Find parametric equations in the parameter $t$, where $0 \leq t \leq 2\pi$, for the circle $x^2 + y^2 = r^2$

Question

My precalculus final exam is tomorrow, but I don't understand how to solve this question. Especially, I don't understand "tracing and direction of $t$".

Find parametric equations in the parameter $t$, where $0 \leq t \leq 2\pi$, for the circle $x^2 + y^2 = r^2$ that is traced as described:

a. Once around, counterclockwise, starting at $(r, 0)$

b. Twice around, counterclockwise, starting at $(r, 0)$

c. Three times around, counterclockwise, starting at $(−r, 0)$

d. Twice around, clockwise, starting at $(0,r)$

e. Three times around, clockwise, starting at $(0,r)$

Answer 1

Let: $$x = r\cos(t)$$ $$y = r\sin(t)$$

Then, $$x^2+y^2 = \left(r\cos(t)\right)^2 + \left(r\sin(t)\right)^2$$ $$ = r^2\left(cos(t)^2 + sin(t)^2\right)$$ $$ = r^2$$ (Since $cos(t)^2 + sin(t)^2 = 1$)

Therefore, $$x = r\cos(t)$$ $$y = r\sin(t)$$ is a parametric description of the circle centered about the origin. This is the standard parameterization.

If you let t = 0 then $x=1$ and $y=0$. As you increase t, you will move along the circle in a counterclockwise sense. You can convince yourself of this by calculating x and y for increasing values of t and plotting them. (Also most graphing calculator will plot parametric equations for you and you will be able to see this as well). This website desmos.com can also plot parametric equations.

With respect to the question of tracing and direction. Imagine that for set of parametric equations given already you held your pencil down and 'traced' the curve described by continuously increasing the value of t. This is of course the circle. And as described above this will happen in a counterclockwise sense. If you replace t by -t in the set of parametric equations you would find that you trace the circle in the clockwise direction, that is in the opposite sense. Again the easiest way to see this is to plot the curve described by the parametric equations.

Each of a, b, c, d, and e can be answered by modifying the original parametric equation for the circle by adding the extra parameters a and b: $$x = r\cos(at+b)$$ $$y = r\sin(at+b)$$

The parameter 'a' controls how 'fast' you trace around the circle and the parameter 'b' controls the starting position. In fact, if you let a = 2, you will trace around the circle twice. If you let a = 3 you will trace around the circle 3 times. Etc. If b = 0 then you will start at the right side of the circle; if you replace b by $\pi/2$ you will start at the top of the circle; if you replace b by $\pi$ you will start at the left side of the circle; or if you replace b by $3\pi/2$, the bottom of the circle.

Answer 2

Some comments and hints:

The basic thing here is that the map $t \mapsto (\cos t, \sin t)$ will trace out the unit circle in an anti clockwise fashion as $t$ ranges from $0$ to $2 \pi$. Note that at $t=0$ the point is $(1,0)$.

Note that the above map is $2\pi$ periodic, so $t \mapsto (\cos (t+2\pi), \sin (t+2\pi))$ will result in exactly the same map.

Note that to go around in the other direction, we can replace $t$ by $-t$, that is $t \mapsto (\cos (-t), \sin (-t))$ will trace out the unit circle in a clockwise fashion as $t$ ranges from $0$ to $2 \pi$. Again, note that at $t=0$ the point is $(1,0)$.

To trace out a circle of radius $r$, we just multiply by $r$. So for a. we have $t \mapsto (r \cos t, r \sin t)$. (Or, perhaps you would rather write $x(t) = r \cos t$, $y(t)=r \sin t$.)

(I have quietly assumed that $r \ge 0$.)

For b., to make the point $(x(t),y(t))$ go around twice we replace $t$ by $2t$ to get $t \mapsto (r \cos (2t), r \sin (2t))$.

For c., we need two changes, to go around three times we need something like $t \mapsto (r \cos (3t), r \sin (3t))$, however this starts at $(r,0)$, so we need to offset the parameters to $\cos,\sin$ so that when $t=0$ the values will result in $(x(0),y(0))= (-r,0)$. Note that a 'shift' of $\pi$ will do the trick, so we see that $t \mapsto (r \cos (3t+\pi), r \sin (3t+\pi))$ will do.

Hopefully that gives you enough 'tricks' to solve d.,e.