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C14 · Unit C4 Contextual derivatives

Related rates: differentiate the relationship

CoreReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 4.4–4.5
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

Choose time-dependent variables, write a geometric relationship, differentiate with respect to time, then substitute the instant’s values. The order matters: substituting a changing radius before differentiation destroys its rate. Include units and signs, and verify the shape or constraint actually remains valid during the motion.

Worked example

For a sphere, V equals 4 pi r cubed /3 gives dV over dt equals 4 pi r squared dr over dt. At r equals 3 cm with dr over dt equals 0.2 cm over s, the volume increases at 7.2 pi cm cubed /s. The radius value alone could not determine this rate without dr over dt.

Explore the mathematics

Related rates: differentiate the relationship

Link a changing radius to the rate of change of circle area.

Setting 3 of 5

Scroll sideways to see the full diagram.

How does a changing radius affect area?xy12345A=πr²; radius increases at 1 unit/time

Predict before continuing

How does a changing radius affect area?

Read the complete explanation
  1. How does a changing radius affect area?
  2. Differentiate the relationship with respect to time, not just r.
  3. dA/dt=2πr·dr/dt; the radius rate is another factor.
  4. At r=4 and dr/dt=1, dA/dt=25.1327 square units/time.
Related rates: differentiate the relationship · visual worked method
Related rates: differentiate the relationship · visual worked methodFor a sphere, V=4πr³/3 gives dV/dt=4πr² dr/dt. At r=3 cm with dr/dt=0.2 cm/s, the volume increases at 7.2π cm³/s. The radius value alone could not determine this rate without dr/dt.Worked example · 1/2For a sphere, V=4πr³/3 givesdV/dt=4πr² dr/dt. At r=3 cm withdr/dt=0.2 cm/s, the volumeincreases at 7.2π cm³/s. TheradiusWorked example · 2/2value alone could not determinethis rate without dr/dt.

For a sphere, V=4πr³/3 gives dV/dt=4πr² dr/dt. At r=3 cm with dr/dt=0.2 cm/s, the volume increases at 7.2π cm³/s. The radius value alone could not determine this rate without dr/dt.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

A negative rate can be physically meaningful: it describes shrinkage or motion in the negative direction.

Investigate and explain

Predict the top’s speed as a ladder approaches horizontal, then examine why a real ladder eventually violates the ideal model.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.