"Let P be the vector-valued function P(t)=(sin⁡t+,cos⁡t). (a)..." was answered by our experts

Sherry Becker

Answered question

2021-12-06

Let P be the vector-valued function

P (t) = (\sin t +, \cos t)

.
(a) What curve is traced by P? Draw a picture of this curve.
(b) Draw the vector

P^{'} (\frac{π}{3})

at the point corresponding to

t = \frac{π}{3}

on the picture you produced in part (a).
(c) Find a vector equation for the line tangent to the curve at

t = \frac{π}{3}

Answer & Explanation

Clara Clark

Beginner2021-12-07Added 17 answers

The equation of the curve is obtained by eliminating the parameter t. The result is the circle center at the origin with radius 1.
(a)In stan dard notation
$P (x (t), y (t)) = (\sin t, \cos t)$
$s o, x = x (t) = \sin t, y = y (t) = \cos$
$\Rightarrow x^{2} + y^{2} = 1, (c i r c l e c e n t e r (0, 0), r a d i u s = 1)$
The curve defined by P(t)
b) The tangent at the point on the circle corresponding to $t = \frac{π}{3}$
The vector equation of the tangent . Here P(t) represents the position vector at any point on the tangent and $\Rightarrow$ denotes the scalar product.

(c)vector equation of the tan gent :
At any $t, P (t) - (\sin \frac{π}{3}, \cos \frac{π}{3})$ is
orthogonal to radius vector $(\sin \frac{π}{3}, \cos \frac{π}{3})$
So, the vector or equation is
$< P (t) - (\frac{\sqrt{3}}{2}, \frac{1}{2}), (\frac{\sqrt{3}}{2}, \frac{1}{2}) \geq 0$

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