Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JerJason

  • 3 years ago

Parametric Equations: Find the equation of the tangent line at the point where the curve crosses itself. x = t^3-6t, y = t^2

  • This Question is Closed
  1. JerJason
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm particularly stuck on how to get the points. In a previous problem I just made a chart where I plugged in values for t in the x and y equation until I got two coordinates that were the same for 2 t-values. When I try to do the same method with this problem, I'm not able to get the same result.

  2. dumbcow
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you meant when x(t) = y(t) ?

  3. dumbcow
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    there are 3 places where they intersect t = -2,0,3

  4. shaan_iitk
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    curve would cross itself at that point where the derivative would cease to exist .. as you see the derivative is 2t/(3t^2 - 6) .. hence for t = +- sqrt(2) derivative ceases to exist ... Hence the point where it crosses itself is (0,2)

  5. shaan_iitk
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no dumbcow he means that a curve in parametric form crosses itself .. what is the equation of the tangent ...

  6. shaan_iitk
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    at the point where it crosses itself...

  7. shaan_iitk
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    also you cannot find the slope at this point ... because the slope is not defined for THIS function at this point ...although you can find the slope of points with parameter slightly greater (or slightly smaller) than +-sqrt(2) and find the slope...

  8. JerJason
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  9. Mashy
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If a curve crosses itself at any point, that point will have 2 tangents right?

  10. JerJason
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    correct.

  11. shaan_iitk
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Mashy ... how will you define the derivative at that point ??? .. As I said you can find the slope by assumint t --> sqrt(2) - ... but at that point derivative won't exist (function cannot have two values for one variable --> function is not defined otherwise ) ...

  12. dumbcow
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[t = \pm \sqrt{y}\] substituting into x(t) gives 2 equations \[x = y \sqrt{y} - 6\sqrt{y}\] and \[x = -y \sqrt{y} + 6\sqrt{y}\] setting them equal will give us the point where the curve crosses itself \[y \sqrt{y} -6\sqrt{y} = -y \sqrt{y} +6\sqrt{y}\] \[2y \sqrt{y} = 12\sqrt{y}\] \[y = 6\] \[x = 6\sqrt{6} - 6\sqrt{6} = 0\] point is (0,6) \[\frac{dx}{dy} = \pm (\frac{3\sqrt{y}}{2} -\frac{3}{\sqrt{y}})\] this is the derivative of the inverse, to find slope of curve at x=0 use: \[f'^{-1}(y) = \frac{1}{f'(x)}\] \[\rightarrow f'(x) = \frac{1}{f'^{-1}(y)}\] plug in y=6 into dx/dy \[dx/dy = \pm(\frac{3\sqrt{6}}{2} -\frac{3}{\sqrt{6}}) = \pm \sqrt{6} \] \[f'(0) = \pm \frac{1}{\sqrt{6}} = \pm \frac{\sqrt{6}}{6}\] tangent lines: \[y = \pm \frac{\sqrt{6}}{6}x + 6\] http://www.wolframalpha.com/input/?i=x+%3D+t%5E3+-6t+%2C+y+%3D+t%5E2

  13. JerJason
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'll have look this over a few times but thank you for your help.

  14. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why would the curve intersect itself where x(t) = y(t)? The only thing I can imagine coming from this is a point graphed that is equal in distance from both axes in the first and third quadrant, e.g. a point on the line y = x, with some arbitrary constant added or subtracted from both sides. I completely understand why the derivative would not exist at the curve's intersection with itself (or another curve), but I fail to understand why x(t) and y(t) being equal means that the graph crosses itself.

  15. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.