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Exercise 6.1 · Q6

Q.The radius of a circle is increasing at the rate of 0.7 cm/s0.7 \text{ cm/s}. What is the rate of increase of its circumference?

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The circumference of a circle is C=2πrC = 2\pi r, so its rate of change is directly proportional to the rate of change of the radius. Given drdt=0.7 cm/s\frac{dr}{dt} = 0.7 \text{ cm/s}, the circumference increases at dCdt=2π⋅0.7=1.4π cm/s\frac{dC}{dt} = 2\pi \cdot 0.7 = 1.4\pi \text{ cm/s}.

This is a classic Related Rates problem — you’re given how fast one quantity (the radius) changes, and you need to find how fast another quantity (the circumference) changes, both with respect to time. The key is that the two quantities are linked by a geometric formula, so differentiating that formula with respect to time gives the relationship between their rates.

Let’s walk through it.

  1. Write the relationship between circumference and radius. For a circle, circumference CC and radius rr are connected by

C=2πr.C = 2\pi r.

This is a direct proportionality — double the radius, double the circumference.

  1. Differentiate both sides with respect to time tt. Since rr changes with time, CC also changes with time. Differentiate implicitly:

dCdt=2πdrdt.\frac{dC}{dt} = 2\pi \frac{dr}{dt}.

Notice that 2π2\pi is a constant, so the rate of change of circumference is simply 2π2\pi times the rate of change of the radius.

  1. Plug in the given rate. You’re told drdt=0.7 cm/s\frac{dr}{dt} = 0.7 \text{ cm/s}. Substitute:

dCdt=2π×0.7=1.4π cm/s.\frac{dC}{dt} = 2\pi \times 0.7 = 1.4\pi \text{ cm/s}.

  1. Interpret the result. …

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