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

  • #1
Matt Jacques
81
0
Hi,

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

Thanks

Matt
 

Answers and Replies

  • #2
mathman
Science Advisor
8,065
541
The derivative is the slope of the tangent line. The atan is thus the angle of the tangent line with the x axis.
 
  • #3
Matt Jacques
81
0
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?
 
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  • #4
On Radioactive Waves
137
0
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:
  • #5
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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.
 
  • #6
Matt Jacques
81
0
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]
 
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