The radius of a cone of height 9 text{ units} | vedclass

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(D) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$. Given $h = 9$,we have $V = \frac{1}{3} \pi r^2 (9) = 3 \pi r^2$. Exact change in vol

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$(1.4832) \pi, (1.44) \pi$

Vedclass · Answer

(D) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$. Given $h = 9$,we have $V = \frac{1}{3} \pi r^2 (9) = 3 \pi r^2$. Exact change in volume: $\Delta V = V(2.12) - V(2) = 3 \pi (2.12)^2 - 3 \pi (2)^2 = 3 \pi (4.4944 - 4) = 3 \pi (0.4944) = 1.4832 \pi$. Approximate change in volume: $dV = \frac{dV}{dr} \Delta r$. Since $V = 3 \pi r^2$,$\frac{dV}{dr} = 6 \pi r$. At $r = 2$ and $\Delta r = 2.12 - 2 = 0.12$,$dV = 6 \pi (2) (0.12) = 12 \pi (0.12) = 1.44 \pi$. Thus,the exact change is $1.4832 \pi$ and the approximate change is $1.44 \pi$.

The radius of a cone of height $9 \text{ units}$ is changed from $2 \text{ units}$ to $2.12 \text{ units}$. The exact change and approximate change in the volume of the cone are respectively:

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