Find the equations of the tangent and normal | vedclass

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Question

(A) Given the curve $x = \cos t$ and $y = \sin t$. At $t = \frac{\pi}{4}$,$x = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $y = \sin(\frac{\pi}{4})

Vedclass · Accepted Answer

Tangent: $x + y - \sqrt{2} = 0$,Normal: $x - y = 0$

Vedclass · Answer

(A) Given the curve $x = \cos t$ and $y = \sin t$. At $t = \frac{\pi}{4}$,$x = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $y = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. The point is $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Now,$\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$. The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{-\sin t} = -\cot t$. At $t = \frac{\pi}{4}$,the slope $m = -\cot(\frac{\pi}{4}) = -1$. The equation of the tangent is $y - \frac{1}{\sqrt{2}} = -1(x - \frac{1}{\sqrt{2}})$,which simplifies to $x + y - \sqrt{2} = 0$. The slope of the normal is $m' = -\frac{1}{m} = -\frac{1}{-1} = 1$. The equation of the normal is $y - \frac{1}{\sqrt{2}} = 1(x - \frac{1}{\sqrt{2}})$,which simplifies to $x - y = 0$.

Find the equations of the tangent and normal to the given curve at the indicated point: $x = \cos t, y = \sin t$ at $t = \frac{\pi}{4}$.

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