http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Mon, 30 Jun 2014 20:33:21 -0500http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/So, I've been trying to work on the homework 13.2, and I'm stuck on both questions 6 and 7. But we'll start with question 6. "6. Find the cosine of the angle between the curves $\langle0,t^2,t\rangle$ and $\langle\cos(\frac{\pi t}{2}),\sin(\frac{\pi t}{2}),t\rangle$ where they intersect." Well I can tell that they intersect at t=1 And from the book on page 336 it states... " $ \cos(\theta) $ = $\mathbf {\frac{\vec r' * \vec s'}{|\vec r'||\vec s'|}}$ = $ \mathbf {\frac {\vec r'}{|\vec r'|} \cdot \frac{\vec s'}{|\vec s'|}} $ " So letting $\vec r$ = $\langle0,t^2,t\rangle$ and $\vec s$ = $\langle \cos(\frac{\pi t}{2}), \sin(\frac{\pi t}{2}),t\rangle$ I end up with $\vec r'$ = $\langle 0,2t,1\rangle$ and $\vec s' $ = $\langle (\frac{-1}{2} \pi \sin(\frac{\pi t}{2})), (\frac 12 \pi \cos(\frac{\pi t}{2})), 1\rangle $ using this to come up with | $ \vec r'$| I get |$\vec r'$| = $\sqrt{0^2 + (2t)^2 + 1^2} $ = $\sqrt{4t^2 +1} $ and |$\vec s'$|= $\sqrt{(\frac{-1}{2}\pi \sin(\frac {\pi t}{2}))^2 + (\frac 12 \pi \cos(\frac{\pi t}{2}))^2 + 1^2) } $ Now when we plug this into the $\cos(\theta)$ equation, do we substitute t=1 for the t in all of the equations to find the answer, or how exactly are we supposed to approach this problem? I end up with this ridiculous answer with a whole bunch of randomness in it that doesn't match up with the back of the book, so I'm beyond confused on this problem, and feel like there is a much easier way than how i'm trying to figure it out. HELLLPPPP! :) *Comment*: When you take the norm of $\vec{s}'$, note that you get a $\sin^2+\cos^2$. That yields a significant simplificatoin that might help.Mon, 30 Jun 2014 14:42:31 -0500http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/?answer=28#post-id-28You were really close! I would strongly recommend plugging in $t = 1$ as soon as you find your derivatives. So our derivatives are: $$ \vec r'(t) = \langle 0, 2t, 1 \rangle $$ $$ \vec s'(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} t\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} t \right), 1 \right\rangle $$ Plugging in $t = 1$: $$ \vec r'(1) = \langle 0, 2(1), 1 \rangle = \langle 0, 2, 1 \rangle $$ $$ \vec s'(t) = \left\langle -\frac{\pi}{2} \sin \left(\frac{\pi}{2} (1)\right), \frac{\pi}{2} \cos \left(\frac{\pi}{2} (1)\right), 1 \right\rangle = \left\langle -\frac{\pi}{2}, 0, 1 \right\rangle $$ Now we shall plug it in to our favorite equation involving the dot product: $$ \cos \theta = \frac{\langle 0, 2, 1 \rangle \cdot \langle -\pi/2, 0, 1 \rangle}{\left|\left| \langle 0, 2, 1 \rangle \right|\right| \left|\left| \langle -\pi/2, 0, 1 \rangle \right|\right|} = \frac{0 * (-\pi/2) + 2 * 0 + 1 * 1 }{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$ That's kinda a crazy fraction... so the book decided to be all clever and simplify. Here is how they simplified. First, multiply by $\frac{2}{2}$. $$ \frac{1}{\sqrt{5} * \sqrt{(\pi^2/4) + 1}} * \frac{2}{2} = \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} $$ This allows us to pull the $2$ *into* the second radical as a $4$ (since $\sqrt{4} = 2$). I also distribute the four under the radical in this step. $$ \frac{2}{2 * \sqrt{5} * \sqrt{(\pi^2/4) + 1}} = \frac{2}{\sqrt{5} * \sqrt{4* ((\pi^2/4) + 1)}} = \frac{2}{\sqrt{5} * \sqrt{\pi^2 + 4}} $$ This is the same thing as: $$ 2 \big/ \sqrt{5} \big / \sqrt{\pi^2 + 4} $$ Which is what the book has! You got the calculus part 100% right... sometimes the algebra can just be a pain in the neck.Mon, 30 Jun 2014 15:11:48 -0500http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/?answer=28#post-id-28http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/?answer=31#post-id-31So, I cant add comments - but this looks great!!Mon, 30 Jun 2014 20:33:21 -0500http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/?answer=31#post-id-31