Question

How to graph this parametric equation by hand?

x = 3sin^3t
y = 3cos^3t

It looks like this: http://www.wolframalpha.com/input/?i=x%3D3*Sin%5E3%5Bt%5D%2C+y%3D3*Cos%5E3%5Bt%5D

I used these values "-pi, -pi/2, 0, pi/2" pi for t.
It alternates 0, -3 -> -3, 0 -> (x,y)

What do I do though?

Answer 1

you may use the sign of dy/dx to figure out whether the graph should be increasing/decreasing in between - may be u may even assume the concavity using just the first derivative as it is a familiar trig function

Answer 2

You can take more values of t between -pi and -pi/2 and between other points.

Answer 3

So anything that is n*pi/2.

Answer 4

n*pi/4*

Answer 5

Right!

Answer 6

generally speaking, graphing a random given parametric is not easy. It's only when the parametric equations are the ones that we are familiar with can we try to graph by hand. We have calculus to help us sketch it but even so, it can take quite amount of effort. And this is when we turn to technology.

However, in this case, the general parametric equation of a *diamond* is
x = a cos^3(t)
y = b cos^3(t), where a and b non zero.

Answer 7

Or n*pi/3 or n*pi/6

Answer 8

What about T then @sourwing

Answer 9

When do I stop finding values then @ShailKumar

Answer 10

when you are satisfied

Answer 11

Start from 0 and stop at 2*pi.That will complete the cycle.

Answer 12

T?

Answer 13

I only get 0, -3 or -3, 0.

Answer 14

typo, i meant
x = a cos^3(t)
y = b sin^3(t)

Answer 15

So you can take t = 0,pi/3,pi/4,pi/6,pi/2,3*pi/2 and so on up to 2pi

Answer 16

try \(\pi/4\)

Answer 17

Oh so I get another point by using pi/4

Answer 18

you don't have to plug in any values for t. By know the its properties, the value of a tells you where how far the vertex is in the x direction and the value of b tells you how far the other vertex is in the y direction

Answer 19

Thank you guys!

Answer 20

I don't know who to give the medal to...

Answer 21

so it looks like this:

Answer 22

That may also be a possibility.