Question

pls help ;( What does this question mean?

Answer 1

Sketch the curve traced out by the equations x=cos t and y=cos^2t, where 0<= t <= pi

Answer 2

well it wants you to make a graph

you can start by plugging in t values to get x and y-coordinates to graph

Answer 3

for example, when t = 0, x = 1 and y = 1, so you would put (1,1) on a graph

Answer 4

still there?

Answer 5

This is what is called a set of parametric equations — each of x and y are determined by an equation that takes a parameter value, t. You vary t and calculate x and y from it. Some things are much more easily expressed this way.

Answer 6

And you can get some surprising results, too. You might not expect to be able to draw a straight line with the cosine function, for example, but

\[y = \cos t\]\[x=\cos t\]will do just that
\[y=\cos t\]\[x=\sin t\]will draw a circle

Answer 7

It's kind of hard to plot for example x= cos(pi/2) and y = cos(pi/2)^2

Answer 8

Just as vocaloid suggested, you make a table of a couple of values that are easy to compute and spread out through the specified range, then plot them.

Answer 9

What is the value of \(\cos(\pi/2)=\)

Answer 10

oh ok 0

Answer 11

okay, and squaring 0 I imagine won't cause too much headache :-)

Answer 12

idk what happened lol i'm so slow

Answer 13

i find this extra confusing
hard to make a table to get enough information to draw a decent graph
i would use technology

Answer 14

omg omg

Answer 15

on the other hand, in your example it is not hard
since \(x=\cos(t)\) you have \(y=x^2\)

Answer 16

like the unit circle lol

Answer 17

regular old parabola

Answer 18

cos from 0 to pi

Answer 19

halfway t= pi/2 that's zero

Answer 20

so that gives you the domain

Answer 21

\[[-1,1]\]

Answer 22

While that is an excellent thing to notice if you already are skilled with parametric equations, the question i have is whether the raison d'être for the problem was to see if you could find an easy way to do it, or to get some practice graphing parametric equations. As the OP clearly needs some work in that department, I'm not sure finding a shortcut around it is ultimately beneficial.

Answer 23

To be honest, I have a test tomorrow. I didn't go to class on monday and my professor went over this stuff. :( That's why I was so lost (I might still be a little bit)

Answer 24

That was in my study guide.

Answer 25

He didn't spend much time explaining parametric equations apparently.

Answer 26

Assuming that you know your unit circle and don't have any trouble computing the values, I'll do the technology bit and make you a table:

\[
\begin{array}{ccc}
t & x & y \\
0 & 1 & 1 \\
\frac{\pi }{6} & \frac{\sqrt{3}}{2} & \frac{3}{4} \\
\frac{\pi }{3} & \frac{1}{2} & \frac{1}{4} \\
\frac{\pi }{2} & 0 & 0 \\
\frac{2 \pi }{3} & -\frac{1}{2} & \frac{1}{4} \\
\frac{5 \pi }{6} & -\frac{\sqrt{3}}{2} & \frac{3}{4} \\
\pi & -1 & 1 \\
\end{array}
\]

Answer 27

and a graph:

Answer 28

ohhhh awesome :D

Answer 29

Thank you! <3

Answer 30

Here's the same graph, except done every \(\pi/24\) instead of \(\pi/6\), and I also drew in the parabola that goes through those points.