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

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

Vedclass · Accepted Answer

$-\frac{4}{17}$

Vedclass · Answer

(C) Given the parametric equations of the curve: $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = \frac{1}{2\sqrt{t}}$$\frac{dy}{dt} = 1 - (-\frac{1}{2})t^{-\frac{3}{2}} = 1 + \frac{1}{2t\sqrt{t}}$Now,calculate the slope of the tangent $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:$\frac{dy}{dx} = \frac{1 + \frac{1}{2t\sqrt{t}}}{\frac{1}{2\sqrt{t}}} = (1 + \frac{1}{2t\sqrt{t}}) 	imes 2\sqrt{t} = 2\sqrt{t} + \frac{1}{t}$.At $t = 4$:$\frac{dy}{dx} = 2\sqrt{4} + \frac{1}{4} = 2(2) + 0.25 = 4 + 0.25 = 4.25 = \frac{17}{4}$.The slope of the normal is given by $-\frac{1}{	ext{slope of tangent}}$:Slope of normal $= -\frac{1}{17/4} = -\frac{4}{17}$.

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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