Parametric Equations: Tangents and Cartesian Equation for a Simple Curve

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SUMMARY

The discussion focuses on the analysis of a curve defined by the parametric equations x = 2 cot(t) and y = 2 sin²(t) for 0 < t ≤ π/2. Participants successfully derived dy/dx and the tangent line at t = π/4, while also finding the Cartesian equation y = 8/(4 + x²). The primary challenge addressed was determining the domain of the curve, which is established as [0, ∞) based on the behavior of the cotangent function as t approaches its limits.

PREREQUISITES
  • Understanding of parametric equations and their derivatives
  • Knowledge of trigonometric functions, specifically cotangent and sine
  • Familiarity with Cartesian equations and their domains
  • Ability to analyze limits and intervals in calculus
NEXT STEPS
  • Study the derivation of dy/dx for parametric equations
  • Learn about tangent lines to curves and their applications
  • Explore the properties of the cotangent function and its graph
  • Investigate domain determination for various types of functions
USEFUL FOR

Students and educators in calculus, mathematicians working with parametric equations, and anyone interested in understanding the relationship between parametric and Cartesian forms of curves.

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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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