# finding the angle

• Mar 11th 2008, 11:21 PM
lllll
finding the angle
I need help finding the angle of the curves $\displaystyle r_1(t) = (t, 1-t, 3+t^2), \ \ r_2(s) = (3-s, s-2, s^2)$ I found that they intersect at $\displaystyle x=1, \ \ y = 0, \ \ z= 4$ , and would the angle between them be $\displaystyle cos^{-1} \frac{(3)(0)+(1)(-2)+(3)(1)}{(\sqrt{1+9})(\sqrt{9+4+1})} = cos^{-1} \frac{1}{\sqrt{140}}=1.48618$ ?
• Mar 12th 2008, 04:13 AM
mr fantastic
Quote:

Originally Posted by lllll
I need help finding the angle of the curves $\displaystyle r_1(t) = (t, 1-t, 3+t^2), \ \ r_2(s) = (3-s, s-2, s^2)$ I found that they intersect at $\displaystyle x=1, \ \ y = 0, \ \ z= 4$ , and would the angle between them be $\displaystyle cos^{-1} \frac{(3)(0)+(1)(-2)+(3)(1)}{(\sqrt{1+9})(\sqrt{9+4+1})} = cos^{-1} \frac{1}{\sqrt{140}}=1.48618$ ?

The intersection point corresponds to t = 1 and s = 2.

So I think you want the angle between the vectors $\displaystyle \left[ \frac{dr_1}{dt}\right]_{t = 1}$ and $\displaystyle \left[ \frac{dr_2}{ds}\right]_{s = 2}$ .....
• Mar 12th 2008, 11:30 AM
lllll
Quote:

So I think you want the angle between the vectors $\displaystyle \left[ \frac{dr_1}{dt}\right]_{t = 1}$ and $\displaystyle \left[ \frac{dr_2}{ds}\right]_{s = 2}$ .....
why would we have to differentiate to find the angle? once you get the derivative you only get:

$\displaystyle r'(t) = (1,-1,2t), \ \ r'(s) = (-1,1,2s)$ and if I put $\displaystyle t=1$ and $\displaystyle s=2$ that gives $\displaystyle r'(t)= (1,-1,2), \ \ r'(s) = (-1,1,4)$. I don't see how I can use those coordinates to find the angle. Would you have to use $\displaystyle cos(\theta) \frac{a \cdot b}{|a||b|}$ ?
• Mar 12th 2008, 12:17 PM
Plato
Quote:

Originally Posted by lllll
why would we have to differentiate to find the angle? once you get the derivative you only get: $\displaystyle r'(t) = (1,-1,2t), \ \ r'(s) = (-1,1,2s)$ and if I put $\displaystyle t=1$ and $\displaystyle s=2$ that gives $\displaystyle r'(t)= (1,-1,2), \ \ r'(s) = (-1,1,4)$. I don't see how I can use those coordinates to find the angle. Would you have to use $\displaystyle cos(\theta) \frac{a \cdot b}{|a||b|}$ ?

The derivative gives the direction of the curve at any time.
The angle between the curves is the angle between the directions.
If $\displaystyle r'(t) = (1,-1,2t), \ \ r'(s) = (-1,1,2s)$ then $\displaystyle r'(1) = (1,-1,2), \ \ r'(2) = (-1,1,4)$ are the two directions.
The angle is $\displaystyle \phi = \arccos \left( {\frac{{ < 1, -1, 2 > \cdot < - 1,1,4 > }}{{\left\| { < 1,-1, 2 > } \right\|\left\| { < -1,1, 4 > } \right\|}}} \right)$.