# Simple parametric equations

1. Nov 30, 2012

### phospho

A curve has parametric equations

x = 2 cot t, y = 2sin² t , 0 < t ≤ pi/2

(a) find an expression for dy/dx in terms of the parameter t
(b) Find an equation of the tangent to the curve at the point where t = pi/4
(c) Find a cartesian equation of the curve in the form y = f(x). State the domain on which the curve is defined.

I've done part a),
I've done part b),
I've partway done part c), I have found the equation (y = 8/(4+x^2)), however I don't know how to find the domain on which the curve is defined, any help?

2. Nov 30, 2012

### LCKurtz

If x = 2 cot t, 0 < t ≤ pi/2, what values of x can you get?

3. Nov 30, 2012

### Staff: Mentor

x as a function of t is defined for all t in (0, $\pi/2$]. What interval constitutes the codomain for x = f(t) = 2 cot(t)?. That will be your domain for your nonparametric equation.

4. Nov 30, 2012

### phospho

Well drawing the cot graph as t approaches pi/2, x approaches 0 ∴ domain will approach infinity? Similar when t is really small the domain will approach 0? so 0<x<∞ or x>0?

5. Nov 30, 2012

### Staff: Mentor

The domain doesn't "approach" anything - it's just an interval that you specify by an inequality or interval notation. (I'm taking shortcuts here - a domain might be the union of several disjoint intervals.)
Well, almost. It's actually 0 ≤ x < ∞, or x ≥ 0. Both of those represent the same interval, which you can also write as [0, ∞).

I don't know if you noticed, but the domain of f(x) = 8/(4 + x2) is all real numbers. The restricted domain you found comes from x(t) = cot(t).

6. Dec 1, 2012

### phospho

okay, thank you

just one question - why does it include 0?

7. Dec 1, 2012

### HallsofIvy

The "the domain on which the y= f(x) is defined" is a set of x values, not t.
t cannot be 0 but x can. $x(\pi/2)= cot(\pi/2)= 0$