The slope of the tangent of the curve left(fr | vedclass

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Question

(A) Given the curve equation: $\left(\frac{x}{31}\right)^n + \left(\frac{y}{1209}\right)^n = 2$ Differentiating both sides with respect to $x$: $\frac

Vedclass · Accepted Answer

-$39$

Vedclass · Answer

(A) Given the curve equation: $\left(\frac{x}{31}\right)^n + \left(\frac{y}{1209}\right)^n = 2$ Differentiating both sides with respect to $x$: $\frac{n}{31} \left(\frac{x}{31}\right)^{n-1} + \frac{n}{1209} \left(\frac{y}{1209}\right)^{n-1} \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{\frac{n}{31} \left(\frac{x}{31}\right)^{n-1}}{\frac{n}{1209} \left(\frac{y}{1209}\right)^{n-1}} = -\frac{1209}{31} \left(\frac{x}{31}\right)^{n-1} \left(\frac{1209}{y}\right)^{n-1}$ Since $1209 / 31 = 39$,we have: $\frac{dy}{dx} = -39 \left(\frac{x}{31}\right)^{n-1} \left(\frac{1209}{y}\right)^{n-1}$ At the point $(31, 1209)$,substitute $x = 31$ and $y = 1209$: $\frac{dy}{dx} = -39 \left(\frac{31}{31}\right)^{n-1} \left(\frac{1209}{1209}\right)^{n-1} = -39(1)^{n-1}(1)^{n-1} = -39$ Thus,the slope of the tangent is $-39$.

The slope of the tangent of the curve $\left(\frac{x}{31}\right)^n + \left(\frac{y}{1209}\right)^n = 2$ at $(31, 1209)$ is:

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