What are the extreme values of the speed for a given parametric curve?

[FrogPad]

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First question:
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This first question is kind of weird. I'm not even sure where to go with it. If anyone has a hint, that would be awesome.

From $\vec A \times \vec B = -\vec B \times \vec A$ deduce $\vec A \times \vec A = 0$

Can it be as simple as:
let $$\vec B = \vec A_0 | \vec A_0 = \vec A$$
thus: $$\vec A \times \vec A_0 = -\vec A_0 \times \vec A = 0$$

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Second question:
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Find the minimum and maximum speed if $x=t+\cos t$, $y=t-\sin t$.

Please allow me to take advantage of the inner space operator for sake of ease while writing the vectors :)

Thus:
$$\vec x = <t+\cos t,t-\sin t>$$
$$\vec v = <1-\sin t, 1-cos t>$$

So speed is computed as: $|\vec v|$. Therefore the largest speed values that can occur are when: $\vec v = <1,2> or <2,1>$ and the lowest speed values that can occur are when $\vec v = <1,0> or <0,1>$.

Is this reasoning even correct with this problem?

 

[StatusX]

For the first one, I'm not sure what A₀ is, but all you have to do is let B=A in the first identity. For the second, you might want to be a little more rigorous. You can find the extrema of the speed by:

$$\frac{d}{dt} |\vec v|^2 = 0$$

And use:$$\frac{d}{dt} |\vec v|^2 = \frac{d}{dt} (\vec v\cdot \vec v)$$

Which will lead you to the equation:

$$\vec v \cdot \vec a = 0$$

 

[FrogPad]

With the $\vec A_0 = \vec A$ I was just trying to show that I was plugging $\vec A$ into the expression. It really wasn't necessary and actually more confusing (I left this out of the homework).

For the second one. Cool Thank you. That's what I wasn't doing. I needed to find the extrema of the speed not of the velocity vector. I appreciate it.