The slope of the normal to the curve x=sqrt{t | vedclass

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(C) Given the parametric equations of the curve are $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$.First,we find the derivatives with respect to $t$:

Vedclass · Accepted Answer

$\frac{-4}{17}$

Vedclass · Answer

(C) Given the parametric equations of the curve are $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}(t^{1/2}) = \frac{1}{2\sqrt{t}}$$\frac{dy}{dt} = \frac{d}{dt}(t - t^{-1/2}) = 1 - (-\frac{1}{2})t^{-3/2} = 1 + \frac{1}{2t^{3/2}}$Now,the slope of the tangent $\frac{dy}{dx}$ is given by $\frac{dy/dt}{dx/dt}$:$\frac{dy}{dx} = \frac{1 + \frac{1}{2t^{3/2}}}{\frac{1}{2\sqrt{t}}} = (1 + \frac{1}{2t^{3/2}}) 	imes (2\sqrt{t}) = 2\sqrt{t} + \frac{2\sqrt{t}}{2t^{3/2}} = 2\sqrt{t} + \frac{1}{t} = \frac{2t\sqrt{t} + 1}{t}$At $t=4$,the slope of the tangent is:$\left(\frac{dy}{dx}ight)_{t=4} = \frac{2(4)\sqrt{4} + 1}{4} = \frac{2(4)(2) + 1}{4} = \frac{16+1}{4} = \frac{17}{4}$The slope of the normal is the negative reciprocal of the slope of the tangent:$	ext{Slope of normal} = -\frac{1}{(\frac{dy}{dx})_{t=4}} = -\frac{1}{17/4} = -\frac{4}{17}$

The slope of the normal to the curve $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

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