Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $2\%$ .

It is given that $\frac{\Delta x}{x} \times 100 =2$ We have, $V = x^{3}$ $\Rightarrow \frac{dV}{dx} = 3x^{2}$ $\therefore \Delta V = \frac{dV}{dx} \times\Delta x$ $\Rightarrow \Delta V = 3x^{2}\,\Delta x$ $\Rightarrow \Delta V = 3x^{2} \times \frac{2x}{100}$ $\Rightarrow \Delta V = 0.06x^{2}$ Thus, the approximate change in volume is $0.06x^3\, m^3$ .

Find the approximate change in the volume $V$ of a

Notes on Application of derivatives

Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by

$\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}$

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

If for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≤ f(x₂); then the function f(x) is known as increasing in this interval.
Likewise, if for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≥ f(x₂); then the function f(x) is known as decreasing in this interval.
The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x₁) < f(x₂) for strictly increasing and f(x₁) > f(x₂) for strictly decreasing.

Read More: Application of Derivatives

Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $2\%$.

The Correct Option is D

Solution and Explanation

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