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posted 2014-06-30 14:42:31 -0600

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! :)

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! :)

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.