The equation of the tangent to the curve y=pi | vedclass

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(D) Given the equation of the curve is $y=\pi e^{\frac{-x}{\pi}}$.To find the point where the curve crosses the $y$-axis, we set $x=0$.Substituting $x

Vedclass · Accepted Answer

$x+y=\pi$

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(D) Given the equation of the curve is $y=\pi e^{\frac{-x}{\pi}}$.To find the point where the curve crosses the $y$-axis, we set $x=0$.Substituting $x=0$ into the equation, we get $y = \pi e^{0} = \pi$.So, the point of contact is $(0, \pi)$.Now, differentiate the equation with respect to $x$ to find the slope of the tangent:$\frac{dy}{dx} = \pi \cdot e^{\frac{-x}{\pi}} \cdot \left(-\frac{1}{\pi}\right) = -e^{\frac{-x}{\pi}}$.At the point $(0, \pi)$, the slope $m$ is:$m = \left. \frac{dy}{dx} \right|_{(0, \pi)} = -e^{0} = -1$.The equation of the tangent line at $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.Substituting the values, we get $y - \pi = -1(x - 0)$.$y - \pi = -x$, which simplifies to $x + y = \pi$.

The equation of the tangent to the curve $y=\pi e^{\frac{-x}{\pi}}$ at the point where it crosses the $y$-axis is

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