Practicing Success
Using differentials the approximate value of $\sqrt{25.2}$, is |
5.01 5.02 5.03 none of these |
5.02 |
Consider the function $y=f(x)=\sqrt{x}$. Let $x=25$ and $x+\Delta x=25.2$. Then, $\Delta x=25.2-25=0.2$ For $x=25$, we have $y=\sqrt{25}=5$ [Putting x = 25 in y = $\sqrt{x}$] Let $d x=\Delta x=0.2$ Now, $y =\sqrt{x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow\left(\frac{d y}{d x}\right)_{x=25}=\frac{1}{2(5)}=\frac{1}{10}$ ∴ $d y =\frac{d y}{d x} d x$ $\Rightarrow d y=\frac{1}{10}(0.2)=0.02 \Rightarrow \Delta y=0.02$ $[∵ \Delta y \cong d y]$ Hence, $\sqrt{25.2}=y+\Delta y=5+0.02=5.02$ |