Understanding Vector Magnitude and Direction in Mechanics

[MathematicalPhysicist]

my question is from klppner's text an introduction to mechanics, on page 26, they suppose a vector A(t) has a constant magnitude A, he states that the direction dA(t)/dt is always perpendicular to A(t), now they define \triangle A=A(t+\triangle t)-A(t), he states that |\triangle A|=2Asin(\frac{\triangle\theta}{2}) i don't understand how did they arrive at the last equation, and don't understand which angle does he refer to. he states that it's defined in a sketch in the book but i can't find where does he state this?

thanks in advance.

 

[neutrino]

I don't have the book with me now, but I think \Delta\theta is the angle between the A(t) and A(t+\Delta t). The equation becomes more accurate as the angle becomes smaller.

 

[quasar987]

\Delta \theta is the angle betweenA(t+\triangle t) and A(t). Since both vectors are of magnitude A, the law of cosines gives

\Delta A^2 = A^2+A^2-2AAcos(\Delta \theta)

And cos(2x) = 1 - 2sin^2(x), so...

 

[benorin]

Ok, Draw a vector for A(t), and then draw another for A(t+\triangle t) with the same length and starting from the same point in space. The angle \triangle\theta is acute angle between these two vectors and \triangle A is the vector that forms the third side of the isosceles triangle having A(t) and A(t+\triangle t) for sides (and it terminates in the same place A(t+\triangle t) does). Construct the altitude of this triangle and consider one of the resulting identical right triangles with one leg being the altitude, one leg is of length \frac{1}{2}\left| \triangle A\right|, and the hypotenuse is of length A. Use trig.

 

[Data]

|\Delta A| = \sqrt{(A(t+\Delta t)-A(t))\cdot(A(t+\Delta t)-A(t))}

= \sqrt{|A(t+\Delta t)|^2 - 2A(t+\Delta t)\cdot A(t) + |A(t)|^2}<br /> = \sqrt{A^2 - 2A^2\cos {\Delta \theta} + A^2}

= \sqrt{2A^2 - 2A^2\cos{\Delta \theta}} = \sqrt{2}A\sqrt{1-\cos{\Delta \theta}}.

Now use the identity \sin^2{\phi} = \frac{1}{2}(1-\cos{2\phi}) to change that into

\sqrt{2}A\sqrt{2\sin^2{\frac{\Delta \theta}{2}}}

= 2A\sin \frac{\Delta \theta}{2}.

Here, \Delta \theta is the angle between A(t+\Delta t) and A(t) and A is the (constant) magnitude of the vector function.

 

[MathematicalPhysicist]

thank you for your help.

 

[MathematicalPhysicist]

by the way, neutrino, you haven't bought the booog, have you?
ive seen in amazon that it's quite coasts a lot.

 

[neutrino]

by the way, neutrino, you haven't bought the booog, have you?
ive seen in amazon that it's quite coasts a lot.

What's that? Some new variant for the word book?

 

[MathematicalPhysicist]

no just big funny typing error.