Missing sketch in K&K Mechanics book

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The discussion revolves around a missing diagram in K&K's "Intro to Mechanics" on page 26, which complicates the understanding of a derivation involving the equation |δA| = 2A sin(δθ/2). Participants clarify that the three vectors form an isosceles triangle, and the missing sketch is crucial for visualizing the relationship between the vectors. They explain that as Δθ and ΔA approach zero, the discrepancies in the diagram resolve, leading to a clearer understanding of the instantaneous derivative. A key point raised is the need for vector ΔA to be perpendicular to A, which is necessary for the derivation to hold true. The discussion concludes with an acknowledgment of the importance of the sketch for clarity.
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Hello guys,

I'm currently reading through K&K Intro to Mechanics book, and I'm on page 26 where I encounter a bit odd derivation.

The authors say:
...

Using the angle ##\delta \theta## defined in the sketch,

##|\delta A| = 2A \sin{\frac{\delta \theta}{2}}##
I'm rather lost on this part. I don't know which sketch corresponds to this equation.
 
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The diagram is indeed missing.

attachment.php?attachmentid=66201&stc=1&d=1391257399.gif


The three vectors form an isosceles triangle. The dashed line bisects it into two right triangles. To get the equation, consider one of the right triangles.

If you take the diagram at the top left of page 26, and fill in the hypotenuse which is ##\vec A (t + \Delta t)##, you get a similar diagram, but with the right angle in a different location. The discrepancy disappears in the limit as ##\Delta \theta \rightarrow 0## and ##\Delta \vec A \rightarrow 0##.
 

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Ok that makes it clear. But it raises another question on why the vector ##\Delta A## is not perpendicular to ##A##? Whereas the book says it is necessary for it to be so because ##A## is not changing in magnitude?
 
In the limit as Δt goes to zero (which is what you need to do in order to have the instantaneous derivative), ΔA becomes perpendicular to both A(t) and A(t+Δt). And of course A(t) and A(t+Δt) become equal to each other.
 
jtbell said:
In the limit as Δt goes to zero (which is what you need to do in order to have the instantaneous derivative), ΔA becomes perpendicular to both A(t) and A(t+Δt). And of course A(t) and A(t+Δt) become equal to each other.


Ah yes I didn't think of that carefully. Thanks for the remark and the sketch. :)
 
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