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!

Solving sin(t^2)-(t^2)=0

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Solve
    sin(t2) - t2 =0. for t

    2. Relevant equations

    None, besides various trig identities. This was actually a dynamics problem where I had to solve for time, and this is simply the equation you get after summing up all the forces.

    3. The attempt at a solution


    Now obviously 0 is a solution, and when I plug it into wolfram alpha I get t=1.04, which I know is the correct answer because my professor told us that in class. However, when I plug it into my TI-89, the only solution that comes out is 0, and when I graph it, it doesn't show the 1.04 solution either.



    So, my question is, is there any way to solve this problem without a numerical solver? And if not, is there any way to plug it into my TI 89 in order to get the correct answer?
     
  2. jcsd
  3. Dec 8, 2012 #2

    Mentallic

    User Avatar
    Homework Helper

    t=1.04 can't possibly be a solution because

    [tex]\sin(t^2)-t^2 < 0[/tex]

    for that value of t. How can you tell? Because the max sin can be is 1, and 1.042>1.
     
  4. Dec 8, 2012 #3

    jedishrfu

    Staff: Mentor

    for small radian values sin(x) ≈ x hence
    and that's probably the best you can do.
     
  5. Dec 8, 2012 #4
    Shoot, I realized that I misstated the problem. It's actually 1.225*sin(t2) - t2 =0

    But yeah, I realize that it still doesn't make much sense, but this is the answer and the graph that wolfram alpha spits out, and I'm just trying to figure out how to replicate that either on paper or on my calculator
     

    Attached Files:

  6. Dec 8, 2012 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why doesn't it make much sense? Your graph clearly shows zeroes near ##\pm 1##. You could work it by hand with your calculator using Newton's method with a starting value of ##x=1##. Make sure your calculator is in radian mode.
     
  7. Dec 8, 2012 #6
    Aaaaand there's my problem. I was doing it in degrees the whole time on my calculator, and couldn't figure out why my graph looked nothing like the wolfram graph. Derp. Thanks!!
     
  8. Dec 8, 2012 #7

    Mentallic

    User Avatar
    Homework Helper

    Common mistake :wink: I've done it enough times myself that it's now on my checklist of things that could have gone wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook