Find the approximate value of (3.968)^{frac{3 | vedclass

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

Question

(A) Let $y = x^{\frac{3}{2}}$.Consider $x = 4$ and $\Delta x = -0.032$.Then,$\Delta y = (x + \Delta x)^{\frac{3}{2}} - x^{\frac{3}{2}} = (3.968)^{\fra

Vedclass · Accepted Answer

$7.904$

Vedclass · Answer

(A) Let $y = x^{\frac{3}{2}}$.Consider $x = 4$ and $\Delta x = -0.032$.Then,$\Delta y = (x + \Delta x)^{\frac{3}{2}} - x^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} - (4)^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} - 8$.So,$(3.968)^{\frac{3}{2}} = 8 + \Delta y$.Now,$dy$ is approximately equal to $\Delta y$,given by $dy = \left(\frac{dy}{dx}\right) \Delta x$.Since $y = x^{\frac{3}{2}}$,we have $\frac{dy}{dx} = \frac{3}{2} x^{\frac{1}{2}}$.Thus,$dy = \frac{3}{2} (4)^{\frac{1}{2}} (-0.032) = \frac{3}{2} (2) (-0.032) = 3 (-0.032) = -0.096$.Therefore,the approximate value is $8 + (-0.096) = 7.904$.

Find the approximate value of $(3.968)^{\frac{3}{2}}$.

Similar Questions

If the percentage error in the radius of a circle is $3\%$,then the percentage error in its area is:

The approximate value of $\sqrt[3]{28}$ is

Find the approximate value of $\sqrt{25.3}$.

The semi-vertical angle of a right circular cone is $30^{\circ}$. If the height of the cone is $6.125 \text{ cm}$,then the approximate value of the volume of the cone (in cubic $\text{cm}$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module