#### Provide solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise  13.1 question 6

Hint: Here we use the basic concept of area and logarithm

Given:$pv^{1.4}$ =constant

Solution: Given as $pv^{1.4}$=constant and the decrease in  $v$ is  $\frac{1}{2}%$

$\Rightarrow$ Thus, we have    $\Delta v=\left(\frac{-1}{2}\right) 100 x v$

So, $\Delta v=-0.005 v$

$\Rightarrow$ Now, $pv^{1.4}$ =constant

taking log on both sides,

$\log \left(p v^{1.4}\right)=\log (\text { constant })$

$\Rightarrow$ Differentiate both sides with respect to v

\begin{aligned} &\frac{d}{d p}(\log p) x \frac{d p}{d v}+\frac{d}{d v}(1.4 \log v)=0 \\\\ &\Rightarrow \frac{d}{d p}(\log p) x \frac{d p}{d v}+1.4 \frac{d}{d v}(\log v)=0 \end{aligned}

\begin{aligned} &\Rightarrow \frac{d}{d x}(\log x)=\frac{1}{x} \\\\ &\Rightarrow \frac{d}{d v}=\frac{-1.4}{v} P \end{aligned}

\begin{aligned} &\Delta p=\left(\frac{-1.4}{v} p\right)(-0.005 x) \\\\ &\Delta p=0.007 p \end{aligned}

$\Rightarrow$ Percentage of error is

$\text { Error }=\frac{0.007 p}{p} \times 100=0.7 \%$