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Information given:
$$r_{1} = (t,t^2,t^3)$$ $$r_{2} = (\sin t,\sin2t,t)$$ These vector intersect at origin $(0,0,0)$

What I need to find: The angle between them.

What I tried: I tried to use the formula $\theta=\cos^{-1}\left(\dfrac{v_1\cdot v_2}{|v_1||v_2|}\right)$ , where $v_1 = r_1, v_2 =r_2$. But when I calculate $v_1.v_2$ I obtain $(t\sin t+t^2\sin {2t} + t^4)$ where $t=0$ so the angle is always 0.

The answer to this problem: around 66$^{\circ}$

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Then the corresponding angle is the angle between two vectors which can be calculated using calculus$$v_1=r_1'=(1,2t,3t^2)\\v_2=r_2'=(\cos t,2\cos 2t,1)$$therefore $$v_1=(1,0,0)\\v_2=(1,2,1)\\\to\theta=\cos^{-1}\left(\dfrac{v_1\cdot v_2}{|v_1||v_2|}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt 6}\right)$$

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  • $\begingroup$ Why did you take the derivative? $\endgroup$ – gimusi Mar 17 '18 at 19:38
  • $\begingroup$ Yea I would like to know too, because this is the good answer $\endgroup$ – xavier corbeil Mar 17 '18 at 19:42
  • $\begingroup$ Well! It is the tangent vector on its corresponding curve which is derived using $$\dfrac{r(x+\Delta x)-r(x)}{\Delta x}$$ $\endgroup$ – Mostafa Ayaz Mar 17 '18 at 19:46
  • $\begingroup$ Also use your intuition on 2D curves $\endgroup$ – Mostafa Ayaz Mar 17 '18 at 19:47
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Let use

$$\cos \theta = \frac{r_1\cdot r_2}{|r_1||r_2|}\implies \theta = \arccos \frac{r_1\cdot r_2}{|r_1||r_2|}$$

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