For some reason I lost the point of intersection (2,0):
I equated the relevant r_{1} and r_{2} components but only got t=0
Eg, consider the lines (x,y)=(t,0), (x,y)=(2t,0).
These lines coincide exactly, so every point on the x-axis is a point of intersection.
Yet the only place where (t,0)=(2t,0) is at the origin.
Excellent. Makes sense.
________________
Find the angle between the two curves at the points of intersection.
I'm having a dumb moment now. Which vectors are we taking the dot product of?
Strange... That's what I did so maybe there's an error in my computation.
But the answer was arccos(0.8)
EDIT: Ouch. I know what I did now.
Last edited by leehuan; 28 Feb 2017 at 12:31 PM.
Last edited by InteGrand; 28 Feb 2017 at 12:32 PM. Reason: Typo
Can be assumed: F=ma
Consider the two metrics and
(you may assume they are metrics).
i) Show that d and δ are not equivalent.
Last edited by QuantumRoulette; 7 Mar 2017 at 10:30 PM. Reason: fixed Tex issues
I assume d(x,y) is supposed to be ||x-y|| and the set is some normed vector space V with norm ||.|| (e.g. R^d). (Please specify more if this is not the intended setting.)
Then these two metrics are not (strongly) equivalent because V is bounded with the delta metric but unbounded with the d metric.
That V is bounded with the delta metric follows immediately from the definition of delta, which must always lie in [0,1). On the other hand d(tx,0)=|t|d(x,0) can be made arbitrarily large for nonzero x.
Note that these two metrics ARE topologically equivalent though, in the sense that convergence in one metric implies convergence in the other. This follows from from the map x->x/(1+x) being a homeomorphism from [0,inf) to [0,1).
Might not be easy to visualise the graph of the function on all of R^2, but you should certainly be able to visualise what it looks like on the slices x=const. or y=const which is all that matters for seeing/proving the nonexistence of the iterated limit. It is an oscillatory expression that oscillates faster as you approach axes. One of the two summands becomes irrelevant as you get close to the axes, so the other one dominates. This thing behaves like (const).sin(1/x), which of course does not converge unless that const is zero.
The boundedness of sine makes it clear that f(x,y) tends to zero as (x,y) tends to zero though.
Long story short: don't be too hasty to form intuitions in analysis, lots of things can behave weirdly...you kind of have to slowly build up a list of things that ARE true (via proof!) rather than assuming innocuous statements are true and ruling these things out as you come across pathological counterexamples.
And when you are looking at functions like this, try to isolate the terms that actually matter for the property you are trying to prove. A large part of analysis is just approximating ugly things by nice things, throwing away small sets on which a function behaves badly, etc etc.
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