Related Rates Problems

In the introduction to related rates, the equation was handed to you: $A = \pi r^2$, $V = s^3$. In real problems you have to build the equation yourself from a picture, using Pythagoras, similar triangles, or a volume formula. This page gives you a strategy that works every time, and walks through the classic setups that appear on the AP exam.

Key ideas

A step-by-step strategy

1. Draw a picture. Label everything that changes with a variable and everything that stays fixed with a number.
2. Write “Know” and “Want” using derivative notation, with signs and units. For example: Know $\dfrac{dx}{dt} = 0.4$ m/s. Want $\dfrac{dy}{dt}$ when $x = 3$.
3. Write an equation linking the variables (Pythagoras, similar triangles, area or volume).
4. Eliminate extra variables if needed. If the equation has a variable whose rate you don’t know and don’t want, use a second relationship (often similar triangles) to replace it.
5. Differentiate both sides with respect to $t$.
6. Substitute the values at the moment in question. Find any missing values from the picture first.
7. Answer in context, with units and the correct sign.

The most important rule from the intro still applies: differentiate first, then substitute the values that are changing.

Three common relationships

• Right triangles (ladders, cars at a corner, planes overhead): $x^2 + y^2 = z^2$.
• Similar triangles (cones, shadows, troughs): ratios of matching sides are equal.
• Volume formulas: cone $V = \dfrac{1}{3}\pi r^2 h$, cylinder $V = \pi r^2 h$, sphere $V = \dfrac{4}{3}\pi r^3$.

Angles are in radians

If an angle is changing, its rate $\dfrac{d\theta}{dt}$ is in radians per unit of time, because the derivative formulas for trig functions only work in radians.

Worked examples

Example 1: A sliding ladder

A $5$ m ladder leans against a wall. The bottom slides away from the wall at $0.4$ m/s. How fast is the top sliding down the wall when the bottom is $3$ m from the wall?

The ladder’s length is fixed at $5$ m; $x$ and $y$ change.

Solution. Let $x$ be the distance from the wall to the bottom and $y$ the height of the top.

Know: $\dfrac{dx}{dt} = 0.4$ m/s. Want: $\dfrac{dy}{dt}$ when $x = 3$.

Equation (Pythagoras): $x^2 + y^2 = 25$. Differentiate:

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

When $x = 3$, $y = \sqrt{25 - 9} = 4$. Substitute:

$$2(3)(0.4) + 2(4)\frac{dy}{dt} = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{2.4}{8} = -0.3$$

The top is sliding down the wall at $0.3$ m/s. (The negative sign means $y$ is decreasing.)

Example 2: Filling a cone

A water tank is an inverted cone (point down) with a top radius of $2$ m and a height of $6$ m. Water is poured in at $3$ m³/min. How fast is the water level rising when the water is $4$ m deep?

The water forms a smaller cone similar to the tank, so $\dfrac{r}{h} = \dfrac{2}{6}$.

Solution. Let $h$ be the water depth and $r$ the radius of the water surface.

Know: $\dfrac{dV}{dt} = 3$ m³/min. Want: $\dfrac{dh}{dt}$ when $h = 4$.

The volume is $V = \dfrac{1}{3}\pi r^2 h$, but we don’t know $\dfrac{dr}{dt}$. Eliminate $r$ using similar triangles:

$$\frac{r}{h} = \frac{2}{6} \quad\Rightarrow\quad r = \frac{h}{3}$$ $$V = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}$$

Differentiate:

$$\frac{dV}{dt} = \frac{\pi h^2}{9}\,\frac{dh}{dt}$$

Substitute $h = 4$:

$$3 = \frac{16\pi}{9}\,\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} = \frac{27}{16\pi} \approx 0.537$$

The water level is rising at $\dfrac{27}{16\pi} \approx 0.537$ m/min.

Example 3: A moving shadow

A person $1.8$ m tall walks away from a $6$ m lamppost at $1.5$ m/s. How fast is the length of their shadow increasing? How fast is the tip of the shadow moving?

Solution. Let $x$ be the person’s distance from the post and $s$ the length of the shadow. The tip of the shadow is $x + s$ from the post.

Know: $\dfrac{dx}{dt} = 1.5$ m/s. Want: $\dfrac{ds}{dt}$.

The light ray from the top of the lamp to the tip of the shadow makes two similar right triangles: the big one (lamppost, height $6$, base $x + s$) and the small one (person, height $1.8$, base $s$):

$$\frac{6}{x + s} = \frac{1.8}{s} \quad\Rightarrow\quad 6s = 1.8x + 1.8s \quad\Rightarrow\quad 4.2s = 1.8x \quad\Rightarrow\quad s = \frac{3}{7}x$$

Differentiate:

$$\frac{ds}{dt} = \frac{3}{7}\frac{dx}{dt} = \frac{3}{7}(1.5) = \frac{9}{14} \approx 0.643$$

The shadow grows at $\dfrac{9}{14} \approx 0.643$ m/s. The tip moves at

$$\frac{d}{dt}(x + s) = 1.5 + \frac{9}{14} = \frac{15}{7} \approx 2.143 \text{ m/s}$$

Notice the answer doesn’t depend on how far away the person is. That happens with shadows because $s$ is a constant multiple of $x$.

Example 4: Two cars approaching a corner

Car A is $0.3$ km east of an intersection, driving west toward it at $60$ km/h. Car B is $0.4$ km north of the intersection, driving south toward it at $80$ km/h. How fast is the distance between the cars changing?

Solution. Let $x$ be car A’s distance from the intersection, $y$ car B’s distance, and $z$ the distance between the cars.

Know: $\dfrac{dx}{dt} = -60$ and $\dfrac{dy}{dt} = -80$ km/h (negative because both distances are shrinking). Want: $\dfrac{dz}{dt}$ when $x = 0.3$, $y = 0.4$.

Equation: $x^2 + y^2 = z^2$. At this moment, $z = \sqrt{0.09 + 0.16} = 0.5$ km. Differentiate:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$$ $$0.3(-60) + 0.4(-80) = 0.5\frac{dz}{dt} \quad\Rightarrow\quad -50 = 0.5\frac{dz}{dt} \quad\Rightarrow\quad \frac{dz}{dt} = -100$$

The distance between the cars is decreasing at $100$ km/h.

Common mistakes

Labelling a changing length with a number. In Example 1, the ladder’s $5$ m is fixed, but the $3$ m is only true at one instant. Label it $x$ and substitute $x = 3$ at the end. If you write $3$ in the picture, its derivative is $0$.

Forgetting to find the missing value. In the ladder problem you need $y = 4$, which isn’t given. Use the original equation (here Pythagoras) to find every value at that instant.

Keeping an extra variable. In the cone, $V = \frac{1}{3}\pi r^2 h$ has two changing variables, and you know neither $\dfrac{dr}{dt}$ nor need it. Use similar triangles to replace $r$ with $\dfrac{h}{3}$ before differentiating.

Wrong signs on given rates. A distance that is shrinking has a negative rate. In Example 4, both cars approach the corner, so $\dfrac{dx}{dt} = -60$ and $\dfrac{dy}{dt} = -80$.

Using the wrong cone ratio. For a cone with top radius $2$ and height $6$, $r = \dfrac{h}{3}$, not $r = 3h$. Check: at full depth $h = 6$, $r$ should be $2$.

Giving the rate with no direction. Say whether the quantity is increasing or decreasing (“sliding down at $0.3$ m/s”), not just "$-0.3$".

Practice

1. (Warm-up) A $10$ m ladder leans against a wall. The bottom is pulled away from the wall at $1$ m/s. How fast is the top sliding down when the bottom is $6$ m from the wall?

Solution

$x^2 + y^2 = 100$. When $x = 6$, $y = 8$.

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Rightarrow\quad 6(1) + 8\frac{dy}{dt} = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{3}{4}$$

The top slides down at $0.75$ m/s.

2. (Warm-up) A conical tank (point down) is $10$ m tall with a top radius of $4$ m. Write the volume of water in the tank in terms of the water depth $h$ only.

Solution

Similar triangles: $\dfrac{r}{h} = \dfrac{4}{10}$, so $r = \dfrac{2h}{5}$.

$$V = \frac{1}{3}\pi\left(\frac{2h}{5}\right)^2 h = \frac{4\pi h^3}{75}$$

3. (Core) Two cars leave the same point at the same time. One drives east at $50$ km/h, the other north at $120$ km/h. How fast is the distance between them increasing after $1$ hour?

Solution

After $1$ hour, $x = 50$, $y = 120$, so $z = \sqrt{2500 + 14400} = 130$ km.

$$z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} \quad\Rightarrow\quad 130\frac{dz}{dt} = 50(50) + 120(120) = 16\,900$$$$\frac{dz}{dt} = 130 \text{ km/h}$$

4. (Core) The tank in question 2 is draining at $2$ m³/min. How fast is the water level falling when the water is $5$ m deep?

Solution

From $V = \dfrac{4\pi h^3}{75}$:

$$\frac{dV}{dt} = \frac{4\pi h^2}{25}\frac{dh}{dt}$$

With $h = 5$ and $\dfrac{dV}{dt} = -2$:

$$-2 = \frac{4\pi(25)}{25}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} = -\frac{1}{2\pi} \approx -0.159$$

The water level is falling at about $0.159$ m/min.

5. (Core) A child $1.5$ m tall walks toward a $4.5$ m lamppost at $1.2$ m/s. How fast is the child’s shadow changing length?

Solution

Let $x$ be the distance to the post and $s$ the shadow length. Similar triangles:

$$\frac{4.5}{x + s} = \frac{1.5}{s} \quad\Rightarrow\quad 4.5s = 1.5x + 1.5s \quad\Rightarrow\quad s = \frac{x}{2}$$

Walking toward the post means $\dfrac{dx}{dt} = -1.2$, so

$$\frac{ds}{dt} = \frac{1}{2}(-1.2) = -0.6$$

The shadow is getting shorter at $0.6$ m/s.

6. (Core) A plane flies horizontally at an altitude of $3$ km and a speed of $600$ km/h, and passes directly over a radar station. How fast is the distance from the plane to the station increasing when that distance is $5$ km?

Solution

Let $x$ be the horizontal distance from the point above the station and $z$ the distance to the station: $x^2 + 3^2 = z^2$. When $z = 5$, $x = 4$.

$$2x\frac{dx}{dt} = 2z\frac{dz}{dt} \quad\Rightarrow\quad 4(600) = 5\frac{dz}{dt} \quad\Rightarrow\quad \frac{dz}{dt} = 480$$

The distance is increasing at $480$ km/h.

7. (Core) A hot-air balloon rises straight up at $3$ m/s. An observer stands $120$ m from the launch point. How fast is the angle of elevation from the observer to the balloon changing when the balloon is $160$ m up? Give your answer in radians per second.

Solution

Let $h$ be the height and $\theta$ the angle of elevation: $\tan\theta = \dfrac{h}{120}$. Differentiate:

$$\sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{120}\frac{dh}{dt}$$

When $h = 160$, the line of sight is $\sqrt{120^2 + 160^2} = 200$ m, so $\cos\theta = \dfrac{120}{200} = 0.6$ and $\sec^2\theta = \dfrac{1}{0.36}$.

$$\frac{d\theta}{dt} = 0.36 \cdot \frac{3}{120} = 0.009$$

The angle is increasing at $0.009$ radians per second.

8. (Challenge) A water trough is $4$ m long. Its ends are triangles (point down) that are $1$ m across the top and $0.5$ m deep. Water flows in at $0.2$ m³/min. How fast is the water level rising when the water is $0.3$ m deep?

Solution

At depth $h$, the water surface is $w$ wide. Similar triangles: $\dfrac{w}{h} = \dfrac{1}{0.5}$, so $w = 2h$.

The water is a prism with a triangular cross-section:

$$V = \frac{1}{2}(2h)(h)(4) = 4h^2 \quad\Rightarrow\quad \frac{dV}{dt} = 8h\frac{dh}{dt}$$$$0.2 = 8(0.3)\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} = \frac{0.2}{2.4} = \frac{1}{12} \approx 0.083$$

The level is rising at $\dfrac{1}{12} \approx 0.083$ m/min.

9. (Challenge) In Example 1, how fast is the area of the triangle formed by the ladder, the wall, and the ground changing when the bottom is $3$ m from the wall? Is it growing or shrinking?

Solution

$A = \dfrac{1}{2}xy$. By the product rule:

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

From Example 1, at this moment $x = 3$, $y = 4$, $\dfrac{dx}{dt} = 0.4$, and $\dfrac{dy}{dt} = -0.3$:

$$\frac{dA}{dt} = \frac{1}{2}\big(0.4(4) + 3(-0.3)\big) = \frac{1}{2}(1.6 - 0.9) = 0.35$$

The area is growing at $0.35$ m²/s.