)t / z = (v_A^2 + v_B^2)t / (\sqrt{v_A^2+v_B^2}\,t) = \sqrt{v_A^2 + v_B^2}$.
The cars separate at exactly $\sqrt{v_A^2 + v_B^2}$ km/h, a constant. This is the magnitude of the velocity vector $(v_B, v_A)$.
(c) The similar-triangles relationship $s = (9/16)x$ is linear. The shadow tip tracks the person’s position proportionally, so differentiating gives a proportional rate relationship. Linearity is preserved under differentiation. This always happens when the constraint between two quantities is a linear (or affine) equation.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
A related-rates problem is an equation that holds at every moment in time, converted by differentiation into a rate equation that also holds at every moment in time. The geometric or physical equation is the constraint; differentiation extracts the rate relationship.
Treat the five-step strategy as a checklist for the thinking process: (1) what are the variables? (2) what ties them together? (3) how do their rates relate? (4) what is the specific moment? (5) does the answer make sense?
Connections
Within MATH161
- Implicit differentiation: Related rates is differentiation of an implicit equation with respect to $t$ instead of $x$. The techniques are identical.
- Chain rule: Every rate relationship comes from the chain rule. The chain rule is what allows differentiation of $z^2 = x^2(t) + y^2(t)$ to give the rate relationship.
- Optimization: Both use derivatives to extract quantitative information from geometric or physical constraints.
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