Question

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (Enter your answer as a comma-separated list of equations. Let x and y be in terms of t.) (a) Once around clockwise, starting at (4, 1). 0 ≤ t ≤ 2π. (b) Two times around counterclockwise, starting at (4, 1). 0 ≤ t ≤ 4π. (c) Halfway around counterclockwise, starting at (0, 5). 0 ≤ t ≤ π.

Answer 1

Answer:

a) $x = 4\cdot \cos t$, $y = 1 + 4\cdot \sin t$, b) $x = 4\cdot \cos t$, $y = 1 + 4\cdot \sin t$, c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$, $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$.

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$, $y = 1 + 4\cdot \sin t$.

b) $x = 4\cdot \cos t$, $y = 1 + 4\cdot \sin t$.

c) $x = 4\cdot \cos t''$, $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$