anonymous
  • anonymous
A bee with velocity vector r'(t) starts out at the origin at t=0 and flies around for T seconds. Where is the bee located at time T if ∫(0-T) r'(u)du = 0? Also, what does the quantity ∫(0-T) ||r'(u)||du represent?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Please help
amistre64
  • amistre64
what does (0-T) represent?
anonymous
  • anonymous
integral from 0 to T

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anonymous
  • anonymous
sorry it's the only way I know how to type it
amistre64
  • amistre64
so those are the limits, ok
amistre64
  • amistre64
do you recall the definition of: \[\int_{a}^{b}f(x)~dx=??\]
amistre64
  • amistre64
the fundamental thrm of calculus is also what it is called
anonymous
  • anonymous
It's F(b) - F(a)
amistre64
  • amistre64
correct, so if F(b)-F(a) = 0, then either a=b, or F has the same value at multiple input levels similar to a periodic function.
amistre64
  • amistre64
for example, cos(pi) - cos(0) = 0 since cos(pi)=cos(0)
amistre64
  • amistre64
but the question is, what does taking the integral of a function tell us?
anonymous
  • anonymous
it can tell you what the area under a curve is
amistre64
  • amistre64
and my example is spose to read cos(2pi) = cos(0), but the site is lagging for me
anonymous
  • anonymous
it's ok
amistre64
  • amistre64
the area under a curve, yes ... but that reflects displacement
amistre64
  • amistre64
ever wonder why in some cases we get a negative area? and in other cases we dont?
anonymous
  • anonymous
Yes because the integral of velocity is displacement
anonymous
  • anonymous
right?
anonymous
  • anonymous
negative displacement ?
amistre64
  • amistre64
when we want to find the area between 2 curves, area is a positive value always ... its an absolute value. And in those cases we have to concern ourselves with the places that one curve crosses over the other.
amistre64
  • amistre64
displacement on the other hand reflects is a vector, it has magnitude and direction.
amistre64
  • amistre64
F(b)-F(a) = 0 assumes that at a and b, we are in the same place. Either a = b, or we have simply ended were we started.
anonymous
  • anonymous
Then the location of the bee is still the origin at time T?
amistre64
  • amistre64
yes, at time T, the bee is simply back in the same place it was at when the time was equal to 0
anonymous
  • anonymous
But why substitute u in for t for the integration? That's the most confusing part. Does that have any effect on the result?
amistre64
  • amistre64
0 and T are values that the variable u can take on. the function itself is dependant on the value of u ...
anonymous
  • anonymous
What is the difference between ∫(0-T) r'(u)du = 0 and ∫(0-T) r'(t)dt = 0 Why did they substitute ?
amistre64
  • amistre64
hmm, im not absolutely sure, but i think it has something to do with what is called a 'dummy' variable - meant to avoid confusion between the overuse of the letter T
anonymous
  • anonymous
oh ok .. well thank you !
amistre64
  • amistre64
good luck

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