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Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (xx) maths textbook solution

Answers (1)

Answer: 0.693

Hint: \Delta y=f(x+\Delta x)-f(x)

            we use this formula

Given:\sqrt{0.48}

Solution:  consider the function y=f(x)=\sqrt{x}

Let

        \begin{aligned} &x=0.49 \\\\ &x+\Delta x=0.48 \end{aligned}

Then,

        \begin{aligned} &\Delta x=-0.01 \\\\ &\Rightarrow \text { for } x=0.49 \\\\ &y=\sqrt{0.19}=0.7 \end{aligned}

Let,

        \begin{aligned} &d x=\Delta x=0.01 \\\\ &\Rightarrow \text { now } y=(x)^{\frac{1}{2}} \\\\ &\frac{d y}{d x}=\frac{1}{2 \sqrt{2 x}} \\\\ &\left(\frac{d y}{d x}\right)_{x=0.49}=\frac{1}{1.4} \end{aligned}

        \begin{aligned} &\Delta y=d y=\frac{d y}{d x} \times d x=\frac{1}{1.4} \times(-0.01) \\\\ &\Delta y=-0.007143 \\\\ &\sqrt{0.48}=y+\Delta y=0.693 \end{aligned}


 

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