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

In summary, we discussed the concept of derivatives and their relationship to tangent lines. We also looked at how to use derivatives to find the angle of a curve with the horizontal. The formula for this is dy/dx = Vy / Vx, where Vy is the vertical speed and Vx is the horizontal speed. Using this formula, we can find the angle of a projectile's path with respect to the horizontal.
  • #1
Matt Jacques
81
0
Hi,

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

Thanks

Matt
 
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  • #2
The derivative is the slope of the tangent line. The atan is thus the angle of the tangent line with the x axis.
 
  • #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?
 
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  • #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 ?
 
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  • #5
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
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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1. What is the formula for finding the angle of a curve in the xy coordinate plane?

The formula for finding the angle of a curve in the xy coordinate plane is arctan(dy/dx), where dy/dx represents the gradient or slope of the curve at a given point.

2. Can I use the same formula for finding the angle of any type of curve?

Yes, the formula for finding the angle of a curve in the xy coordinate plane can be used for any type of curve, including straight lines, circles, and more complex curves.

3. What is the significance of finding the angle of a curve in the xy coordinate plane?

Knowing the angle of a curve at a given point can provide insight into the direction and rate of change of the curve. It can also be useful in various applications, such as engineering and physics.

4. How do I calculate the gradient or slope of a curve for the formula?

The gradient or slope of a curve can be calculated by finding the derivative of the curve's equation with respect to the variable (usually x) and plugging in the x-value of the point you want to find the angle of.

5. Are there any other methods for finding the angle of a curve in the xy coordinate plane?

Yes, there are other methods such as using the arc length formula or finding the angle between tangents at two points on the curve. However, the arctan(dy/dx) formula is the most commonly used and easiest method.

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