Parametric Equations: Tangents and Cartesian Equation for a Simple Curve

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Homework Help Overview

The discussion revolves around a curve defined by parametric equations: x = 2 cot t and y = 2sin² t, with the parameter t constrained to the interval (0, π/2]. Participants are tasked with finding the derivative dy/dx, the equation of the tangent at a specific point, and the Cartesian equation of the curve, including its domain.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the process of finding dy/dx and the tangent line, with some expressing uncertainty about determining the domain of the Cartesian equation derived from the parametric equations. Questions arise regarding the behavior of x as t approaches the boundaries of its interval.

Discussion Status

Some participants have made progress on the initial parts of the problem but are seeking clarification on the domain of the Cartesian equation. There is an ongoing exploration of the relationship between the parameter t and the resulting values of x, with multiple interpretations being considered.

Contextual Notes

Participants note that while t cannot equal 0, the corresponding value of x can be 0 when t approaches π/2. This distinction is central to understanding the domain of the function derived from the parametric equations.

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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?
 
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phospho said:
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?

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

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.
 
Mark44 said:
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.

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?
 
phospho said:
Well drawing the cot graph as t approaches pi/2, x approaches 0 ∴ domain will approach infinity?
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.)
phospho said:
Similar when t is really small the domain will approach 0? so 0<x<∞ or x>0?
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).
 
Mark44 said:
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).

okay, thank you

just one question - why does it include 0?
 
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
 

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