Abgle Between Curves at a Point

Suppose two curves meet at a point. We can find the angle between the curves by considering the gradients of the tangents to the two curves at the point.
Example:  
\[C_1 :x^2+y^2=25, \; C_2:y=x^2-5\]
.
To find intersection points solve the simultaneous equations
\[x^2+y^2=25\]
  (1)
\[x^2-y=5\]
  (2)
(2)-(1) gives  
\[y^2+y=20 \rightarrow y^2+y-20=0 \rightarrow(y+5)(y-4)=0 \rightarrow y=-5, \; 4\]
.
Take  
\[y=4\]
  then  
\[x^2=4+5=9 \rightarrow x=-3, \; 3\]
.
The gradient function for  
\[C_1:x^2+y^2=25\]
  is  
\[\frac{dy}{sx}- \frac{x}{y}\]
  so at  
\[(3,4)\]
  the gradient is  
\[m= \frac{dy}{dx} |_{(3,4)}=- \frac{3}{4}\]
.
The gradient function for  
\[C_2:y=x^2-5\]
  is  
\[\frac{dy}{dx}=2x\]
  so at  
\[(3,4)\]
  the gradient is  
\[m=\frac{dy}{dx} |_{(3,4)} = 6\]
.
We can take tangent vectors to be  
\[\mathbf{u} = \begin{pmatrix}3\\ -4 \end{pmatrix}\]
  and  
\[\mathbf{v} = \begin{pmatrix}1\\ 6 \end{pmatrix}\]
.
The angle between the curves is
\[cos \theta = \frac{ \mathbf{u} \cdot \mathbf{v}}{ | \mathbf{u} | | \mathbf{v} | }= \frac{\begin{pmatrix}3\\ -4 \end{pmatrix} \cdot \begin{pmatrix}1\\ 6 \end{pmatrix}}{ \sqrt{3^2+(-4)^2} \sqrt{1^2+6^2}}= \frac{3 \times 1+ (-4) \times 6}{5 \sqrt{37}}= - \frac{21}{5 \sqrt{37}}\]

Then  
\[\theta = cos^{-1}( - \frac{21}{5 \sqrt{37}})=133.67^o \]
.

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