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Exercise 6.1 · Q1

Q.Find the rate of change of the area of a circle with respect to its radius rr when

(a) r=3r = 3 cm
(b) r=4r = 4 cm
Punjab PsebTextbookSubjective· 2mImportance★★★★★
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The rate of change of area with respect to radius is the derivative dA/dr=2πrdA/dr = 2\pi r. At r=3r = 3 cm, it is 6π6\pi cm²/cm; at r=4r = 4 cm, it is 8π8\pi cm²/cm.

The question asks for the rate of change of the area of a circle with respect to its radius. That phrase "rate of change" is a direct signal to use a derivative. When one quantity changes as another changes, the instantaneous rate of change is the derivative of the first with respect to the second.

Here, the area AA depends on the radius rr through the familiar formula A=πr2A = \pi r^2. So the rate of change of AA with respect to rr is simply dAdr\frac{dA}{dr}. This derivative tells us how fast the area grows (in square centimeters) for each tiny increase in radius (in centimeters), at a specific value of rr.

Let’s work through it.

  1. Write the relationship. The area of a circle is

A=πr2.A = \pi r^2.

  1. Differentiate with respect to rr. Since π\pi is a constant,

dAdr=π⋅2r=2πr.\frac{dA}{dr} = \pi \cdot 2r = 2\pi r.

This is the general formula for the rate of change of area with respect to radius. Notice it is not constant — it grows linearly with rr. That makes intuitive sense: if you increase the radius of a large circle by 1 cm, you add a much bigger ring of area than if you increase the radius of a tiny circle by the same amount.

  1. Evaluate at the given values.

    (a) For r=3r = 3 cm:

dAdr∣r=3=2π(3)=6π cm2/cm.\left.\frac{dA}{dr}\right|_{r=3} = 2\pi (3) = 6\pi \text{ cm}^2/\text{cm}.

(b) For r=4r = 4 cm:

dAdr∣r=4=2π(4)=8π cm2/cm.\left.\frac{dA}{dr}\right|_{r=4} = 2\pi (4) = 8\pi \text{ cm}^2/\text{cm}.

Tip

A neat way to check: the derivative 2πr2\pi r is exactly the circumference of the circle. That’s not a coincidence — if you increase the radius by a tiny amount drdr, the added area is a thin ring of length 2πr2\pi r and thickness drdr, so the added area per unit drdr is 2πr2\pi r. This geometric insight matches the calculus result perfectly.

Watch out

A common mistake is to treat the rate of change as the change in area itself (like πr2\pi r^2) rather than the derivative. The question asks for the rate of change, not the area. Always look for the phrase "rate of change" and reach for differentiation.

✓Final answer

The rate of change of area with respect to radius is 6π6\pi cm²/cm at r=3r = 3 cm, and 8π8\pi cm²/cm at r=4r = 4 cm.

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