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Let $r_1(t)=t^2i+tj+3t^3k$ and $r_2(t)=(t-1)i+(1/4)t^2j+(5-t)k$. First show that the graphs of $r_1$ and $r_2$ intersect at $(1,1,3)$. Second find an equation for the tangent line to $r_1$ at $(1,1,3)$. And hird, find $\cos(\theta)$, there $\theta$ is the acute angle between the tangent lines to $r_1$ and $r_2$ at $(1,1,3)$.

tracy
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  • @PeterWoolfitt Can you help me here too? – tracy Mar 05 '15 at 23:17
  • How far have you gotten? Can you do the first part at least? How does one find the tangent line to a graph at a point? Is there some formula you know that would let you calculate the cosine of the angle between $r_1$ and $r_2$ at $(1,1,3)$? –  Mar 05 '15 at 23:27
  • @Bye_World Can you just work out the intersection for me, I can do the rest after that – tracy Mar 05 '15 at 23:52
  • OK. That's the easy part though. –  Mar 05 '15 at 23:58

1 Answers1

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Part $1)$

If there is an intersection of $r_1$ and $r_2$, then the point of intersection must be a solution to $r_1(t) = ...$ and $r_2(s) = ...$. So let's make sure it is.

$$r_1(t) = (1,1,3) = (t^2,t,3t^3) \implies \begin{cases} 1=t^2 \\ 1=t \\ 3=3t^3\end{cases}$$ Clearly this has a solution: $t=1$.

$$r_2(s) = (1,1,3) = (s-1, (1/4)s^2, 5-s) \implies \begin{cases} 1=s-1 \\ 1=(1/4)s^2 \\ 3 = 5-s\end{cases}$$ Clearly this has the solution $s=2$.

So, because the graphs of $r_1$ and $r_2$ pass through the same point (at two different times $t=1$ and $s=2$), that point $(1,1,3)$ is a point of intersection of the two graphs.