The angle at which the curve y = K e^{Kx} int | vedclass

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(B) The curve is given by $y = K e^{Kx}$.To find the slope of the tangent at the point of intersection with the $y$-axis,we set $x = 0$.At $x = 0$,$y

Vedclass · Accepted Answer

$\cot^{-1}(k^2)$

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(B) The curve is given by $y = K e^{Kx}$.To find the slope of the tangent at the point of intersection with the $y$-axis,we set $x = 0$.At $x = 0$,$y = K e^{K(0)} = K$.The derivative is $\frac{dy}{dx} = K \cdot K e^{Kx} = K^2 e^{Kx}$.At $x = 0$,the slope $m = \left. \frac{dy}{dx} ight|_{x=0} = K^2 e^0 = K^2$.Let $\psi$ be the angle the tangent makes with the positive $x$-axis. Then $	an \psi = m = K^2$.The angle $	heta$ that the curve makes with the $y$-axis is the complement of the angle $\psi$ that the tangent makes with the $x$-axis,so $	heta = \frac{\pi}{2} - \psi$.Thus,$	an 	heta = 	an(\frac{\pi}{2} - \psi) = \cot \psi = \frac{1}{	an \psi} = \frac{1}{K^2}$.Therefore,$	heta = 	an^{-1}(\frac{1}{K^2}) = \cot^{-1}(K^2)$.Comparing this with the given options,the correct option is $B$.

The angle at which the curve $y = K e^{Kx}$ intersects the $y$-axis is

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