What is the length of the normal at any point | vedclass

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(B) Given the curve equation: $y = \frac{a}{2}(e^{x/a} + e^{-x/a}) = a \cosh(\frac{x}{a})$.Find the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = a \cdo

Vedclass · Accepted Answer

$y^2/a$

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(B) Given the curve equation: $y = \frac{a}{2}(e^{x/a} + e^{-x/a}) = a \cosh(\frac{x}{a})$.Find the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = a \cdot \frac{1}{a} \sinh(\frac{x}{a}) = \sinh(\frac{x}{a})$.The formula for the length of the normal is $L_n = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.Substitute the values:$L_n = y \sqrt{1 + \sinh^2(\frac{x}{a})}$.Using the identity $1 + \sinh^2(	heta) = \cosh^2(	heta)$:$L_n = y \sqrt{\cosh^2(\frac{x}{a})} = y \cosh(\frac{x}{a})$.Since $y = a \cosh(\frac{x}{a})$,we have $\cosh(\frac{x}{a}) = \frac{y}{a}$.Therefore,$L_n = y \cdot (\frac{y}{a}) = \frac{y^2}{a}$.

What is the length of the normal at any point on the curve $y = \frac{1}{2}a(e^{x/a} + e^{-x/a})$?

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