The distance, from the origin, of the normal to the curve, $x = 2 cos t + 2 t sin t, y = 2 sin t - 2 t cos t$ at $t = \frac{π}{4}$ , is

2

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4

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\sqrt{2}

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2 \sqrt{2}

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Solution

The correct option is A $2$
$x = 2 cos t + 2 t sin t$
$\frac{d x}{d t} = 2 t cos t + 2 sin t - 2 sin t = 2 t cos t$
$y = 2 sin t - 2 t cos t$
$\frac{d y}{d t} = 2 cos t + 2 t sin t - 2 cos t = 2 t sin t$
$\frac{d y}{d x} = \frac{2 t sin t}{2 t cos t} (t = \frac{π}{4})$
$\frac{d y}{d x} = 1$

$\frac{d x}{d y} = - 1$ for slope of normal.
$x (\frac{π}{4}) = 2 cos (\frac{π}{4}) + 2. \frac{π}{4} sin (\frac{π}{4}) = \sqrt{2} (1 + \frac{π}{4})$
$y (\frac{π}{4}) = 2 sin (\frac{π}{4}) - 2. \frac{π}{4} cos (\frac{π}{4}) = \sqrt{2} (1 - \frac{π}{4})$
Equation of normal with given point
$x + y - 2 \sqrt{2} = 0$
$d = \frac{(2 \sqrt{2)}}{\sqrt{2}} = 2$

Suggest Corrections

Similar questions

x = 2 cos t + 2 t sin t, y = 2 sin t - 2 t cos t, at t = \frac{π}{4}

x = 2 cos t + 2 t sin t,

y = 2 sin t - 2 t cos t

t = \frac{π}{4}

x = 2 cos t + 2 t sin t, y = 2 sin t - 2 t cos t, at t = \frac{π}{4}

t = \frac{π}{2}

x = 2 sin t, y = 2 cos t

x = 2 cos t - cos 2 t

y = 2 sin t - sin 2 t

y

(t = \frac{π}{2})

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