ooops...sorry
The question is determining the curvature of a curve defined by a vector valued function
K = curvature = ll T'(x) ll / ll r'(x) ll = ( ll r'(x) * r"(x) ll ) / ll r'(x) ll ^3
Stephen,
That's still slowly sinking in - I think I need more coffee
To be completely honest my brain is stuck on how we go from here:
= P((X≥2) and (X≥1))/P(X≥1)
to here:
= P(X≥2)/P(X≥1).
I know that's 101 somewhere but my brain just won't work - it's been something that's been bugging me...
Proof: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2) for y=f(x)
Prove: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2)
r(x)= xi + f(x)j = <x , f(x)>
r'(x)= 1i + f'(x)j= <1, f'(x)>
T(x) = r'(x)/llr'(x)ll
= <1, f'(x)> / ((1^2+(f'(x))^2)^1/2)
This is where I start to get even more lost:
T'(x)...
Hi guys,
I can't get my head around this, if anyone could help that would be great.
"A robotic assembly line contains 20 stations. Suppose that the probability
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Thank you chiro,
Your response helped me greatly and solidified my (lack of) understanding of a cdf.
However, in the interest of discussion/further learning only I do have take you up on one point:
The definition of a cdf is in fact a splitting of the probabilities. From wikipedia...
Hi Guys,
Long time reader first time poster...
This simple question has stumped me all day and I think I've finally cracked it! I'm hoping someone can confirm that, or tell me how wrong I am - either is fine :)
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