Help with a vector proof about perpedndicular unit vectors

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  • #1
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Prove that if a(t) is a unit vector whose components are functions of t, then the vector b(t) = da(t)/dt is perpendicular to a(t)



here i have the parameters that make a(t) a unit vector:

1 = f(t)^2 + g(t)^2




i know of the example sinxi + cosxj = a(t)
but that is the only one i can get to work. I need a general proof though so:

a(t) = f(t)^2 i + g(t)^2 j

b(t) = f '(t) i + g '(t) j

(dot) product of a(t) and b(t) is:

f(t)*f '(t) + g(t)*g '(t)

i tried to take the derivative of a(t) and solve for both f'(t) and g'(t)

1 = f(t)^2 + g(t)^2

0 = 2*f(t)*f'(t) + 2*g(t)*g'(t)

i treated this like one big equation and solved each for g'(t) and f'(t)

f ' (t) = [-g(t)*g ' (t)]/ f(t)

g ' (t) = [-f(t)*f ' (t)]/ g(t)

when i carry out the dot product i still dont get zero, i need some advice, am i even on the right track.

the only other way i could think of was to setting each component equal to zero to solve for the derivative of the functions, but i did not know if this was mathematically legal

ex:

2*g(t)*g'(t) = 0

g'(t) = 0
 
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Answers and Replies

  • #2
cepheid
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Prove that if a(t) is a unit vector whose components are functions of t, then the vector b(t) = da(t)/dt is perpendicular to a(t)



here i have the parameters that make a(t) a unit vector:

1 = f(t)^2 + g(t)^2
Use this expression to express g(t) in terms of f(t). That will help greatly.
i know of the example sinxi + cosxj = a(t)
but that is the only one i can get to work. I need a general proof though so:

a(t) = f(t)^2 i + g(t)^2 j

b(t) = f '(t) i + g '(t) j



I don't understand the part in red. Why have you squared the functions? They should not be squared if your definition of a is to be consistent with your definition of b. The rest of what you have doesn't make sense either. See if following my advice above helps.
 
  • #3
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Use this expression to express g(t) in terms of f(t). That will help greatly.
did you mean this one: 1 = f(t)^2 + g(t)^2

so f(t) = (1 - g(t)^2)^1/2

and g(t) = (1- f(t)^2)^1/2

then the derivative of the above:

f ' (t) = -[g(t)*g ' (t)] / [1-g(t)^2]
g ' (t) = -[f(t)*f ' (t)] / [1-f(t)^2]

so then if i take the dot product

[f(t) * f ' (t)] + [ g(t) * g ' (t) ]

(1 - g(t)^2)^1/2 * -[g(t)*g ' (t)] / [1-g(t)^2] + (1- f(t)^2)^1/2 * -[f(t)*f ' (t)] / [1-f(t)^2]

simplified :

-[g(t)*g ' (t)] -[f(t)*f ' (t)]

still dont understand it, where is my error?? :confused:
 
  • #4
cepheid
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did you mean this one: 1 = f(t)^2 + g(t)^2
Yes.

so f(t) = (1 - g(t)^2)^1/2

and g(t) = (1- f(t)^2)^1/2
You just need one of these. The whole point is to eliminate one by expressing it in terms of the other. For example, if you express g in terms of f, then henceforth everything is in terms of f and you don't need g anymore. Do you understand?

then the derivative of the above:

f ' (t) = -[g(t)*g ' (t)] / [1-g(t)^2]
g ' (t) = -[f(t)*f ' (t)] / [1-f(t)^2]
Here is the problem. This differentiation is wrong. You need to use the chain rule since:

dg/dt = dg/df * df/dt

What is g(f), and what is its derivative with respect to f? That is the first step. The second step is to multiply that by df/dt.

Also, again, you only have to compute ONE of the above derivatives.
 
  • #5
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got it finished, thanks. i wasnt thinking of eliminating terms correctly, like you said
 

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