Find parametric equations for the path of a particle that moves along the circle described by $x^2 + (y - 1)^2 = 16$ 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 $(4, 1)$. $0 \leq t \leq 2\pi$.
(b) Four times around counterclockwise, starting at $(4, 1)$. $0 \leq t \leq 8\pi$.
(c) Halfway around counterclockwise, starting at $(0, 5)$. $0 \leq t \leq \pi$.

Neetesh Kumar | January 3, 2025

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This is the solution to Math 1c
Assignment: 10.1 Question Number 9
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Step-by-step solution:

Understanding the Circle and Its Parametrization

The given circle is described by the equation:

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

This is the equation of a circle centered at $(0, 1)$ with a radius of $4$.

The general parametric equations for a circle are:

• $x = r\cos(t) + h$
• $y = r\sin(t) + k$

where $(h, k)$ is the center of the circle, and $r$ is the radius.

In this case, $h = 0$, $k = 1$, and $r = 4$. The parametric equations become:

• $x = 4\cos(t)$
• $y = 4\sin(t) + 1$

The direction (clockwise or counterclockwise) is determined by how $t$ progresses.

Part (a) Once Around Clockwise

For clockwise motion, the angle decreases with $t$.

Thus, we use $-t$ in the cosine and sine functions.

The parametric equations are:

• $x = 4\cos(-t)$
• $y = 4\sin(-t) + 1$

Using trigonometric identities ($\cos(-t) = \cos(t)$ and $\sin(-t) = -\sin(t)$):

• $x = 4\cos(t)$
• $y = -4\sin(t) + 1$

Answer for (a):

$x = 4\cos(t), y = -4\sin(t) + 1$ for $0 \leq t \leq 2\pi$.

Part (b) Four Times Around Counterclockwise

For counterclockwise motion, we use the regular parametric equations.

Since the particle goes around four times, the range of $t$ becomes $0 \leq t \leq 8\pi$.

The parametric equations are:

• $x = 4\cos(t)$
• $y = 4\sin(t) + 1$

Answer for (b):

$x = 4\cos(t), y = 4\sin(t) + 1$ for $0 \leq t \leq 8\pi$.

Part (c) Halfway Around Counterclockwise

Starting at $(0, 5)$ corresponds to the topmost point of the circle, which occurs when $t = \pi/2$.

From there, the particle moves counterclockwise halfway around the circle.

The range of $t$ is adjusted so that $t = 0$ corresponds to $(0, 5)$.

To achieve this, we shift the parameter $t$ by $\pi/2$:

• $x = 4\cos(t - \pi/2)$
• $y = 4\sin(t - \pi/2) + 1$

Using trigonometric identities ($\cos(t - \pi/2) = \sin(t)$ and $\sin(t - \pi/2) = -\cos(t)$):

• $x = -4\sin(t)$
• $y = 4\cos(t) + 1$

Answer for (c):

$x = -4\sin(t), y = 4\cos(t) + 1$ for $0 \leq t \leq \pi$.

Final Answers

(a) $\boxed{x = 4\cos(t), y = -4\sin(t) + 1}$

(b) $\boxed{x = 4\cos(t), y = 4\sin(t) + 1}$

(c) $\boxed{x = -4\sin(t), y = 4\cos(t) + 1}$

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