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Exercise 8.8 · Q1

Q.Find the rate of change of the area of a circle with respect to its radius rr when r=5r = 5 cm.

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✓ Free question

Differentiate the area of a circle with respect to its radius and evaluate at r=5r=5 cm.

Area of a circle of radius rr:

A=πr2A = \pi r^2

with power rule ddr(rn)=nrn−1\dfrac{d}{dr}(r^n) = nr^{n-1}.

  1. Write the area formula.

A=πr2A = \pi r^2

  1. Differentiate AA with respect to rr.

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

  1. Substitute the given value r=5r=5 cm.

dAdr∣r=5=2π(5)=10π\left.\frac{dA}{dr}\right|_{r=5} = 2\pi (5) = 10\pi

  1. Numeric value. Using π≈3.1416\pi \approx 3.1416:

10π≈10×3.1416=31.416 cm2/cm10\pi \approx 10 \times 3.1416 = 31.416 \ \text{cm}^2/\text{cm}

  1. Self-check. Dimensionally, AA is in cm2^2 and rr in cm, so dAdr\frac{dA}{dr} has units cm2^2/cm == cm, consistent with a rate of area-change per unit radius. Also, dAdr=2πr\frac{dA}{dr}=2\pi r is exactly the circumference of the circle at that radius — a well-known geometric check (a thin outer ring of width drdr has area ≈2πr dr\approx 2\pi r\,dr).
✓Final answer

dAdr∣r=5 cm=10π≈31.42 cm2/cm\dfrac{dA}{dr}\Big|_{r=5\text{ cm}} = 10\pi \approx 31.42\ \text{cm}^2/\text{cm}

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