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?
$\endgroup$ 42 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$You wrote dot product of 2 vectors as a vector. It should be a scalar number.
u′(t)⋅v(t) should be a scalar.
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