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Need solution for RD Sharma maths class 12 chapter Differentials Errors and Approximations exercise 13.1 question 7

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Answer: 2k%

Hint: total surface area of the cone is given by s=\pi r^{2}+\pi r l

Given: \Delta x=\left(\frac{k}{100}\right) \times(x)

            \Delta x=0.01 k x

Solution: Suppose x be the height of the cone and \Delta x be the change in the value of x

⇒ Thus we have \Delta x=\left(\frac{k}{100}\right) \times x

                         \Delta x=0.01 k x

⇒ suppose us assume that radius, the semi height and semi vertical angle of the come to be r, 1 and respectively as shown                        

from above figure, using trigonometry

\tan d=\frac{O B}{O A}=\frac{r}{x}

⇒ we have also

\cos x=\frac{O A}{A B}=\frac{x}{l}

1=\frac{x}{\cos x}

⇒ Now the total surface area of the cone is given by

s=\pi x^{2}+\pi r l

⇒ from the above, we have r=x \tan x \text { and } l=x \sec x

\Rightarrow s=\pi x^{2}+\pi(x \tan x \cdot x \sec x)

⇒ Differentiate s with respect to x

\begin{aligned} &\frac{d s}{d x}=\frac{d}{d x}\left[\pi x^{2} \tan x(\tan x+\sec x)\right] \\\\ &\Rightarrow \frac{d s}{d x}=\pi \tan x(\tan x+\sec x) \frac{1}{d x}\left(x^{2}\right) \end{aligned}

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\\\ &\Rightarrow \frac{d s}{d x}=2 \pi x \tan x(\tan x+\sec x) \end{aligned}

\begin{aligned} &\Rightarrow \text { so, } \Delta x=\left(\frac{d y}{d x}\right) \Delta x \\\\ &\Rightarrow \text { Here } \frac{d s}{d x}=2 \pi x(\tan x+\sec x) \text { and } \Delta x=0.01 k x \\\\ &\Rightarrow \Delta s=(2 \pi x \tan (x)[\tan (x)+\sec (x)](0.01 k x) \end{aligned}

⇒ Percentage of increase in S

\begin{aligned} &\text { Increase }=\frac{\Delta s}{s} \times 100 y \\\\ &=100+\times 0 . .03 k \pi x^{2} \tan x[\tan x+\sec x] \\\\ &\pi x^{2} \tan x(\tan x+\sec x) \end{aligned}

\begin{aligned} &\text { Increase }=0.02 k \times 100 \\\\ &=2 k \% \end{aligned}

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