Question

#Quality Question

#Applicaton of Derivative.


A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at the rate of 50 cm^3/min. When the thickness of ice is 5 cm then the rate(in cm/min) at which of the thickness of ice decreases is ​

Answer 1

$\purple{ \qquad \qquad \underline{ \pmb{ \mathbb{ \maltese \: \: \:ANSWER = \dfrac{1}{18\pi} }}}}$

$\huge\bold{ \mathfrak{Given \: that}}$

A spherical iron ball of 10 cm and radius is coated with the layer of ice of uniform thickness,

Let the thickness is x cm.

$\huge\bold{ \mathfrak{then}}$

$\mathcal Volume \: \: of \: \: the \: \: ball \: \: is \\ \\ V = \frac{4}{3} \pi(10 + x) {}^{3 }$

On differentiating both side w.r.t t we get,

$\dfrac{dV}{dt} = 4\pi(1 0 + x) {}^{2} \dfrac{dx}{dt}$

where t is time in min

$\huge\bold{ \mathfrak{Given }} \\ \\ \frac{dV}{dt } = - 50 {cm}^{3} / min$

$\huge\bold{ \mathfrak{Now}}$

When x = 5cm then

$\bf - 50 = 4\pi(10 + 5) {}^{2} \dfrac{dx}{dt} \\ \\ \implies \sf \: \frac{dx}{dt} = - \frac{50}{4\pi(225)} \\ \\ \implies \bf \dfrac{dx}{dt} - \frac{1}{18\pi}$

Negative sign indicates that the thickness of ice layer decreases with time.

Answer 2

Let the thickness is x cm.

So,

$Volume \: \: of \: \: the \: \: ball \: \: is \\ \\ V = \frac{4}{3} \pi(10 + x) {}^{3 }$

Differentiating both side w.r.t t

$\dfrac{dV}{dt} = 4\pi(1 0 + x) {}^{2} \dfrac{dx}{dt}$

${ {Given }} \\ \\ \frac{dV}{dt } = - 50 {cm}^{3} / min$

So

When x = 5cm then

$\ - 50 = 4\pi(10 + 5) {}^{2} \dfrac{dx}{dt} \\ \\ \implies \sf \: \frac{dx}{dt} = - \frac{50}{4\pi(225)} \\ \\ \implies \ \dfrac{dx}{dt} - \frac{1}{18\pi}$