Parametric Curves: Solving & Approximating

  • Thread starter Thread starter misogynisticfeminist
  • Start date Start date
  • Tags Tags
    Curves Parametric
Click For Summary
SUMMARY

The discussion focuses on solving and approximating parametric curves, specifically the curve defined by x = 2/t and y = 1 - 2t. The tangent line at t = -2 is established as y = 4x + 9, and the method to find additional intersection points with the curve is outlined. Additionally, the approximation of sec 61 degrees is addressed, emphasizing the conversion to radians and the application of the linear approximation formula f(x0 + h) ≈ f(x0) + h f'(x0).

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of tangent lines and their equations
  • Familiarity with trigonometric functions and their approximations
  • Ability to convert degrees to radians
NEXT STEPS
  • Learn how to find intersection points of parametric curves and lines
  • Study the process of converting angles from degrees to radians
  • Explore linear approximation techniques in calculus
  • Investigate the properties and applications of secant functions
USEFUL FOR

Mathematicians, calculus students, and anyone interested in the analysis and approximation of parametric curves and trigonometric functions.

misogynisticfeminist
Messages
370
Reaction score
0
1. I am given a curve defined parametrically by [tex]x= 2/t , y=1-2t[/tex] i have found the equation of tangent at t=-2 to be y=4x+9, they have asked whether it cuts the curve again. how do i find that, since i don't know the original equation of the curve and can't solve them simultaneously.

2. Also they have asked to find an approximation for sec 61. I have used [tex]\frac{\delta y}{\delta x}= \frac{dy}{dx}[/tex], but i did not get the answer, 2.0605. How do i get about doing it?
 
Physics news on Phys.org
misogynisticfeminist said:
1. I am given a curve defined parametrically by [tex]x= 2/t , y=1-2t[/tex] i have found the equation of tangent at t=-2 to be y=4x+9, they have asked whether it cuts the curve again. how do i find that, since i don't know the original equation of the curve and can't solve them simultaneously.
Intersect the line with the curve and see if you find any other intersection points besides t = -2. Since y = 4x+9, using the parametric equation for x, y = 4(2/t) + 9. Now substitute in the parametric equation for y and solve for t.

For the second question, I'm not really sure what you mean.
 
You should first convert 610 into radians (most work in calculus use radians, instead of degrees), i.e:
[tex]61 ^ \circ = \frac{61 \pi}{180} \mbox{ rad}[/tex]
Since you already have:
[tex]\sec \left( \frac{60 \pi}{180} \right) = \sec \left( \frac{\pi}{3} \right) = 2[/tex]
You can use
f(x0 + h) ≈ f(x0) + h f'(x0) (h ≈ 0) to solve your problem.
Can you go from here?
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K