The length of the subnormal at any point on t | vedclass

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Question

(C) Given the curve $y = (\frac{x}{2024})^k$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = k(\frac{x}{2024})^{k-1} \cdot \frac{1}{2024}

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) Given the curve $y = (\frac{x}{2024})^k$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = k(\frac{x}{2024})^{k-1} \cdot \frac{1}{2024} = \frac{k x^{k-1}}{(2024)^k}$. The length of the subnormal is given by the formula $|y \frac{dy}{dx}|$. Substituting the values,we get $|(\frac{x}{2024})^k \cdot \frac{k x^{k-1}}{(2024)^k}| = |\frac{k x^{2k-1}}{(2024)^{2k}}|$. For the length of the subnormal to be constant,it must be independent of $x$. This implies that the exponent of $x$ must be $0$,so $2k - 1 = 0$. Solving for $k$,we get $k = \frac{1}{2}$.

The length of the subnormal at any point on the curve $y = (\frac{x}{2024})^k$ is constant if the value of $k$ is

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