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Finding the range of a rational function

  1. Jan 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A curve is given by the parametric equations
    ##x=t^2 +3##
    ##y=t(t^2+3)##
    Find dy/dx in terms of t and show that (dy/dx)^2 >=9

    2. Relevant equations
    Parametric derivatives

    3. The attempt at a solution
    Using the chain rule, I arrived at ##\frac{dy}{dx}=\frac{3}{2}(\frac{t^2+1}{t})##
    However, when I squared both sides to get ##(\frac{dy}{dx})^2##, I was unable to prove that (dy/dx)^2>=9. I know I need to find the range, however I'll need to graph the function, which proves to be tedious. Is there any workaround to this?
     
  2. jcsd
  3. Jan 24, 2015 #2

    PeroK

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    What did you try? Isn't it just some simple quadratic algebra?
     
  4. Jan 24, 2015 #3
    I was thinking to graph the function, though I'll be dealing with quartic functions.
     
  5. Jan 24, 2015 #4

    PeroK

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    Some quartics look like quadratics, wouldn't you say?
     
  6. Jan 24, 2015 #5
    Yup, but how do I begin though?
     
  7. Jan 24, 2015 #6

    PeroK

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    If ##(dy/dx)^2 < 9## what does that give you?
     
  8. Jan 24, 2015 #7
    There are no values of t that satisfies the inequality?
     
  9. Jan 25, 2015 #8

    PeroK

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    Yes, but you've got to be prepared to do the algebra that shows why.
     
  10. Jan 25, 2015 #9
    So how do I begin?
     
  11. Jan 25, 2015 #10

    Mark44

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    Your problem is equivalent to showing that ##(\frac{t^2 + 1}{t})^2 \ge 4##.
    Presumably you are in a calculus class. How do you find the minimum value of a function?
     
    Last edited: Jan 25, 2015
  12. Jan 25, 2015 #11
    Derive and set the value of the derivative to 0?
     
  13. Jan 25, 2015 #12

    Mark44

    Staff: Mentor

    Yes, differentiate it and set the derivative to 0. But how will you know whether you have a minimum or maximum?

    Now show us something...
     
  14. Jan 25, 2015 #13
    But first, why did you say ((t^2+1)/t)^2 >4?

    Use the first derivative test for critical points?
     
  15. Jan 25, 2015 #14

    Mark44

    Staff: Mentor

    I changed what I wrote slightly, to ##(\frac{t^2 + 1}{t})^2 \ge 4##
    What do you think? Take a stab at it.
     
  16. Jan 25, 2015 #15
    But what makes you think why ((t^2 +1)/t))^2 >=4?

    The sign of the slopes changes?
     
  17. Jan 25, 2015 #16

    Mark44

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    In your first post you have
    Isn't the inequality on the right equivalent to the one I wrote?
     
  18. Jan 25, 2015 #17
    Wasn't it should be >=9?
     
  19. Jan 26, 2015 #18

    Mark44

    Staff: Mentor

    You tell me.
    1. Start with (dy/dx)^2 >=9.
    2. Simplify.

    This is post 18 in the thread. So far I haven't seen any work from you. What's stopping you?
     
  20. Jan 26, 2015 #19
    I have two inequalities, 3/2((t^2 +1)/t) <=-3 or ((t^2 +1)/t) >=2. Then how should I proceed from here?
     
  21. Jan 26, 2015 #20

    Mark44

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    ##(dy/dx)^2 \ge 9##
    ##\Rightarrow (3/2)^2(\frac{t^2 + 1}{t})^2 \ge 9##
    ##\Rightarrow (\frac{t^2 + 1}{t})^2 \ge 4##
    This last inequality is equivalent to ##\frac{t^2 + 1}{t} \ge 2## or ##\frac{t^2 + 1}{t} \le -2##
    Since this is what you need to show, you can't assume that it's true. You can use calculus techniques to find the minimum and values of the rational expression.
     
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