1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tough Derivative need help

  1. Dec 1, 2006 #1

    PMP

    User Avatar

    Would you tell me what is:

    [tex]\frac{d\sqrt{(v_0cos \theta t)^2 + (v_0 \sin \theta t - 1/2gt^2)^2}}{dt}[/tex]

    Please. Thanks in advance.
     
  2. jcsd
  3. Dec 1, 2006 #2
    It's a simple application of the chain rule. What have you done so far?
     
  4. Dec 1, 2006 #3

    PMP

    User Avatar

    I have just basic knowledge of derivatives, this exercise is one difficult for my level, so I would just like to know if I derivated this correctly. Would you help me, neutrino?
     
  5. Dec 1, 2006 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Hint:
    Simplify the radicand first!
     
  6. Dec 1, 2006 #5
    Post what you have done so that we can check if it is done correctly.

    Treat [tex]v_0cos \theta t)^2 + (v_0 \sin \theta t - 1/2gt^2)^2[/tex] as a single function of t, say g, under the square root. If the whole function is denoted by f, then [tex]f = \sqrt{g}[/tex]. Now use the chain rule [tex]\frac{df}{dt} = \frac{df}{dg}\frac{dg}{dt}[/tex]
     
  7. Dec 1, 2006 #6

    PMP

    User Avatar

    Thanks arildno, I have already done that, and have my result. I really just want to see if it is correct. Is it:

    [tex]2v_0^2t +g^2t^3 - 3v_0\sin \theta gt^2[/tex] ?

    The denominator doesen't matter because I will equal that derivative to zero.
     
  8. Dec 1, 2006 #7
    That is right.
     
  9. Dec 1, 2006 #8

    PMP

    User Avatar

    Ok, thank you all of you. My basic derivative technique works. :)

    How would I find the maximum value of theta, such that the this derivative is positive for t belonging to the interval [0, 2v0sin(theta)/g]?
     
  10. Dec 1, 2006 #9

    PMP

    User Avatar

    No ideas? :)
     
  11. Dec 1, 2006 #10

    PMP

    User Avatar

    Knock, knock!!
     
  12. Dec 1, 2006 #11
    Who's there? :P

    If you want to maximise the above derivative wrt to theta, then differentiate it wrt to theta and set it zero.
     
  13. Dec 1, 2006 #12

    PMP

    User Avatar

    That person already ran way. :rofl:

    I should differentiate wrt to theta [tex]2v_0^2t +g^2t^3 - 3v_0\sin \theta gt^2[/tex] or the initial [tex]\sqrt{(v_0cos \theta t)^2 + (v_0 \sin \theta t - 1/2gt^2)^2}[/tex] ?

    Because if it is to differentiate the above I would get [tex]\cos \theta = 0[/tex], right? And it is not the correct result.
    Thanks.
     
    Last edited by a moderator: Dec 1, 2006
  14. Dec 1, 2006 #13
    Depends on which function you want to maximise.
     
  15. Dec 1, 2006 #14

    PMP

    User Avatar

    This is a projectile position vector and I need to find theta such that the magnitude of the position vector during the projectile motion is always increasing. So I did dr/dt>=0. Got the expression above, but now if I differentite this wrt to theta I will not get the correct result. The deravite is [itex]-3v_0\cos\theta gt^2[/itex]. Right?
     
  16. Dec 1, 2006 #15

    PMP

    User Avatar

    neutrino, knock, knock...
     
  17. Dec 1, 2006 #16

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    My turn, my turn.... who's there...? :wink:
    Is [itex]\theta[/itex] your launch angle?
     
  18. Dec 1, 2006 #17

    PMP

    User Avatar

    Yes, it is. Thanks for being prepared to help.
     
  19. Dec 1, 2006 #18

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No problem, then why may I ask you to state the complete problem as given in your text.
     
  20. Dec 1, 2006 #19

    PMP

    User Avatar

    This not in a text. But it is like this: What is the maximum lauch angle for a projectile be always going away from me? That is, for the magnitude of the position vector be always increasing in time. I don't want you to solve the problem, just to say if I am in the right way. Thanks in advance!
     
  21. Dec 1, 2006 #20

    PMP

    User Avatar

    I guess. The initial conditions are:

    x_0=0
    y_0=0

    But that is the magnitude of the postion vector. Is it right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Tough Derivative need help
  1. Need help on a tough ? (Replies: 4)

Loading...