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Second derivative chain rule

  1. Mar 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Question has been attached to topic.

    2. Relevant equations
    Chain rule.

    3. The attempt at a solution
    $$\frac {dy}{dt} . \frac{dt}{dx} = \sqrt{t^2+1}.cos(π.t)$$
    $$\frac{d^2y}{dt^2}.(\frac{dt}{dx})^2 = 2 $$
    $$\frac{d^2y}{dt^2}.(t^2+1).cos^2(π.t)= 2 $$ and for the t=3/4,
    $$\frac{d^2y}{dt^2}.\frac{25}{16}.\frac{1}{2} = 2 $$
    $$\frac{d^2y}{dt^2} = \frac{64}{25}$$
    $$\frac{dy}{dt} = \frac{8}{5}$$

    I count the dt\dx as the function itself because it is the previous status of the function, I mean the function in the problem statement is a result of dt/dx.

    Is my solution correct? Is my approach correct? If not , where am I wrong and how to solve?

    Thank you!
     

    Attached Files:

  2. jcsd
  3. Mar 17, 2017 #2
    Check your arithmetic on the last step. Also note that cos(3π/4) is negative but there's a sign ambiguity for (dt/dx), if you're getting it from your second equation.
     
  4. Mar 18, 2017 #3

    haruspex

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    Can you justify ##\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left(\frac{dt}{dx}\right)^2##?
    I tried it with x=tm, y=tn and it did not seem to work.
     
  5. Mar 18, 2017 #4
    Last edited: Mar 18, 2017
  6. Mar 18, 2017 #5

    haruspex

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    For typing ease, I'll use dot for d/dt and ' for d/dx.
    ##\dot y=y'\dot x##
    Differentiating and applying the product rule:
    ##\ddot y=\dot x \frac d{dt}y'+y'\ddot x##
    ##=\dot x \frac {dx}{dt}y"+y'\ddot x##
    ##=\dot x^2y"+y'\ddot x##.
    That checks out with my example.
     
  7. Mar 18, 2017 #6
    Looks good. I can insist to myself that I've learned from the experience.
     
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