anonymous
  • anonymous
Find the unit tangent, principal normal,and wrtie an equation in x,y,z for the osculating plane at the point on the curve that corresponds to the indicated value of t. 35. r(t) = i + 2tj +t^2k ; t =1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
the normal to a curve is the gradient of the function
amistre64
  • amistre64
which sounds odd when the brits call gradient a slope ...
amistre64
  • amistre64
but we have a vector function, not a surface right

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anonymous
  • anonymous
yes r(t) is a vector
amistre64
  • amistre64
we still need r', and then unit it
amistre64
  • amistre64
then the derivative of the unit tangent produces the normal ....
amistre64
  • amistre64
crossing the tangent and normal gives us the ... I know they call it a B vector, curve maybe?
anonymous
  • anonymous
so the unit tangent is r'(t)/ its magnitude?
amistre64
  • amistre64
yes r = 1 i + 2t j +t^2 k r' = 0 i + 2 j +2t k |r'| = sqrt(4+4t^2) = 2sqrt(1+t^2)
amistre64
  • amistre64
for some reason, the normal has to be calculated using the unit tangent ...
anonymous
  • anonymous
ok, so we include the t when calculating the magnitude
amistre64
  • amistre64
of course
anonymous
  • anonymous
|r'| = sqrt(4+4t^2) = 2sqrt(1+t^2) how did you take out the 2?
amistre64
  • amistre64
\[T = 0~ i + \frac{2}{2\sqrt{1+t^2~}}~ j +\frac{2t}{2\sqrt{1+t^2}}~k\] well, factor out the 4, and by now im well aware that sqrt4 = 2
anonymous
  • anonymous
oh i see now.
amistre64
  • amistre64
sqrt(4+4t^2) sqrt[4(1+t^2)] sqrt(4) sqrt(1+t^2) 2sqrt((1+t^2)
anonymous
  • anonymous
ok. so then we take the derivitive of this to find the normal? or am i misunderstanding
amistre64
  • amistre64
N = T'/|T'|
amistre64
  • amistre64
taking the derivative, that right
anonymous
  • anonymous
so how would you go from T(t) = 1/2sqrt(1+t^2)(2j +2tk) to its derivitive
amistre64
  • amistre64
\[T = 0~ i + \frac{2}{2\sqrt{1+t^2~}}~ j +\frac{2t}{2\sqrt{1+t^2}}~k\] \[T' = 0~ i + \left(\frac{1}{\sqrt{1+t^2~}}\right)'~ j +\frac12ln|\sqrt{1+t^2}~|~k\] that j part looks maybe like a tangent?
amistre64
  • amistre64
nah, ln(sec+tan) i think
amistre64
  • amistre64
lol, im integrating
anonymous
  • anonymous
yah you had me confused for a second.
anonymous
  • anonymous
so basically need the derivitive of 1/sqrt(1+t^2) and derivitive of t/sqrt(1+t^2)
amistre64
  • amistre64
taking the derivative .... \[T' = 0~ i - \frac{4t}{(1+t^2)^{3/2}~}~ j +(...)~k\] yeah
amistre64
  • amistre64
ive got to run, but im sure you can manage from here :) take the normal, and anchor it ti r(t), at t=1 to define the plane
anonymous
  • anonymous
ok thank you

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