2
$\begingroup$

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: $\Bigg\langle\frac{-\sin(t) + 2t\cos(t)}{2\sqrt{t}}, 1, \frac{2}{3t^{1/3}}\Bigg\rangle$

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

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

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

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

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

  • 3
    Yes, you can do it that way, and, yes, at least for this problem it looks like it will go faster.2012-05-25

2 Answers 2

4

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.

  • 0
    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).2012-05-25
1

You wrote dot product of 2 vectors as a vector. It should be a scalar number.

u′(t)⋅v(t) should be a scalar.

  • 1
    Does this solve the OP's question as a whole? Please elaborate.2017-04-29