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The rate of change (in cm2/s\mathrm{cm}^{2} / \mathrm{s} ) of the total surface area of a hemisphere with respect to radius rr at r=1.3313 cmr=\sqrt[3]{1.331} \mathrm{~cm} is :

Solution

✅ Correct Option: 2

The total surface area of a hemisphere includes both the curved surface and the circular base:

A=2πr2+πr2=3πr2A = 2\pi r^2 + \pi r^2 = 3\pi r^2


To find the rate of change, differentiate with respect to rr:

dAdr=ddr(3πr2)=3π⋅ddr(r2)=3π⋅2r=6πr\dfrac{dA}{dr} = \dfrac{d}{dr}(3\pi r^2) = 3\pi \cdot \dfrac{d}{dr}(r^2) = 3\pi \cdot 2r = 6\pi r


Evaluate at r=1.3313r = \sqrt[3]{1.331}:

1.3313=1131033=1110=1.1\sqrt[3]{1.331} = \sqrt[3]{\dfrac{11^3}{10^3}} = \dfrac{11}{10} = 1.1 cm


Substitute into our formula:

dAdr=6πr=6π(1.1)=6.6π\dfrac{dA}{dr} = 6\pi r = 6\pi(1.1) = 6.6\pi cm2^2/s


Therefore, the rate of change of the total surface area of the hemisphere with respect to radius at r=1.3313r = \sqrt[3]{1.331} cm is 6.6π6.6\pi cm2^2/s.

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