Using differentials,find the approximate valu | vedclass

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

Question

(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} =

Vedclass · Accepted Answer

$0.677$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = \left(\frac{17}{81}ight)^{\frac{1}{4}} - \left(\frac{16}{81}ight)^{\frac{1}{4}} = \left(\frac{17}{81}ight)^{\frac{1}{4}} - \frac{2}{3}$.Thus,$\left(\frac{17}{81}ight)^{\frac{1}{4}} = \frac{2}{3} + \Delta y$.Using the differential $dy \approx \Delta y$,we have $dy = \frac{dy}{dx} \Delta x = \frac{1}{4} x^{-\frac{3}{4}} \Delta x$.Substituting $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$:$dy = \frac{1}{4 \left(\frac{16}{81}ight)^{\frac{3}{4}}} 	imes \frac{1}{81} = \frac{1}{4 	imes \frac{8}{27}} 	imes \frac{1}{81} = \frac{27}{32} 	imes \frac{1}{81} = \frac{1}{32 	imes 3} = \frac{1}{96} \approx 0.0104$.Therefore,the approximate value is $\frac{2}{3} + 0.0104 = 0.6667 + 0.0104 = 0.6771 \approx 0.677$.

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

Similar Questions

If the radius of a sphere is measured as $9 \text{ cm}$ with an error of $0.03 \text{ cm}$,then find the approximate error in calculating its volume. (in $\pi \text{ cm}^3$)

Find the approximate value of $(15)^{\frac{1}{4}}$.

The angle $A$ of $\triangle ABC$ is found by measurement to be $67 \frac{1}{2}^{\circ}$ and the area of $\triangle ABC$ is calculated from the measurements of $b, c, A$. In measuring $A$,an error of $9 \text{ min}$ is made. Then the percentage error in the area of the triangle is

Electric current is measured by a tangent 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 it,then the percentage error in the current is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module