If x^k + y^k = a^k (a, k 0) and frac{dy}{dx | vedclass

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(B) Given equation: $x^k + y^k = a^k$ Differentiating with respect to $x$: $k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{k x^{k-1}

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Given equation: $x^k + y^k = a^k$ Differentiating with respect to $x$: $k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{k x^{k-1}}{k y^{k-1}} = -\frac{x^{k-1}}{y^{k-1}} = -(\frac{x}{y})^{k-1}$ $\frac{dy}{dx} = -(\frac{y}{x})^{-(k-1)} = -(\frac{y}{x})^{1-k}$ So,$\frac{dy}{dx} + (\frac{y}{x})^{1-k} = 0$ Comparing this with the given equation $\frac{dy}{dx} + (\frac{y}{x})^{\frac{1}{3}} = 0$: $1 - k = \frac{1}{3}$ $k = 1 - \frac{1}{3} = \frac{2}{3}$

If $x^k + y^k = a^k$ $(a, k > 0)$ and $\frac{dy}{dx} + (\frac{y}{x})^{\frac{1}{3}} = 0$,then $k$ has the value

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