# Thread: Angle between 2 curves

1. ## Angle between 2 curves

Here's another one that he didn't go over & I am totally lost on.
Determine the angle between the two curves to the nearest degree for the curves at each point of intersection. Note they are not orthogonal trajectories.

a. y=x^2 and y=(x-2)^2
b. x^2 - y^2=3 and x^2 - 4x+y^2 +3=0

Thanks again!

2. a)

$\displaystyle y = x^2$

$\displaystyle y = (x-2)^2$

Point of intersection:

$\displaystyle x^2 = (x-2)^2$
$\displaystyle -4x + 4 = 0$
$\displaystyle x = 1 \Rightarrow y = 1$

Derivative:
$\displaystyle 1. y' = 2x$
$\displaystyle 2. y' = 2x - 4$

$\displaystyle 1. y'(1) = 2 = k_1$
$\displaystyle 2. y'(1) = -2 = k_2$

$\displaystyle tan \phi = |\frac{k_1 - k_2}{1+ k_1k_2}| = |\frac{2+2}{1-4}| = \frac {4}{3}$

$\displaystyle \phi = 53^0 8'$

3. Yeah, he didn't go over that at all. Thanks for the help with a. I am trying to figure out b the same way but I am pretty sure I'm not going to be getting that one on my own.

4. b)

$\displaystyle 1. x^2 - y^2 = 3 \Rightarrow y=\sqrt{x^2 -3}$
$\displaystyle 2. x^2 - 4x + y^2 + 3 = 0 \Rightarrow y = \sqrt{-x^2+4x-3}$

point of intersection:

$\displaystyle x^2 - 3 = -x^2 +4x -3$
$\displaystyle 2x^2 = 4x$
$\displaystyle x_1 = 0$ -> not suitable
$\displaystyle x_2 = 2 \Rightarrow y_1 = 1, y_2 = -1$

Derivatives:

1st curve: $\displaystyle 2x - 2y \cdot y' = 0$
$\displaystyle y' = \frac {x}{y} = \frac{x}{\sqrt{x^2-3}}$

2nd curve: $\displaystyle 2x -4 + 2y \cdot y' = 0$
$\displaystyle y' = \frac{-x+2}{y} = \frac{-x+2}{\sqrt{-x^2+4x3}}$

from the first curve we get 2 k's.
$\displaystyle k_1 = f'_1(2) = \frac {x}{y} = \frac{x}{\sqrt{x^2-3}} = \frac {2}{1} = 2$
$\displaystyle k_2 = f'_2(2) = \frac {x}{y} = \frac{x}{\sqrt{x^2-3}} = \frac {2}{-1} = -2$

from the second curve:
$\displaystyle k_3 = f'_3(2) = {-x+2}{\sqrt{-x^2+4x3}} = \frac {0}{\pm 1} = 0$

$\displaystyle tan \Phi = |\frac {k_{1,2} - k_3}{1+k_{1,2}k_3}| = 2$
$\displaystyle \Phi = 63^0 26'$

$\displaystyle k_{1,2}$ means you can take any of the 2 k's and the result will be the same..

5. hahahaha yeah. Wasn't even close to getting "b" right by myself. Thank you so very, very much for the help, even though I don't understand it!

6. what is it that you don't understand?

i can explain

7. Originally Posted by metlx
what is it that you don't understand?

i can explain
The POI for B. How do you get x1 and x2 from the problem 2x^2=4x?
If I am not mistaken, you are assigning x1 to 2x^2 and x2 to 4x.

8. $\displaystyle x^2 - 3 = -x^2 + 4x - 3$
$\displaystyle 2x^2 - 4x = 0$

$\displaystyle x(2x-4) = 0$
$\displaystyle x_1 = 0 ; x_2 = 2$

But $\displaystyle x_1 = 0$does not fit into the first equation $\displaystyle x^2 - y^2 = 3$

$\displaystyle 0^2 - y^2 = 3$
$\displaystyle 3 + y^2 = 0$

and since anything squared within $\displaystyle R$ cannot be negative then you can assume that the above equation is not right..