Related Rates
In short: Related rates problems connect the speeds of two or more changing quantities through an equation, then differentiate with respect to time to find an unknown rate from a known one.
AP Calculus ABCalculusAP Calculus AB Unit 4
What is Related Rates?
In related rates problems, several quantities change together and an equation ties them to each other. A ladder slides, so the height of its top and the distance of its base both change, yet the ladder length stays fixed.
The idea is to differentiate that equation with respect to time \(t\). Every variable gets a derivative, such as \(\frac{dx}{dt}\), and the chain rule links them. Substitute the numbers only after differentiating, never before, because the quantities are moving.
How to spot it
- The wording mentions how fast something is growing, shrinking, sliding or filling at a given instant.
- Two or more measurements, such as radius and volume, change together over time, which is the signature of related rates.
- A rate is supplied (inches per second) and a different rate is requested at a specific moment.
- A geometric relationship, like the Pythagorean theorem or a volume formula, connects the quantities.
The method
- For related rates, draw a picture, name each changing quantity with a variable, and list the known and wanted rates with units.
- Write an equation linking the variables that is true at every moment, with no numbers that change.
- Differentiate both sides with respect to \(t\), using the chain rule on every variable.
- Substitute the values at the instant in question and solve for the unknown rate; include units and sign.
Related Rates: solved example
Problem
A 25-foot ladder leans against a wall. Its base is 7 feet from the wall and is being pulled away at 1.5 feet per second. How fast is the top of the ladder sliding down?
Solution
The equation \(x^2 + y^2 = 625\) holds at all times. When \(x = 7\): \(y = \sqrt{625 – 49} = 24\).
Differentiate: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\), so \(\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{7}{24}(1.5) = -\frac{7}{16}\).
The top slides down at \(\frac{7}{16} = 0.4375\) feet per second; the negative sign means the height is decreasing.
Try it
Question 1
A ripple on a pond is a circle whose radius grows at 3 inches per second. How fast is the enclosed area growing when the radius is 10 inches?
Show the answer
\(A = \pi r^2\) gives \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(10)(3) = 60\pi\) square inches per second, about 188.5.
Question 2
The edge \(s\) of an ice cube is growing at 0.5 inch per minute. How fast is its volume changing when \(s = 4\) inches?
Show the answer
\(V = s^3\) gives \(\frac{dV}{dt} = 3s^2\frac{ds}{dt} = 3(16)(0.5) = 24\) cubic inches per minute.
Related Rates vs. Derivative
Related rates are not a plain derivative question. A plain derivative asks how one quantity changes as another changes. In these problems the independent variable is always time and several quantities are linked, so you differentiate an equation with respect to \(t\). A frequent mistake is plugging in 7 or 24 before differentiating, which freezes values that actually vary.
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