The approximate change in the volume of a cub | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The volume of a cube $(V)$ with side $x$ is given by $V = x^{3}$.The approximate change in volume $(dV)$ is given by the formula $dV = \left(\frac

Vedclass · Accepted Answer

$0.09$

Vedclass · Answer

(A) The volume of a cube $(V)$ with side $x$ is given by $V = x^{3}$.The approximate change in volume $(dV)$ is given by the formula $dV = \left(\frac{dV}{dx}ight) \Delta x$.First,differentiate $V$ with respect to $x$: $\frac{dV}{dx} = 3x^{2}$.Given that the side increases by $3 \%$,the change in side $\Delta x = 3 \% 	ext{ of } x = 0.03x$.Now,substitute these values into the formula for $dV$:$dV = (3x^{2}) 	imes (0.03x)$$dV = 0.09x^{3} 	ext{ m}^{3}$.Thus,the approximate change in the volume of the cube is $0.09x^{3} 	ext{ m}^{3}$.

The approximate change in the volume of a cube of side $x$ metres caused by increasing the side by $3 \%$ is (in $x^{3} \text{ m}^{3}$)

Similar Questions

Given $f(x) = x^3 - 4x$,if $x$ changes from $2$ to $1.99$,then the approximate change in the value of $f(x)$ is

If $f(x)=3x^{2}+15x+5$,then the approximate value of $f(3.02)$ is (in $.66$)

Using differentials,find the approximate value of $\left(\frac{17}{81}\right)^{\frac{1}{4}}$.

Using differentials,find the approximate value of $(33)^{-\frac{1}{5}}$.

Use differentials to approximate $\sqrt{36.6}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module