1. Limited time only! Sign up for a free 30min personal 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!

How to find angle of a curve in xy coordinate plane?

  1. Aug 14, 2003 #1
    Hi,

    I think it is atan of the derivative of the curve, is that right?

    Thanks

    Matt
     
  2. jcsd
  3. Aug 14, 2003 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    The derivative is the slope of the tangent line. The atan is thus the angle of the tangent line with the x axis.
     
  4. Aug 14, 2003 #3
    Thanks for the quick reply. Yeah, that is what I thought. However, I have a followup question.

    For the vertical position of a projectile the formula is:

    y = initialheight + VyT - .5(9.8)T^2

    The derivative of that is

    d/dx = Vy - 9.8T

    I did a sample problem, Vi=4, angle = 30º, initial height = 1.2

    Total Time: .739 seconds
    Height of Max time: .202 seconds
    Distance: 2.56 m

    if I take atan( Vy - 9.8T) when t = 0 the angle comes up as 63º, not 30º which it should be.

    Hmm?
     
    Last edited: Aug 14, 2003
  5. Aug 15, 2003 #4
    I started to work out the problem, but you didn't state it quite clearly....

    Do you mean V0x is the initial Vi in the direction of vector i ?
     
    Last edited: Aug 15, 2003
  6. Aug 15, 2003 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    dy/dx is the arctan of the angle the curve makes with the horizontal.

    dy/dt has no geometric meaning (in an xy-coordinate system).

    If Vi= 4 is the initial speed and the projectile is fired at a 30 degree angle to the horizontal, then the initial vertical speed is Vy= 4 cos(30)= 4(1/2)= 2 and the initial horizontal speed is
    Vx= 4 sin(30)= 4([sqrt](3)/2)= 2 [sqrt](3).

    The vertical speed would be given by dy/dt= 2- 9.8t and the horizontal speed (assuming gravity is the only force) by
    dx/dt= 2[sqrt]3. At t= 0, dy/dt= 2 and dx/dt= 2[sqrt](3).
    By the chain rule, dy/dx= (dy/dt)/(dx/dt)= 2/2[sqrt](3)= 1/[sqrt](3)= [sqrt](3)/3, the tangent of 30 degrees.
     
  7. Aug 15, 2003 #6
    Thanks, Hallsofivy. You created a good thorough reply. However, I think you accidentally swapped cosine and sine in Vx and Vy.

    Anyway, for the benifit of searches, here is the formula for the angle:

    http://homepage.mac.com/jjacques2/angle.gif [Broken]
     
    Last edited by a moderator: May 1, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook