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Exercise 8.2 · Q5

Q.The trunk of a tree has diameter 3030 cm. During the following year, the circumference grew 66 cm.

(i) Approximately, how much did the tree's diameter grow?
(ii) What is the percentage increase in area of the tree's cross-section?
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Diameter 3030 cm gives radius r0=15r_0=15 cm. The circumference C=2πrC=2\pi r growing by 66 cm pins down drdr; propagate that through the diameter (2r2r) and the cross-sectional area A=πr2A=\pi r^2.

Step 1. Find drdr from the circumference growth. C=2πr⇒dC=2π drC=2\pi r\Rightarrow dC=2\pi\,dr. Given dC=6dC=6: dr=62π=3π≈0.9549dr=\dfrac{6}{2\pi}=\dfrac3\pi\approx0.9549 cm.

Step 2. Part (i): growth in diameter. Diameter =2r=2r, so d(diameter)=2 dr=2(3π)=6π≈1.9099d(\text{diameter})=2\,dr=2\left(\dfrac3\pi\right)=\dfrac6\pi\approx1.9099 cm.

Step 3. Part (ii): change in cross-sectional area. A(r)=πr2⇒dA=2πr drA(r)=\pi r^2\Rightarrow dA=2\pi r\,dr. At r0=15r_0=15: dA=2π(15)(3π)=2(15)(3)=90dA=2\pi(15)\left(\dfrac3\pi\right)=2(15)(3)=90 cm2^2 (the π\pi cancels cleanly). …

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