Question

Find the tangent and normal to the curve $y=2^{\frac{-x}{3}}$ at the point where the curve meets the Y-axis

Answer 1

Given $y=2^{\frac{-x}{3}}$ ,

$\frac{dy}{dx}=ln\left(2\right)\left(2^{\frac{-x}{3}}\right)\frac{d}{dx}\left(\frac{-x}{3}\right)$ Applying exponential functional rule,

$=-\frac{ln2}{3.2^{x/3}}$

The point where at the curve meets y axis, we substitute $x=0$ in the first equation gives $y=1$

The slope `m' of the curve at the point $(0,1)$ is given as $-\frac{ln\left(2\right)}{3}$.

The equation of tangent at $(x_1,y_1)$ is

$(y-y_1)=m(x-x_1)$

The equation of normal at $(x_1,y_1)$ is

$(y-y_1)=-\frac{1}{m}(x-x_1)$, $-1/m$ is the slope of normal

the equation at the points $(0,1)$ for slope $-\frac{ln\left(2\right)}{3}$.

$y-1=-\frac{ln\left(2\right)}{3}x$ (Equation of tangent)

$y-1=\frac{3}{ln\left(2\right)}x$ (Equation of normal)