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

The approximate value of $\cot ^{-1}(1.001)$ is

The approximate value of $\sin(60^{\circ} 0^{\prime} 10^{\prime \prime})$ is (given that $\sqrt{3}=1.732, 1^{\circ}=0.0175^{c}$):

There is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere. When the radius is $10 \text{ cm}$, the percentage error in the volume of the sphere is

Electric current $(I)$ is measured by a galvanometer, where the current is proportional to the tangent of the angle $( \theta)$ of deflection. If the deflection is read as $45^{\circ}$ and an error of $1\%$ is made in reading the angle, the percentage error in the current is:

The radius and the height of a right circular solid cone are measured as $7 \text{ ft}$ each. If there is an error of $0.002 \text{ ft}$ for every foot in measuring them,then the error in the total surface area of the cone (in $\text{sq. ft}$) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module