Find the cosine of the angle between two curves and also find where they intersect

Question

Find the cosine of the angle between the curves $\left\langle 0,t^2,t \right\rangle$ and $\left\langle \cos\left(\pi \frac{t}{2}\right), \sin \left(\pi \frac{t}{2} \right), t \right\rangle$ where they intersect.

To solve this I first found the point of intersection by setting each of the respective components equal to each other.

0ドル = \cos \left(\frac{\pi u}{2}\right)$ solving this equation gives 1ドル=u$

$t^2 = \sin \left(\frac{\pi u}{2} \right)$ solving this equation gives 1ドル=u$

$t = u$ solving this equations gives 1ドル=u$

Putting this together I can see that my point of intersection is $( 0,1,1)$ Now to find the cosine of the angle between those curves I know I need the derivatives of those two vectors:

$\langle 0,2t,1 \rangle$ and $\left\langle -\sin\left(\frac{\pi}{2}\right), \cos\left(\frac{\pi}{2}\right), 1 \right\rangle$

From here, though, I'm not sure how to calculate the cosine of the angle between the two curves. Any hints?

• mr_e_man: The second derivative (that is, the first derivative of the second curve) needs factors of $\pi/2$ outside of $\sin$ & $\cos$.
• mr_e_man: Also your derivation of the intersection point is wrong (there are many solutions to $\cos\theta=0$), though the final result $t=u=1$ is right.
• CiaPan: If you prefix the trigonometric functions' names with a backslash, they will become LaTeX/MathJax symbols, which will make them to render in an upright font with appropriate spacing (m \cos 2x → $m \cos 2x$) instead of a blob of italic, variable-like letters (m cos 2x → $m cos 2x$).

Answer 1

Once you have found the derivative vectors, at the intersection point $(0,2,1)$ and $(-1,0,1)$ render them unitary.
Thereafter their dot product will be the cosine of the angle between the two curves, at the intersection point.

Answer 2

The dot product of two vectors, u, v, is give by $u\cdot v= |u||v|cos(\theta)$ where $\theta$ is the angle between the two vectors. At t= 1, the two tangent vectors are <0, 1, 1> and <-1, 0, 1>.

• nullByteMe: I thought the tangent vectors were the derivatives of the given vectors?