Unit 4.4 – Introduction to Related Rates
AP® Calculus AB & BC | Real-World Problems Where Everything Changes at Once
Related rates problems involve two or more variables that change with respect to time. The rate at which one quantity is changing is "related" to the rate another quantity changes. You'll use implicit differentiation and the chain rule to connect these rates.
🔗 Related Rates: The Main Idea
- Variables are functions of time: \(x(t)\), \(y(t)\), etc.
- You are given a relationship between the variables, such as \(F(x, y, z) = 0\).
- Differentiating both sides with respect to \(t\) finds how the rates are connected:
\[ \frac{dx}{dt},\ \frac{dy}{dt},\ \frac{dr}{dt}, \dots \]
Key: Use the chain rule & implicit differentiation. Every variable that is a function of time gets differentiated with respect to time (e.g., \(x^2\) becomes \(2x \frac{dx}{dt}\)).
🚦 General Steps for Related Rates
- Draw and label a diagram (pictures are extremely helpful!).
- Assign variables (with units!) for each quantity that changes over time.
- Write an equation that relates the variables to each other.
- Differentiate both sides with respect to time \(t\) – remember the chain rule for every variable!
- Plug in the known values for the quantities and their rates of change at the specific instant in question.
- Solve for the desired rate.
- State your final answer with correct units and in the context of the problem.
🗂️ Classic Related Rates Formulas
- Circle Area: \(A = \pi r^2\) → \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
- Sphere Volume: \(V = \frac{4}{3}\pi r^3\) → \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)
- Volume of Cylinder: \(V = \pi r^2 h\) → \(\frac{dV}{dt}=2\pi r h\frac{dr}{dt}+\pi r^2\frac{dh}{dt}\)
- Pythagorean Theorem: \(x^2 + y^2 = h^2\) → \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2h\frac{dh}{dt}\)
- Rectangle Area: \(A = lw\) → \(\frac{dA}{dt} = l \frac{dw}{dt} + w \frac{dl}{dt}\)
💡 Short Notes, Tips & Tricks
- Only substitute numerical values after you have differentiated. The only exception is for a quantity that is constant for all time.
- Use the Pythagorean theorem for right triangles; solve for any missing side lengths at the specific instant before solving for the rate.
- Label everything clearly: what are your variables, units, and known rates?
- Chain rule is key: every time you differentiate a variable (like \(r\)), you must multiply by its rate (like \(\frac{dr}{dt}\)).
- Rates of quantities that are decreasing (draining, shrinking) should be negative.
📖 Worked Examples
Example 1: Leaky Water Tank (Cylinder)
The volume is \(V = \pi r^2 h\). Differentiating with respect to time gives:
\(\frac{dV}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}\)
Often, the radius is constant (\(\frac{dr}{dt}=0\)), so the formula simplifies to \(\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}\).
\(\frac{dV}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}\)
Often, the radius is constant (\(\frac{dr}{dt}=0\)), so the formula simplifies to \(\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}\).
Example 2: Expanding Circle
The area is \(A = \pi r^2\).
If you are given \(\frac{dr}{dt}\) and a value for \(r\) at a specific instant, you can find \(\frac{dA}{dt}\) by differentiating: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).
If you are given \(\frac{dr}{dt}\) and a value for \(r\) at a specific instant, you can find \(\frac{dA}{dt}\) by differentiating: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).
Example 3: Ladder Sliding Down a Wall
If a ladder of length \(h\) has its base \(x\) feet from a wall and its top \(y\) feet up the wall, then \(x^2 + y^2 = h^2\).
Differentiating gives: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\) (since the ladder's length \(h\) is constant, \(\frac{dh}{dt}=0\)).
Differentiating gives: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\) (since the ladder's length \(h\) is constant, \(\frac{dh}{dt}=0\)).
📝 Practice Problems
Try These Yourself:
- A spherical balloon’s radius is increasing at a rate of \(1.5\) cm/min. How fast is the volume changing when \(r = 10\) cm?
- A rectangle has a constant width of \(4\) cm, but its length increases at \(3\) cm/sec. Find the rate of change of the area, \(\frac{dA}{dt}\).
- Two cars leave the same point at the same time. One travels east at \(60\) km/h, and the other travels north at \(80\) km/h. How fast is the distance between them changing after \(2\) hours?
Answers (Setup):
- \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \implies 4\pi (10^2)(1.5) = 600\pi\) cm³/min.
- \(A = wl \implies \frac{dA}{dt} = w \frac{dl}{dt} + l\frac{dw}{dt}\). Since width is constant, \(\frac{dw}{dt}=0\), so \(\frac{dA}{dt} = 4 \times 3 = 12\) cm²/sec.
- Let \(x\) be distance east, \(y\) be distance north, \(z\) be distance between them. \(z^2=x^2+y^2\). After 2h, \(x=120\), \(y=160\), \(z=200\). Differentiate to get \(2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}\) and solve for \(\frac{dz}{dt}\).
✏️ AP® Exam Tips for Related Rates
- Draw a diagram and define all variables (with units) first! This is a crucial first step.
- Differentiation comes before substitution for any quantity that is changing over time.
- Keep units in every step and in your final answer. Points are often lost here.
- State your answer with the appropriate sign (positive for increasing, negative for decreasing).
- Explicitly show your differentiation step in FRQs for full credit.