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Differential Calculus Question

  1. Jul 31, 2015 #1
    If z = e ^ (xy ^ 2), x = tcost, and y = tsint compute dx / dt for t = pi / 2

    I kind of lost in this difficult question pls help

    I tried putting down the xy but using ln

    lnz = xy^2

    Product rule? Or what. This is my first time encountering this kind of question
    Last edited: Jul 31, 2015
  2. jcsd
  3. Jul 31, 2015 #2
    Are you sure that you need to calculate ##\frac{dx}{dt}##? If so, then the first and 3rd equations are irrelevant - just calculate ##\frac{d}{dt} (t cost)## using the product rule, and plug in ##\frac π {2}## for ##t## into the derivative.
  4. Jul 31, 2015 #3
    Hmm that make sense
    but in my hw it is dx/dt idk if that is a correction

    I also think it is a dz/dt
  5. Jul 31, 2015 #4
    For dz/dt, why don't you substitute the values of x and y in terms of t for the expression of z first?
  6. Jul 31, 2015 #5
    hmmmm I tried this


    Product rule?
  7. Jul 31, 2015 #6


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    If the question is just "what is dx/dt", then the answer is just d(t cos(t))/dt= cos(t)- t sin(t), by the product rule.

    But perhaps the problem was to find dz/dt and you are asked to find dx/dt as part of that.

    By the "chain rule", [itex]dz/dt= (\partial z/\partial x)(dx/dt)+ (\partial z/\partial y)(dy/dt)[/itex]

    Since [itex]z= e^{xy^2}[/itex], [itex]\partial z/\partial x= y^2e^{xy^2}[/itex] and [itex]\partial z/\partial y= 2xy e^{xy^2}[/itex]

    As before, [itex]dx/dt= cos(t)- t sin(t)[/itex] and, since [itex]y= t sin(t)[/itex], [itex]dy/dt= sin(t)+ t cos(t)[/itex].
  8. Jul 31, 2015 #7
    There are two ways to proceed from here. Are you familiar with logarithmic differentiation? If yes, then just differentiate the left side with respect to t using chain rule, and the right side in the regular product rule manner.

    If not, then convert this back into exponential form and use chain rule in combination with the product rule by letting the exponent equal to a new intermediate variable, such as u.
  9. Jul 31, 2015 #8
    I think I can do the Logarithmic Differentiation

    d/dx(lnu) = 1/u * du/dx ??
  10. Jul 31, 2015 #9
    This is true for the LHS, except that instead of u there will be a z, and instead of x we will have a t - there is no need to introduce a new variable for differentiating the LHS.
  11. Jul 31, 2015 #10
    Hmm can you lead me there? I think I can understand if I can grasp more of it thank you!
  12. Jul 31, 2015 #11
    You already did that. The equation becomes $$\frac 1 {z} \frac{dz}{dt} = \frac{d}{dt} (\frac{t^4 (sin 2t)^2}{4})$$ , and you can proceed from here by applying the product rule.
    Last edited: Jul 31, 2015
  13. Jul 31, 2015 #12
    I'll chain rule the RHS?
    I'll took the constant out first so
    I don't know if that's what you meant
  14. Jul 31, 2015 #13
    Oh, my bad. I meant use the product rule.
  15. Jul 31, 2015 #14

    I'll just product rule this then substitue pi/2?
  16. Jul 31, 2015 #15
  17. Jul 31, 2015 #16
    I got 0.3448 is it right?
  18. Jul 31, 2015 #17
    Are you sure that works out to give 0.3448 ?
  19. Aug 1, 2015 #18
    So did you get an answer?
  20. Aug 1, 2015 #19
    I got 0.3448 I don't know if the answer is correct but at least I understand the concept which is more important so thank you mates!
  21. Aug 1, 2015 #20
    EDIT: The answer is close, but not 0.3448 . Can show the derivative that you got?
    Last edited: Aug 1, 2015
  22. Aug 1, 2015 #21
    I think your answer is wrong . How then , can you say this ?
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