Question

The slope of the tangent to the curve y = be^−x/a where it crosses y-axis is ______________.

Answer 1

The equation of given curve is

$y=b{e}^{-\frac{x}{a}}$ .....(1)

It crosses the y-axis at the point, where x = 0.

Putting x = 0 in (1), we get

$y=b{e}^{-\frac{0}{a}}$

$\Rightarrow y=b{e}^0$

$\Rightarrow y=b$

So, the coordinates of the point where the given curve crosses y-axis is (0, b).

$y=b{e}^{-\frac{x}{a}}$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=b{e}^{-\frac{x}{a}}\times \left(-\frac{1}{a}\right)$

$\Rightarrow \frac{dy}{dx}=-\frac{b}{a}{e}^{-\frac{x}{a}}$

∴ Slope of the tangent at (0, b) $={\left(\frac{dy}{dx}\right)}_{\left(0,b\right)}$ $=-\frac{b}{a}{e}^{-\frac{0}{a}}$ $=-\frac{b}{a}{e}^0$ $=-\frac{b}{a}$

Thus, the slope of tangent to the given curve $y=b{e}^{-\frac{x}{a}}$ where it crosses y-axis i.e. (0, b) is $-\frac{b}{a}$.


The slope of the tangent to the curve $y=b{e}^{-\frac{x}{a}}$ where it crosses y-axis is $\underline{-\frac{b}{a}}$.