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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$x+y=\pi$

Vedclass · Answer

(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

Similar Questions

If the normal drawn at a point $P$ on the curve $3y = 6x - 5x^3$ passes through $(0,0)$,then the positive integral value of the abscissa of the point $P$ is

The equation of the normal to the curve $xy + 7 = 0$ is $Ax + By + C = 0$. Which of the following is true?

What are the lengths of the intercepts on the coordinate axes made by the tangent to the curve $y = 2x^2 + 3x - 2$ at the point $(1, 3)$?

If the tangent at a point $P$ on the curve $y=4x^4+x$ is perpendicular to the tangent to the same curve at $(0,0)$,then the point $P$ is

If the normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\left(\frac{3 \pi}{4}\right)^{C}$ with the positive $X$-axis,then $f^{\prime}(3)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module