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I have to find the derivative of the dot-product of two vectors using the product rule. It took me an hour, checked every component and double checked, and then when I check it on Wolfram, of course it is wrong.

I have two vectors: $u(t) = \langle-\sqrt{2}sin(t), t, t^{2/3}\rangle$ and $v(t) = \langle-\sqrt{2}sin(t), cos^{2}(t), t^{-1/3}\rangle$

Since by product rule $\frac{d}{dt}[u(t) \cdot v(t)] = u'(t) \cdot v(t) + u(t) \cdot v'(t)$, I need to differentiate each vector before finding the sum-of-products by finding the component of each derivative.

For $\vec{u'(t)}$ I found: $$\Bigg\langle\frac{-sin(t) + 2tcos(t)}{2\sqrt{t}}, 1, \frac{2}{3t^{1/3}}\Bigg\rangle$$

For $\vec{v'(t)}$ I found: $$\Bigg\langle\frac{-sin(t) + 2tcos(t)}{2\sqrt{t}}, -2sin(t)cos(t), \frac{-1}{3t^{4/3}}\Bigg\rangle$$

Finding $\vec{u'(t)} \cdot v(t)$: $$\Bigg\langle\frac{sin^{2}(t)}{2} + tsin(t)cos(t), cos^{2}(t), \frac{-2}{3t^{2/3}}\Bigg\rangle$$

Finding $\vec{u(t)} \cdot v'(t)$: $$\Bigg\langle\frac{sin^{2}(t)}{2} + tsin(t)cos(t), -tsin(2t), \frac{-1}{3t^{2/3}}\Bigg\rangle$$

Finally, after finding the products, this is what I calculated for the sum: $$\Bigg\langle\sin^{2}(t) + 2tsin(t)cos(t), cos^{2}(t)-tsin(2t), \frac{-1}{t^{2/3}}\Bigg\rangle$$

But apparently, according to Wolfram, this is wrong. Where did I go astray here?

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Or, maybe they solved using different trig identities?? –  Dylan May 25 '12 at 0:44
Probably....... –  Gastón Burrull May 25 '12 at 0:45
Is it possible to find the derivative of the dot-product by first finding the dot product of the two vectors and then taking the derivative of the result? –  Dylan May 25 '12 at 0:58
I only ask because the method I used above took a long time and was very tedious. The latter approach, if possible, seems like it could be faster. –  Dylan May 25 '12 at 0:59
Yes, you can do it that way, and, yes, at least for this problem it looks like it will go faster. –  Gerry Myerson May 25 '12 at 1:01
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1 Answer

up vote 2 down vote accepted

The dot product returns a scalar, i.e. a real number. The derivative of this real-valued function is again a real-valued function. Thus, you should be looking for a real-valued solution, rather than the vector-valued solution that you've produced.

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I see. Do you mean to just add the values of the resulting elements in the vector? I guess I forgot that part. I wasn't thinking about it because the vector elements are functions of t. –  Dylan May 25 '12 at 5:05
Well, that's the most obvious thing; if you're asked for apples and you give oranges, then there's a problem. Also, though, I think there are some differentiation errors. Unless, when you wrote $\sqrt{2}\sin(t)$, you rather meant $\sqrt(t)\sin(t)$? –  Mark McClure May 25 '12 at 5:10
Yes. The differentiation should have read $-\sqrt{t}sin(t)$. As for obviousness ... it loses its obviousness when you are trying really hard to make sure the differentiation and all of the other operations in the algorithm are correct. Losing the forest for the trees I guess you could say. After correcting everything, the scalar I got was zero. So these vectors are orthogonal to each other (I probably got it wrong though, knowing me). –  Dylan May 25 '12 at 5:20
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