If $\alpha$ is the angle between two curves, find $\cos (\alpha)$ in terms of parametric representation of the curves. If $\alpha$ is the angle between two curves, find $\cos (\alpha)$ in terms of parametric representation of the curves.  Answer in book:

Is this answer correct?  I keep getting a minus sign between the top terms.  
Outline of my approach:
We draw a horizontal at the point of intersection.  By simple geometry the angle between the tangents is $\alpha=\theta + \phi$ where $\theta$ and $\phi$ are the angles between the $x$ axis and the tangent to the curves.  Taking cosine both sides, and using the fact that 
$\cos (t)=\pm \frac{\dot{x}}{\sqrt{\dot{x}^2 + \dot{y}^2}}$ and 
$\sin (t)=\pm \frac{\dot{y}}{\sqrt{\dot{x}^2 + \dot{y}^2}}$ as given in book for angle $t$ between $x$ axis and tangent to curve.  I find that I always get the negative after using the additional formula for $\cos$ after expansion.  

Assume that
$$\gamma:\quad u\mapsto{\bf z}(u)=\bigl(x(u),y(u)\bigr)$$
is the parametric representation of a curve  $\gamma\subset{\mathbb R}^2$. Then for any parameter value the vector ${\bf z}'(u)=\bigl(x'(u),y'(u)\bigr)$, if $\ne{\bf 0}$, is a tangent vector to $\gamma$ at the point ${\bf z}(u)$. In the problem at hand two curves $\gamma_1$, $\gamma_2$ are given, and it is assumed that for two particular parameter values $u$, resp. $v$, the curves intersect at a common point ${\bf z}_1(u)={\bf z}_2(v)$. In order to find the angle of intersection we have to make use of the following fact:
Given two nonzero vectors ${\bf a}$, ${\bf b}\in{\mathbb R}^n$ the cosine of the enclosed angle $\alpha$ is given by
$$\cos\alpha={{\bf a}\cdot{\bf b}\over|{\bf a}|\>|{\bf b}|}\ ,$$
whereby the dot denotes the scalar product. The latter is computed as follows:
$${\bf a}\cdot{\bf b}=\sum_{i=1}^n a_i\,b_i\ .$$
Now apply this to the two vectors ${\bf z}_1'(u)$, $\>{\bf z}_2'(v)$.
