The radius of the base of a cone is increasin | vedclass

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(D) Let $r$ be the radius and $h$ be the altitude of the cone. The slant height $l$ is given by $l = \sqrt{r^2 + h^2}$.Given: $\frac{dr}{dt} = 3 	ext

Vedclass · Accepted Answer

$54 \pi 	ext{ cm}^2/	ext{min}$

Vedclass · Answer

(D) Let $r$ be the radius and $h$ be the altitude of the cone. The slant height $l$ is given by $l = \sqrt{r^2 + h^2}$.Given: $\frac{dr}{dt} = 3 	ext{ cm/min}$ and $\frac{dh}{dt} = -4 	ext{ cm/min}$.The lateral surface area $S$ of a cone is $S = \pi rl = \pi r \sqrt{r^2 + h^2}$.At $r = 7$ and $h = 24$,$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 	ext{ cm}$.Differentiating $S$ with respect to $t$:$\frac{dS}{dt} = \pi \left( \frac{dr}{dt} \sqrt{r^2 + h^2} + r \cdot \frac{1}{2\sqrt{r^2 + h^2}} \cdot (2r \frac{dr}{dt} + 2h \frac{dh}{dt}) ight)$.$\frac{dS}{dt} = \pi \left( 3 \cdot 25 + 7 \cdot \frac{1}{25} \cdot (7 \cdot 3 + 24 \cdot (-4)) ight)$.$\frac{dS}{dt} = \pi \left( 75 + \frac{7}{25} \cdot (21 - 96) ight) = \pi \left( 75 + \frac{7}{25} \cdot (-75) ight)$.$\frac{dS}{dt} = \pi (75 - 7 \cdot 3) = \pi (75 - 21) = 54 \pi 	ext{ cm}^2/	ext{min}$.

The radius of the base of a cone is increasing at the rate $3 \text{ cm/min}$ and the altitude is decreasing at the rate $4 \text{ cm/min}$. The rate at which the lateral surface area is changing,when the radius is $7 \text{ cm}$ and altitude is $24 \text{ cm}$,is:

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