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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

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

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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