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Another parametric equation

  1. Mar 27, 2005 #1
    [tex]x = cos(t)^7[/tex]
    [tex]y= 8sin(t)^2[/tex]

    Find [tex]\frac{d^2y}{dx^2}[/tex] expressed as a function of [tex]t[/tex]

    [tex]\frac{d^2y}{dx^2}[/tex] = _________

    well second derivative for y is [tex]\frac{d^2y}{dt} = (16*cos(2t))[/tex]


    dx/dt = (-7*cos(t)^6*sin(t))

    so dx^2 = ((-7*cos(t)^6*sin(t)))^2 right?

    so...[tex]\frac{d^2y}{dx^2}[/tex] = [tex]\frac{(16*cos(2t))}{(-7*cos(t)^6*sin(t))^2} [/tex] right?
     
  2. jcsd
  3. Mar 27, 2005 #2
    [tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]

    [tex] \frac{d^2y}{dx} = \frac{\frac{d^2y}{dt}}{\frac{d^2x}{dt}} [/tex]

    [tex] x = cos(t)^7. y= 8sin(t)^2 [/tex]

    [tex] \frac{dx}{dt} = -7cos(t)^6sin(t) [/tex]

    [tex] \frac{dy}{dt} = 16sin(t)cos(t) [/tex]

    I'll let you derive the rest.
     
  4. Mar 27, 2005 #3
    Remember [tex] \frac{d^2y}{dx} = \frac{d\frac{dy}{dx}}{dx} [/tex]

    Aka the second derivative.
     
  5. Mar 27, 2005 #4
    so it wants the second derivative of both y and x? then divide y/x? but it's dx^2 not d^2x, or is it the same? well if you want the second derivative of both, it should look like this, but it's not correct:

    [tex]\frac{(16*cos(2t))}{(-7/2*cos(t)^5*((7*cos(2t))-5))}[/tex]
     
  6. Mar 27, 2005 #5
    by the chain rule,

    [tex]\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}[/tex]

    differentiate directly for [itex] \frac{dy}{dx}[/itex] and implicitly for [itex]\frac{dt}{dx}.[/itex]

    You can then just differentiate again (implicitly) for [itex] \frac{d^2y}{dx^2}[/itex]

    I get

    [tex] \frac{d^2y}{dx^2} = \frac{80}{49} \sec^{12}{t}[/tex]
     
  7. Mar 27, 2005 #6
    i dont get how you got 80/49*sec(t)^12 which is the correct answer.

    dy/dx = [tex]\frac{(8*2*sin(t)*cos(t))}{(-7*cos(t)^6*sin(t))}[/tex] (this is correct, it was the first question asked and i got it correct)

    so are you telling me to now differentiate that function from above agian? if i do that, i get a different answer from you
     
  8. Mar 27, 2005 #7
    Your derivative is correct. It simplifies to

    [tex] \frac{dy}{dx} = \frac{-16}{7}\sec^5{t}[/tex]

    from there, differentiate implicitly wrt [itex]x[/itex] for the second derivative. If you try and don't get the right answer, post your work and I'll help~
     
    Last edited: Mar 27, 2005
  9. Mar 27, 2005 #8
    I think your getting confused with mixing the chain rule + implicit differentiation. Just take it one step at a time. The answer you posted above is correct and indeed simplifies to what Data told you. Now take the numerator, differentiate it with respect to T, do the same with the denominator, and divide the two. This is dy/dx.

    It may be harder to see, but both my method and datas method are two ways to do the same problem. His is more efficient, but mine has easier steps.
     
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