Question

Graph the Plane Curve defined by: x=3cos(t)−t+5, y=sin(t), for t in [0,3π]

Answer 1

Answer:

Please see attached image for the graph

Step-by-step explanation:

Give different values of "t" in radians, selecting those that can facilitate the calculation of sine and cosine. The values need to be in the interval from zero to $3\,\pi$.

Notice that for t = 0 we get :

$x=3\,cos(0)-0+5=3+5=8\\y=sin(0)=0$

This is our initial point (8,0)

We can evaluate for $t=\frac{\pi}{2}$ which also gives an easy to calculate value:

$x=3\,cos (\frac{\pi}{2} )-\frac{\pi}{2} +5=3*0-\frac{\pi}{2} +5=5-\frac{\pi}{2} \\y=sin(\frac{\pi}{2} )=1$

This point is the maximum reached by the curve on the first quadrant.

Another easy point is for $t=\pi$, which gives:

$x=3\,cos(\pi)-\pi+5=-3\,-\pi +5=2-\pi\\y=sin(\pi)=0$

This is the crossing of the x-axis that we see to the left of the origin of coordinates.

The crossing of the x axis that appears then on the right of the origin of coordinates is when we use $t=2\,\pi$ (another simple to calculate point)

The minimum value reached by the graph (on the fourth quadrant) is obtained when we use $t=\frac{3\,\pi}{2}$

The maximum observed for the graph on the second quadrant is obtained when we use $t=\frac{5\,\pi}{2}$.

And finally, the last point we obtained is when $t=3\, \pi$ which is the point where the graph stops on the left :

$x=3\,cos(3\pi)-3\pi+5=-3\,-3\pi +5=2-3\pi\\y=sin(3\pi)=0$