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4.5 - Solving Related Rates Problems

From Setup to Solution

Last lesson we learned to set up related rates problems. Now we’ll solve them by substituting known values and finding the unknown rates.

Example: Pure Algebra Problem

Given the relationship y2=3x4+6xy^2 = 3x^4 + 6x, find dydt\frac{dy}{dt} when x=1x = 1 and dxdt=2\frac{dx}{dt} = 2.
Step 1: Differentiate with respect to time
ddt[y2]=ddt[3x4+6x]\frac{d}{dt}[y^2] = \frac{d}{dt}[3x^4 + 6x]2ydydt=12x3dxdt+6dxdt2y\frac{dy}{dt} = 12x^3\frac{dx}{dt} + 6\frac{dx}{dt}
Step 2: Identify what we’re looking for
We want dydt\frac{dy}{dt}. That’s our unknown.
Step 3: Find the missing value
We know x=1x = 1 and dxdt=2\frac{dx}{dt} = 2, but we don’t know yy. Use the original equation:
y2=3(1)4+6(1)=3+6=9⇒y=±3y^2 = 3(1)^4 + 6(1) = 3 + 6 = 9 \quad \Rightarrow \quad y = \pm 3
Note on signs: in context problems, you’ll know whether the quantity is positive or negative based on the situation. For abstract problems with no context, the problem should tell you which sign to use. This one doesn’t, so we’ll assume y>0y > 0, which gives y=3y = 3.
Step 4: Substitute all known values
2(3)dydt=12(1)3(2)+6(2)2(3)\frac{dy}{dt} = 12(1)^3(2) + 6(2)6dydt=24+126\frac{dy}{dt} = 24 + 126dydt=366\frac{dy}{dt} = 36dydt=6\boxed{\frac{dy}{dt} = 6}

The Five-Step Process for Related Rates

Step 1: Draw a Picture

Sketch the situation and label all the changing quantities with variables.

Step 2: Make a List ⭐ NEW

List all the known and unknown quantities and rates: - Known rates: given derivatives like dxdt=5\frac{dx}{dt} = 5 - Unknown rates: what you’re solving for, so circle this - Known quantities: given values like x=3x = 3 - Unknown quantities: you may need to find these using the original equation

Step 3: Relate the Variables in an Equation

Write an equation connecting all the variables, using geometry formulas, physics principles, or whatever fits.

Step 4: Differentiate with Respect to Time

Apply ddt\frac{d}{dt} to both sides, using the chain rule and the product or quotient rules as needed.

Step 5: Substitute and Solve ⭐ NEW

Plug in all the known values and solve for the unknown rate.

Critical Rule: When to Substitute

⚠️ Important: only substitute constants before differentiating.
Constants (substitute before differentiating): - “The ladder is 15 feet long,” since the length doesn’t change - “Flying at a constant altitude of 10,000 feet,” since the height doesn’t change - Fixed dimensions or unchanging quantities
Variables (substitute after differentiating): - Distances that change as objects move - Dimensions that grow or shrink - Any quantity that has a rate of change
Why this matters: if you substitute a changing value before differentiating, you’re treating it as a constant. That eliminates its derivative term and gives you a wrong answer.

Example 1: Expanding Rectangle

Problem: The width of a rectangle is increasing at 2 cm/s and the length is increasing at 3 cm/s. At what rate is the area increasing when the width is 4 cm and the length is 5 cm?

Step 1: Draw and Label

4.5 - Solving Related Rates Problems

Step 2: Make a List

Known rates: - dWdt=2\frac{dW}{dt} = 2 cm/s, the width is increasing - dLdt=3\frac{dL}{dt} = 3 cm/s, the length is increasing
Unknown rate: - dAdt=?\frac{dA}{dt} = ? ← FIND THIS
Known quantities: - W=4W = 4 cm - L=5L = 5 cm

Step 3: Relate the Variables

Area of a rectangle:
A=L⋅WA = L \cdot W

Step 4: Differentiate with Respect to Time

⚠️ Product rule required, since we have two variables multiplied together.
dAdt=dLdt⋅W+L⋅dWdt\frac{dA}{dt} = \frac{dL}{dt} \cdot W + L \cdot \frac{dW}{dt}
Common mistake: writing dAdt=dLdt⋅dWdt\frac{dA}{dt} = \frac{dL}{dt} \cdot \frac{dW}{dt} is wrong. You have to use the product rule properly.

Step 5: Substitute and Solve

dAdt=(3)(4)+(5)(2)=12+10=22\frac{dA}{dt} = (3)(4) + (5)(2) = 12 + 10 = 22
Units: area is in cm² and time is in seconds, so:
dAdt=22 cm2/s\boxed{\frac{dA}{dt} = 22 \text{ cm}^2\text{/s}}
Interpretation: the area is increasing at 22 square centimeters per second.
Common unit mistake: writing “22 cm/s²” is wrong, since that would be acceleration. The correct units are cm²/s, which is area per time.

Example 2: Police Chase Continued

Problem: A police car is 0.6 miles north of an intersection, traveling toward it at 60 mph. A speeding car is 0.8 miles east of the intersection. At that instant, radar shows the distance between the cars is increasing at 20 mph. How fast is the speeding car traveling?

Step 1: Draw and Label

4.5 - Solving Related Rates Problems

Step 2: Make a List

Known quantities: - y=0.6y = 0.6 miles, the police car’s distance - x=0.8x = 0.8 miles, the speeding car’s distance - z=?z = ?, we need to find this
Known rates: - dydt=−60\frac{dy}{dt} = -60 mph, negative because it’s moving toward the intersection - dzdt=20\frac{dz}{dt} = 20 mph, the distance between the cars is increasing
Unknown rate: - dxdt=?\frac{dx}{dt} = ? ← FIND THIS, the speeding car’s speed

Step 3: Relate the Variables

Pythagorean theorem:
x2+y2=z2x^2 + y^2 = z^2

Step 4: Differentiate with Respect to Time

2xdxdt+2ydydt=2zdzdt2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}

Step 5: Substitute and Solve

First, find the missing value zz:
z2=(0.8)2+(0.6)2=0.64+0.36=1⇒z=1 milez^2 = (0.8)^2 + (0.6)^2 = 0.64 + 0.36 = 1 \quad \Rightarrow \quad z = 1 \text{ mile}
Now substitute everything:
2(0.8)dxdt+2(0.6)(−60)=2(1)(20)2(0.8)\frac{dx}{dt} + 2(0.6)(-60) = 2(1)(20)1.6dxdt−72=401.6\frac{dx}{dt} - 72 = 401.6dxdt=1121.6\frac{dx}{dt} = 112dxdt=70 mph\boxed{\frac{dx}{dt} = 70 \text{ mph}}
Interpretation: the speeding car is going 70 miles per hour at that moment.

Example 3: Inflating Balloon

Problem: A spherical balloon is being inflated. When the radius is 6 cm, the radius is increasing at 0.5 cm/s. How fast is the volume increasing at that moment?

Step 1: Draw and Label

4.5 - Solving Related Rates Problems

Step 2: Make a List

Known quantities: r=6r = 6 cm
Known rates: drdt=0.5\frac{dr}{dt} = 0.5 cm/s
Unknown rate: dVdt=?\frac{dV}{dt} = ? ← FIND THIS

Step 3: Relate the Variables

Volume of a sphere:
V=43πr3V = \frac{4}{3}\pi r^3

Step 4: Differentiate with Respect to Time

dVdt=43π⋅3r2⋅drdt⇒dVdt=4πr2drdt\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} \quad \Rightarrow \quad \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Step 5: Substitute and Solve

dVdt=4π(6)2(0.5)=4π(36)(0.5)=72π≈226.2\frac{dV}{dt} = 4\pi (6)^2 (0.5) = 4\pi (36)(0.5) = 72\pi \approx 226.2dVdt=72π cm3/s≈226.2 cm3/s\boxed{\frac{dV}{dt} = 72\pi \text{ cm}^3\text{/s} \approx 226.2 \text{ cm}^3\text{/s}}
Interpretation: the volume is increasing at about 226.2 cubic centimeters per second at that moment.

Example 4: Draining Cone

Problem: Water is draining from a conical tank at a rate of 5 ft³/min. The cone has a height of 10 ft and a radius of 4 ft. How fast is the water level dropping when the water is 6 ft deep?
Note: this one needs similar triangles to relate rr and hh.

Step 1: Draw and Label

4.5 - Solving Related Rates Problems

Step 2: Make a List

Known quantities: - Tank dimensions: height = 10 ft, radius = 4 ft, both constants - h=6h = 6 ft, the current water depth
Known rates: - dVdt=−5\frac{dV}{dt} = -5 ft³/min, negative because it’s draining
Unknown rate: - dhdt=?\frac{dh}{dt} = ? ← FIND THIS

Step 3: Relate the Variables

Volume of a cone:
V=13πr2hV = \frac{1}{3}\pi r^2 h
Problem: we have two variables, rr and hh, but only one rate in the problem.
Solution: use similar triangles to get rid of one variable. From the tank dimensions, the ratio stays constant:
rh=410=25⇒r=2h5\frac{r}{h} = \frac{4}{10} = \frac{2}{5} \quad \Rightarrow \quad r = \frac{2h}{5}
Substitute into the volume formula:
V=13π(2h5)2h=13π⋅4h225⋅hV = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^2 h = \frac{1}{3}\pi \cdot \frac{4h^2}{25} \cdot hV=4πh375V = \frac{4\pi h^3}{75}

Step 4: Differentiate with Respect to Time

dVdt=4π75⋅3h2⋅dhdt⇒dVdt=4πh225dhdt\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \cdot \frac{dh}{dt} \quad \Rightarrow \quad \frac{dV}{dt} = \frac{4\pi h^2}{25} \frac{dh}{dt}

Step 5: Substitute and Solve

−5=4π(6)225dhdt-5 = \frac{4\pi (6)^2}{25} \frac{dh}{dt}−5=144π25dhdt-5 = \frac{144\pi}{25} \frac{dh}{dt}dhdt=−5⋅25144π=−125144π\frac{dh}{dt} = \frac{-5 \cdot 25}{144\pi} = \frac{-125}{144\pi}dhdt≈−0.276 ft/min\boxed{\frac{dh}{dt} \approx -0.276 \text{ ft/min}}
The water level is dropping at about 0.276 feet per minute, and the negative sign confirms it’s decreasing.

Example 5: Drone Continued

Problem: An observer on the ground is watching a drone. At the moment the drone is 40 m away horizontally and 30 m above the ground, it is flying away from the observer at 5 m/s and climbing at 2 m/s. At what rate is the angle of elevation from the observer to the drone changing at that moment?

Step 1: Draw and Label

4.5 - Solving Related Rates Problems

Step 2: Make a List

Known quantities: - x=40x = 40 m, the horizontal distance - y=30y = 30 m, the height of the drone
Known rates: - dxdt=5\frac{dx}{dt} = 5 m/s, positive because the drone is moving away - dydt=2\frac{dy}{dt} = 2 m/s, positive because the drone is climbing
Unknown rate: - dθdt=?\frac{d\theta}{dt} = ? ← FIND THIS

Step 3: Relate the Variables

The side opposite θ\theta is yy and the side adjacent to θ\theta is xx:
tan⁡θ=yx\tan\theta = \frac{y}{x}

Step 4: Differentiate with Respect to Time

⚠️ Quotient rule required, since yy and xx are both changing. The left side needs the chain rule because θ\theta changes with time.
sec⁡2θ⋅dθdt=xdydt−ydxdtx2\sec^2\theta \cdot \frac{d\theta}{dt} = \frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2}

Step 5: Substitute and Solve

First, find the missing value sec⁡2θ\sec^2\theta:
The line of sight is the hypotenuse:
402+302=2500=50 m\sqrt{40^2 + 30^2} = \sqrt{2500} = 50 \text{ m}cos⁡θ=4050=45⇒sec⁡2θ=(54)2=2516\cos\theta = \frac{40}{50} = \frac{4}{5} \quad \Rightarrow \quad \sec^2\theta = \left(\frac{5}{4}\right)^2 = \frac{25}{16}
Now substitute everything:
2516⋅dθdt=(40)(2)−(30)(5)402=80−1501600=−7160\frac{25}{16} \cdot \frac{d\theta}{dt} = \frac{(40)(2) - (30)(5)}{40^2} = \frac{80 - 150}{1600} = -\frac{7}{160}dθdt=−7160⋅1625=−7250\frac{d\theta}{dt} = -\frac{7}{160} \cdot \frac{16}{25} = -\frac{7}{250}dθdt=−0.028 rad/s\boxed{\frac{d\theta}{dt} = -0.028 \text{ rad/s}}
Interpretation: the angle of elevation is decreasing at 0.028 radians per second at that moment. The drone is climbing, but it’s also moving away, and right now that makes yx\frac{y}{x} (which is tan⁡θ\tan\theta) smaller. So the observer has to look a little lower, and the negative sign tells you the angle is decreasing.

Common Mistakes to Avoid

  • ❌ Forgetting the product rule. A=LWA = LW differentiates to dAdt=dLdt⋅W+L⋅dWdt\frac{dA}{dt} = \frac{dL}{dt} \cdot W + L \cdot \frac{dW}{dt}, not dLdt⋅dWdt\frac{dL}{dt} \cdot \frac{dW}{dt}
  • ❌ Substituting variables before differentiating. Only substitute constants before taking ddt\frac{d}{dt}
  • ❌ Wrong sign on rates. If a distance is decreasing, its rate should be negative
  • ❌ Incorrect units. For area rates, use units like cm²/s, not cm/s², since that’s acceleration
  • ❌ Forgetting to find missing values. If you need a quantity that isn’t given, use the original equation to find it
  • ❌ Not using the original equation. When you’re missing a quantity like yy or zz, go back to the original relationship
  • ❌ Mixing up what you’re solving for. Always circle or clearly mark the unknown you’re finding
  • ❌ Using degrees for angles. Rates of change of angles come out in radians per unit of time, so keep your calculator in radian mode

Strategy Summary

  1. Read carefully. Identify what’s given and what you’re finding
  1. Draw and label. A visual keeps you from getting confused
  1. List everything. Known and unknown quantities and rates
  1. Write the relationship. Geometry, physics, or other formulas
  1. Differentiate correctly. Chain rule, product rule, quotient rule as needed
  1. Find missing values. Use the original equation if you need to
  1. Substitute after differentiation. Plug in all the known values
  1. Solve for the unknown. Algebra to isolate the rate you want
  1. Check units. Make sure they make sense for the context
  1. Interpret. State the answer in context

Key Takeaways

  • Five-step process: draw, list, relate, differentiate, substitute and solve
  • Make a list: write down all the known and unknown rates and quantities, and circle what you’re finding
  • Constants only: substitute constants before differentiating, and variables after
  • Product rule matters. Don’t forget it when variables are multiplied
  • Find missing values using the original equation
  • Sign conventions: negative rates mean decreasing, positive mean increasing
  • Units matter. Area rates are (length)²/time, not length/time²
  • Similar triangles help. Use proportions to eliminate extra variables
  • Angles: relate them with right triangle trig (like tan⁡θ=yx\tan\theta = \frac{y}{x}), and give angle rates in radians
  • Work systematically. Follow the steps in order to avoid mistakes
  • Practice makes perfect. The more problems you do, the more patterns you’ll recognize

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