Related Rates
Link two changing quantities by an equation, differentiate the equation with respect to time, and solve for the rate you cannot measure directly.
What a learner can do afterwards
- Set up a relation between two quantities before differentiating anything
- Differentiate a geometric formula with respect to time
- Substitute the instantaneous values only after differentiating, and say why
1 · Read
Some problems give you two quantities that change with time. A balloon has a growing radius and a growing volume. The two rates are linked because one equation links the two quantities. Your job has four moves. Name each changing quantity with its units. Write the equation that links them. Differentiate both sides with respect to time. Substitute the instant values only after that.
A circle has radius 8 cm, growing at 3 cm per second. Area and radius obey A equals pi r squared. Differentiate with respect to time and the chain rule gives dA dt equals 2 pi r dr dt. Now substitute r equals 8 and dr dt equals 3. The area grows at 48 pi square cm per second.
A sphere obeys V equals 4 thirds pi r cubed. Differentiate first and you get dV dt equals 4 pi r squared dr dt. Only now may you substitute numbers such as r equals 3. If you substitute too early, the radius freezes into a constant and its rate silently vanishes. Finish with units, since cubic cm per second tells a different story from square cm per second.
A 10 m ladder slides down a wall. Its foot sits 6 m from the wall and slides out at 2 m per second. Foot and height obey x squared plus y squared equals 100, so y equals 8. Differentiate to get 2 x dx dt plus 2 y dy dt equals 0. Substitute and solve: dy dt equals negative 1.5 m per second. The minus sign says the top slides down.
List each quantity with its units before you start. That bookkeeping turns a confusing paragraph into a routine calculation. And when your answer comes out negative, trust it: the quantity is shrinking.
Write the link, differentiate with respect to time, then substitute the instant values.
More to read
4.1 Related Rates
Related rates problems link two or more quantities that change with time. You start with a geometry equation that ties them together, such as area equals pi times r squared for a circle. Then you differentiate both sides with respect to time, which brings the chain rule into play and connects the individual rates. The golden rule is to plug in the snapshot values only after differentiating. Substituting too early turns a changing quantity into a constant and its rate silently vanishes. Finish by stating the rate with units, since square units per second and cubic units per second tell different stories.
Key Concepts
Every related rates solution follows the same spine. Draw a picture, name each changing quantity, write the equation that relates them, differentiate with respect to time, and only then substitute the known instant values. The chain rule is the engine: each term gains a rate factor such as dr dt or dh dt. Common relations are Pythagoras for ladders, similar triangles for cones and shadows, and volume formulas for balloons and tanks. A negative answer is meaningful, it says the quantity is shrinking, so never discard a minus sign.
Analyzing problems involving related rates
Think of a related rates problem as a story about linked change. A circle example makes it concrete: the radius grows at 3 cm per second, and when the radius is 8 cm you want the rate of change of the area. You relate area to radius, differentiate to get dA dt equals 2 pi r dr dt, and then insert the numbers to get 48 pi square cm per second. Listing each quantity with its units before you start keeps expressions, values, and units matched up.
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14 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.