# Thread: eliminating t in a paramentric equation t (stewart pg 652, example #2)

1. ## eliminating t in a paramentric equation t (stewart pg 652, example #2)

What curve is represented by the parametric equations x=cost and y=sint
0<=t<=2pie ?

The solution says the following: "If we plot points, it appears that the curve is a circle. We can confirm this by eliminating t.
Observe that x^2 + y^2 = cos^2t + sin^2t = 1 "

But by my reasoning, if we eliminate t we would be left with y = sin (arccos x)

heres how i eliminated t:
x = cos t
t = arccos x
then after inserting into the y equation i'm left with: y = sin (arccos x)

I of course understand that we have equation of circle and the trig identity, but I dont understand how they're arriving at the identify by eliminating t??
-tanks!

What curve is represented by the parametric equations x=cost and y=sint
0<=t<=2pie ?

The solution says the following: "If we plot points, it appears that the curve is a circle. We can confirm this by eliminating t.
Observe that x^2 + y^2 = cos^2t + sin^2t = 1 "

But by my reasoning, if we eliminate t we would be left with y = sin (arccos x)

heres how i eliminated t:
x = cos t
t = arccos x
then after inserting into the y equation i'm left with: y = sin (arccos x)

I of course understand that we have equation of circle and the trig identity, but I dont understand how they're arriving at the identify by eliminating t??
-tanks!

x = cost
y = sint

Consider a right triangle with t as one of the acute angles.
cost = x = x/1 --> the adjacent leg is x, the hypotonuse is 1
sint = y = y/1 --> the oposite leg is y, the hypotouse is 1

from this we can use the pythagorean theorm to find x and y:
a^2 + b^2 = c^2
x^2 + y^2 = 1^2 = 1

That's how we relate x and y. doing it the other way (solving for t) won't work out very well in this situation.

What curve is represented by the parametric equations:
. . x = cosθ and y = sinθ, . 0 < θ <
Your approach is correct . . . just keep going.

It began with: .x = cosθ

Since sin²θ + cos²θ .= .1, we have: .sin²θ + x² .= .1
. . . - . . - . . . . . ._____ . . . - . . - . . . . . . . . _____
Hence: .sinθ .= .√1 - x² . . . . θ .= .arcsin(√1 - x²)

Substitute into the other equation:
. . . . . . . . . . . . . . . . . . . . _____
. . y .= .sinθ .= .sin[arcsin(√1 - x²)]

. . . - . . - . . . . . . . . _____
And we have: .y .= .√1 - x² . . . . x² + y² .= .1

4. Originally Posted by Soroban

Your approach is correct . . . just keep going.

It began with: .x = cosθ

Since sin²θ + cos²θ .= .1, we have: .sin²θ + x² .= .1
. . . - . . - . . . . . ._____ . . . - . . - . . . . . . . . _____
Hence: .sinθ .= .√1 - x² . . . . θ .= .arcsin(√1 - x²)

Substitute into the other equation:
. . . . . . . . . . . . . . . . . . . . _____
. . y .= .sinθ .= .sin[arcsin(√1 - x²)]

. . . - . . - . . . . . . . . _____
And we have: .y .= .√1 - x² . . . . x² + y² .= .1

Soroban reminded me of another way to look at this problem.

Since we are trying to get an equation with only x's and y's, let's look back at the relationships we have:

x = cos(theta)
y = sin(theta)

I can square each equation to get:

x^2 = cos^2(theta)
y^2 = sin^2(theta)

But recall that cos^2(theta) = 1 - sin^2(theta), so the x^2 equation becomes:

x^2 = 1 - sin^2(theta)

And we know that sin^2(theta) = y^2, so we get:

x^2 = 1 - y^2
x^2 + y^2 = 1

Notice that to get the "x" equation in terms of only x and y we needed to change the cos(theta) into a sin(theta) so that we could replace the sin(theta) with y. The only way to do this is to square both sides of each equation.