- #1
Ocata
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Homework Statement
Basically, I'm looking at the property that says if the magnitude of a vector valued function is constant, then the vector function dotted with it's derivative will be zero. But I'm stuck towards the end because the proof I found online seems to skip a step that I'm not certain about.
r(t) = <x(t),y(t) >
r(t) ⋅ r(t) = <x(t),y(t) > ⋅ <x(t),y(t) >
= [itex] x(t)^{2} + y(t)^{2} = \sqrt{x(t)^{2} + y(t)^{2}}^{2}[/itex]
= [itex] ll<x(t),y(t)>ll^{2} = ll r(t)ll^{2} [/itex] = constant (suppose)
that is,
r(t) ⋅ r(t) = constant
Then, the derivative of r(t) ⋅ r(t):
[itex] \frac{d}{dt}r(t) ⋅ r(t) = \frac{d}{dt} c [/itex]
r'(t) ⋅ r(t) + r(t) ⋅ r'(t) = 0
After this is where I'm stuck.
The proofs I've seen online then say:
2r(t) ⋅ r'(t) = 0
r(t) ⋅ r'(t) = 0
But, algebraically, can this only be true if:
r'(t) ⋅ r(t) + r(t) ⋅ r'(t) = r'(t) ⋅ r(t) + r'(t) ⋅ r(t) ?
I know that the dot product is commutative such that:
v ⋅ u = u ⋅ v
But, does the commutative property for the dot product extend to the product rule for dot product of vector valued functions?
Homework Equations
v ⋅ u = u ⋅ v
The Attempt at a Solution
since r'(t) = < x'(t),y'(t)>
then if commutative property is true, then:
r'(t) ⋅ r(t) + r(t) ⋅ r'(t) = r'(t) ⋅ r(t) + r'(t) ⋅ r(t)
The only way I can believe this to be true is if I prove it some how.
I will try to break it down to components and rearrange the terms in green so that that they resemble the portion in blue. Not sure if it will work, but I'll give it a try..
<x'(t),y'(t) > ⋅ <x(t),y(t) > + <x(t),y(t) > ⋅ <x'(t),y'(t)> = <x'(t),y'(t) > ⋅ <x(t),y(t) > + <x'(t),y'(t) > ⋅ <x(t),y(t) >
(x'(t)x(t) + y'(t)y(t)) + (x(t)x'(t) + y(t)y'(t)) = (x'(t)x(t) + y'(t)y(t)) + (x'(t)x(t) + y'(t)y(t))
Now, I suppose it is reasonable that: x(t)x'(t) = x'(t)x(t)
and so, x(t)x'(t) + y(t)y'(t) = x'(t)x(t) + y'(t)y(t)
Then,
(x'(t)x(t) + y'(t)y(t)) + (x(t)x'(t) + y(t)y'(t)) = (x'(t)x(t) + y'(t)y(t)) + (x'(t)x(t) + y'(t)y(t))
can be written as:
(x'(t)x(t) + y'(t)y(t)) + (x(t)x'(t) + y(t)y'(t)) = (x'(t)x(t) + y'(t)y(t)) + (x(t)x'(t) + y(t)y'(t))
And thus:
<x'(t),y'(t) > ⋅ <x(t),y(t) > + <x(t),y(t) > ⋅ <x'(t),y'(t)> = <x'(t),y'(t) > ⋅ <x(t),y(t) > + <x(t),y(t) > ⋅ <x'(t),y'(t)>
r'(t) ⋅ r(t) + r(t) ⋅ r'(t) = r'(t) ⋅ r(t) + r(t) ⋅ r'(t)In Conclusion, since dot product is commutative by itself, it is also true that it is commutative when the dot product exists within larger statement such as a product rule, that is:
r'(t) ⋅ r(t) + r(t) ⋅ r'(t) = r'(t) ⋅ r(t) + r'(t) ⋅ r(t)
And therefore,
It can be written that
r'(t) ⋅ r(t) + r(t) ⋅ r'(t) = r'(t) ⋅ r(t) + r'(t) ⋅ r(t) = 2(r'(t) ⋅ r(t))Is this correct reasoning?
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