Help with a vector proof about perpedndicular unit vectors

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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cepheid
Staff Emeritus
Gold Member
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.

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??

cepheid
Staff Emeritus
Gold Member
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.

got it finished, thanks. i wasnt thinking of eliminating terms correctly, like you said