Question

A sphérical báll of salt is dissolving in water in such a manner that the ráte of décrease of the volume at any iñstant is propôrtional to the surface. Prove that the radius is decréasing at a coñstant rate.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that radius of the spherical ball be r units.

We know,

Volume (V) and Surface Area (S) of sphere of radius r is given by

$\boxed{\sf{ \: \: V \: = \: \frac{4}{3} \: \pi \: {r}^{3} \: \: }}$

and

$\boxed{\sf{ \: \: S \: = \: 4 \: \pi \: {r}^{2} \: \: }}$

Now, According to statement, It is given that

The ráte of décrease of the volume at any iñstant is propôrtional to the surface area.

So,

$\rm \: \dfrac{dV}{dt} = - k \: S$

[ - ve sign is because V is decreasing]

where k is constant of proportionality and k > 0.

So, on substituting the values of S and V, we get

$\rm \: \dfrac{d}{dt}\bigg(\dfrac{4}{3}\pi \: {r}^{3} \bigg) = - k \: (4\pi \: {r}^{2})$

$\rm \: \dfrac{4}{3}\pi \: ({3r}^{2})\dfrac{dr}{dt} = - k \: (4\pi \: {r}^{2})$

$\rm \: 4\pi \: ({r}^{2})\dfrac{dr}{dt} = - k \: (4\pi \: {r}^{2})$

$\rm\implies \:\dfrac{dr}{dt} \: = - \: k \:$

$\rm\implies \:r \: decreases \: with \: constant \: rate.$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Spherical ball of radius 'r'.

$\sf\dfrac{d V}{d t}=-k A$ [- ve sign because volume is decreasing]

Where 'V' Is Volume And 'A' Is Surface Area.

$\leadsto\bf V=\dfrac{4}{3} \pi r^{3}, \leadsto A=4 \pi r^{2}$

$\leadsto\bf{\dfrac{d V}{d t}=\dfrac{4}{3} \pi \dfrac{d}{d t}\left(r^{3}\right)=\dfrac{4 \pi}{3} 3 r^{2} \dfrac{d r}{d t}=4 \pi r^{2} \dfrac{d r}{d t} }$

$\[ \begin{aligned} \bf 4 \leadsto\pi r^{2} \dfrac{d r}{d t} & \bf=-k\left(4 \pi r^{2}\right) \\ \leadsto \bf \cfrac{d r}{d t} & \bf=-k \end{aligned} \]$

Hence radius is decreasing at a constant rate.