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Find the angle between the curves $v = 2u + 1, v = -2u +1$ on a surface with the first quadratic form: $E = 2, F = 1, G = 4$. I know I should probably use the $cos(\theta)=\frac{T_1(0)\cdot T_2(0)}{|T_1|\cdot|T_2|}$ however I'm unsure how to work in the first quadratic information without having more information about the surface. Thanks in advance for any help!

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The first curve is $r(t)=(t,2t+1)$ with tangent $T_1=(1,2)$ and the second is $\alpha(s)=(s,-2s+1)$ with tangent $T_2=(1,-2)$, so $$\cos \theta=\frac{2*1*1+1*1*-2+1*2*1+4*2*-2}{\sqrt{2*1*1+2(1*1*2)+4*2*2}\cdot\sqrt{2*1*1+2(1*1*-2)+4*-2*-2}} \\=\frac{2-2+2-16}{\sqrt{2+4+16}\cdot\sqrt{2-4+16}}$$

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  • $\begingroup$ That really helped me understand it. Thanks so much for your help! $\endgroup$
    – user116900
    Mar 5 '14 at 6:45

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