The radius of the base of a cone is increasing at the rate of 3 cm/minute
Question:

The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude 24 cm is

(a) $54 \pi \mathrm{cm}^{2} / \mathrm{min}$

(b) $7 \pi \mathrm{cm}^{2} / \mathrm{min}$

(c) $27 \mathrm{~cm}^{2} / \mathrm{min}$

 

(d) none of these

Solution:

Let $r$ be the radius, $h$ be the height and $S$ be the lateral surface area of the cone at any time $t .$

Given : $\frac{d r}{d t}=3 \mathrm{~cm} / \min$ and $\frac{d h}{d t}=-4 \mathrm{~cm} / \min$

Here,

$l^{2}=h^{2}+r^{2}$

$\Rightarrow l=\sqrt{(24)^{2}+(7)^{2}}$

$\Rightarrow l=\sqrt{625}$

$\Rightarrow l=25$

$S=\pi r l$

$\Rightarrow S^{2}=(\pi r l)^{2}$

$\Rightarrow S^{2}=\pi^{2} r^{2}\left(h^{2}+r^{2}\right)$

$\Rightarrow S^{2}=\pi^{2} r^{4}+\pi^{2} h^{2} r^{2}$

$\Rightarrow 2 S \frac{d S}{d t}=4 \pi^{2} r^{3} \frac{d r}{d t}+2 \pi^{2} r^{2} h \frac{d h}{d t}+2 \pi^{2} h^{2} r \frac{d r}{d t}$

$\Rightarrow 2 \pi r l \frac{d S}{d t}=2 \pi^{2} r h\left[\frac{2 r^{2}}{h} \frac{d r}{d t}+r \frac{d h}{d t}+h \frac{d r}{d t}\right]$

$\Rightarrow 25 \frac{d S}{d t}=24 \pi\left[\frac{2(7)^{2}}{24} \times 3-7 \times 4+24 \times 3\right]$                [Given : $r=7, h=24$ ]

$\Rightarrow 25 \frac{d S}{d t}=24 \pi\left[\frac{49}{4}-28+72\right]$

$\Rightarrow 25 \frac{d S}{d t}=24 \pi\left[\frac{49+288-112}{4}\right]$

$\Rightarrow \frac{d S}{d t}=24 \pi\left[\frac{225}{100}\right]$

$\Rightarrow \frac{d S}{d t}=24 \pi(2.25)$

$\Rightarrow \frac{d S}{d t}=54 \pi \mathrm{cm}^{2} / \mathrm{sec}$

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