Cross product of vector functions
$${\bf r}'(t) \times {\bf r}''(t) = \langle 5 \cos t, 5 \cos t, 4 \sin t \rangle \times \langle 5 \sin t, 5 \sin t, 4 \cos t \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ 5 \cos t & 5 \cos t & 4 \sin t \\ 5 \sin t & 5 \sin t & 4 \cos t \end{vmatrix}$$ $$= \begin{vmatrix} 5 \cos t & 4 \sin t \\ 5 \sin t & 4 \cos t \end{vmatrix} {\bf i}+ \begin{vmatrix} 5 \cos t & 4 \sin t \\ 5 \sin t & 4 \cos t \end{vmatrix} {\bf j}+ \begin{vmatrix} 5 \cos t & 5 \cos t \\ 5 \sin t & 5 \sin t \end{vmatrix} {\bf k} = \langle (5\cos t)(4\cos t)(5\sin t)(4\sin t), \left( (5\cos t)(4 \cos t)(5\sin t)(4\sin t) \right), (5\cos t)(5\sin t)(5\sin t)(5\cos t) \rangle$$ $$=\langle 20\cos^2 t +20\sin^2 t, 20\cos^2 t20\sin^2 t, 25\sin t\cos t+25\sin t \cos t\rangle = \langle 20, 20, 0 \rangle$$
Using $20\cos^2 t +20\sin^2 t = 20(\cos^2 t+\sin^2 t) = 20(1) = 20$ etc.
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Jeremy Rowler
Updated on August 01, 2022Comments

Jeremy Rowler over 1 year
I was trying to make sense of a problem when I stumbled upon this on yahoo answers. I was just wondering if it was correct. If it is, can you please maybe explain why?
${\bf r}'(t) = \langle 5 \cos t, 5 \cos t, 4 \sin t \rangle$
${\bf r}''(t) = \langle 5 \sin t, 5 \sin t, 4 \cos t \rangle$.
${\bf r}'(t) \times {\bf r}''(t) = \langle 20, 20, 0 \rangle$.

zerosofthezeta about 10 yearsI'm learning this exact same thing in class. Are you in my class??

Jeremy Rowler about 10 yearsI don't know.... U of M?


Jeremy Rowler about 10 yearsBut shouldn't it be <40,40,0>

Jonathan Aronson about 10 yearsno, 5 * 4 = 20.

Jeremy Rowler about 10 yearsI didn't see the way the cos and sin cancel out. But now I see, it makes a lot more sense now