Review for final Exam

Tue, 29 Jul 2014 18:56:25 -0500

I was looking back over Exam 1 and question number ten is still giving me a problem. Can someone please explain the right path that I should have taken to get the right answer? The question was Let $p(t)=<1+3t, 2+4t>$ and let $q(t)=<3-t, -4+3t>$. Find the point of intersection between the paths parametrized by p and q. What I attempted was to solve for u and t $$1=3t+3-u \rightarrow -2+3t=-u \rightarrow (2-3t)=u$$ $$2+4t=-4-3u$$ $$2+4t=-4-3(2-3t)$$ $$6+4t=-6+9t$$ $$12=5t$$ $$t=12/5$$ but then I don't know what to do complete the problem after that.

Answer by Christina for Review for final Exam

Tue, 29 Jul 2014 20:01:54 -0500

This is how I solved this problem: I left $\vec{p}(t)$ in terms of $t$ and changed the parameter of $\vec{q}$ to $s$. I then set the corresponding components of each equal to each other. First the x terms: $$1+3(t) = 3-s$$ Then solve for s $$s=-3t+2$$ Now the y terms: $$2+4t = -4+3s$$ Here I plugged in my s value: $$2+4t=-4+3(-3t+2)$$ And now everything is in terms of t, so solve for t: $$2+4t=-4-9t+6$$ $$4t=9t$$ $$t=0$$ Now I use this t value to solve my original x term equality: $$1+3(0)=3-s$$ Solve for s: $$s=2$$ Now, go back to the original paths and plug in our values, this is the point that the paths will cross: $$\vec{p}(0) = \langle1+0, 2+0\rangle$$ $$=\langle1,2\rangle$$ $$\vec{q}(2) = \langle3-2,-4+3(2)\rangle$$ $$=\langle1,2\rangle$$ Point of intersenction: $\langle1,2\rangle$ As far as part b, do the motions collide? I answered no, because the paths cross this point at different time values.

Line Integrals

Review for final Exam

Answer by Christina for Review for final Exam