I've understood that if we are given a curve $R(x,y,z)=xi + yj + zk$, $\frac{d}{dt}R(x,y, z) $ is a tangent vector to the curve. If we differentiate the tangent vector again, we obtain a vector that is normal to the curve (As it is normal to the tangent vector). Similarly, if R(x,y,z) defines the position of the body, $\frac{d}{dt}R(x,y, z) $ should give the equation of the velocity vector V(x,y,z) of the particle, which is tangential to the position curve. Using the above concept, shouldn't the acceleration, $\frac{d}{dt}V(x,y, z) $ just be normal to the position curve? Why do we manipulate $\frac{d}{dt}R(x,y, z) $ as $\frac{d}{ds}R(x,y, z) \frac{d}{dt}s $ (where s is the length of curve) and then differentiate it to arrive at two components of acceleration?
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Under arc-length parametrization, curves will have unit length. – qqo Jan 22 '17 at 08:10
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I understand that, but why should acceleration be an exception to the fact that derivative of tangent is a normal? – b_boundary Jan 22 '17 at 08:27
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Hint: think of a point accelerating in a straight line (e.g. an object falling downwards). What happens if you differentiate the velocity vector?
Anonymous
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The acceleration should be parallel to the velocity vector? Physically I know it's true, but mathematically I have only understood that differentiating the tangent vector(velocity in this case) should give it's normal, which doesn't tell me why acceleration should be in any other direction – b_boundary Jan 22 '17 at 08:30
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You know how to differentiate a vector, right? You write it as a modulus times a unit vector. Then apply the usual rules of differentiation to the product: derivative of the modulus times the unit vector plus modulus times the derivative of the unit vector. It's the derivative of a unit vector (or in general, of a constant modulus vector) that is orthogonal to the vector, not the derivative of an arbitrary vector. – Anonymous Jan 22 '17 at 08:36
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