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Partial differentiation question rocket trajectory

  1. Jan 4, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem and my attempt are attached

    2. Relevant equations

    Chain rule for partial differentiation perhaps

    And basic algebra
    3. The attempt at a solution
    I'm unsure of how to approach this but I differentiated all the expression at the top.
     

    Attached Files:

  2. jcsd
  3. Jan 4, 2015 #2
    Can you try and differentiate the square root [itex] \sqrt{x(t)^2 + y(t)^2 + z(t)^2} [/itex] to [itex] x(t)^2 + y(t)^2 + z(t)^2 [/itex] first and work from there?
     
  4. Jan 4, 2015 #3

    Mark44

    Staff: Mentor

    No. The three coordinate functions x(t), y(t), and z(t) are functions of t alone, and you're asked to find dr/dt.

    Calculate x2 + y2 + z2, then take the square root, then differentiate with respect to t.
    In your work it looks like you're trying to integrate both sides. Try to resist that urge.
     
  5. Jan 4, 2015 #4
    hi

    How can I differentiate this?! Is it just
    (Terms)^1/2 and differentiate using chain rule?

    Also how would I be able to get rid of the t's at the end?
     

    Attached Files:

  6. Jan 4, 2015 #5
    I've differentiated it using the chain rule, looks horrible

    But now I need to get rid of the t's
     
  7. Jan 4, 2015 #6

    Mark44

    Staff: Mentor

    Yes.
    You don't get rid of them.
     
  8. Jan 4, 2015 #7

    Mark44

    Staff: Mentor

    Please post your work here in the form, not as an image, and especially not as an image that is turned on its side.
     
  9. Jan 4, 2015 #8
    The question wanted it in terms of the constants, the topic I'm doing involves partial differentiation it seems weird that there is no partial differentiation
     

    Attached Files:

  10. Jan 4, 2015 #9
    Yeah sorry, sure.
     
  11. Jan 4, 2015 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, it is just (Terms)1/2 and differentiate using chain rule .

    You don't get rid of the time. The resulting expression for speed is a function of time, t .
     
  12. Jan 4, 2015 #11

    Stephen Tashi

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    Science Advisor

    [itex] \frac{dr}{dt} [/itex] is a function of [itex] t [/itex] so you don't want to get rid of the t's. The t's might cancel out in some problems, but, in general, we expect a function of t to have t's in it.
     
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