How Do I Find the Derivative of This Vector Function?

megr_ftw
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Homework Statement


Find the derivative of the vector function
r(t)=ta X (b+at)
where a=<4,5,2>, b=<1,-3,2>, and c=<4,3,1>


Homework Equations





The Attempt at a Solution


I know how to take the derivative and everything but the way this question is worded confuses me!
I'm assuming the X means cross product? but it may just mean multiply. Do I plug in the values of a,b,c, and then do what with all the t's?
 
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I'm pretty sure X means cross product. People generally don't use X to mean ordinary multiplication at the calculus level.

Yes, substitute the values for a, b, and c, and then carry out the cross product. You'll end up with either (...)i + (...)j + (...)k or <..., ..., ...>, both of which will have terms with t in them. To get r'(t), just take the derivative of each of the three components.
 
should I distribute the t to the a values?
 
Yes. t is a scalar, so ta = <4t, 5t, 2t>. at is the same as ta.
 
Don't carry out the product! Just use the product rule for vector multiplication:
\vec{f}= \vec{a}t\times (\vec{b}+\vec{c}t)

so \vec{f}&#039;= \vec{a}\times (\vec{b}+ \vec{c}t)+ \vec{a}t \times \vec{c}= \vec{a}\times\vec{b}+ 2\vec{a}\times\vec{c}t.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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