Question

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

Answer 1

the normal to a curve is the gradient of the function

Answer 2

which sounds odd when the brits call gradient a slope ...

Answer 3

but we have a vector function, not a surface right

Answer 4

yes r(t) is a vector

Answer 5

we still need r', and then unit it

Answer 6

then the derivative of the unit tangent produces the normal ....

Answer 7

crossing the tangent and normal gives us the ... I know they call it a B vector, curve maybe?

Answer 8

so the unit tangent is r'(t)/ its magnitude?

Answer 9

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)

Answer 10

for some reason, the normal has to be calculated using the unit tangent ...

Answer 11

ok, so we include the t when calculating the magnitude

Answer 12

of course

Answer 13

|r'| = sqrt(4+4t^2)
= 2sqrt(1+t^2)
how did you take out the 2?

Answer 14

\[T = 0~ i + \frac{2}{2\sqrt{1+t^2~}}~ j +\frac{2t}{2\sqrt{1+t^2}}~k\]
well, factor out the 4, and by now im well aware that sqrt4 = 2

Answer 15

oh i see now.

Answer 16

sqrt(4+4t^2)

sqrt[4(1+t^2)]

sqrt(4) sqrt(1+t^2)

2sqrt((1+t^2)

Answer 17

ok. so then we take the derivitive of this to find the normal? or am i misunderstanding

Answer 18

N = T'/|T'|

Answer 19

taking the derivative, that right

Answer 20

so how would you go from T(t) = 1/2sqrt(1+t^2)(2j +2tk) to its derivitive

Answer 21

\[T = 0~ i + \frac{2}{2\sqrt{1+t^2~}}~ j +\frac{2t}{2\sqrt{1+t^2}}~k\]
\[T' = 0~ i + \left(\frac{1}{\sqrt{1+t^2~}}\right)'~ j +\frac12ln|\sqrt{1+t^2}~|~k\]
that j part looks maybe like a tangent?

Answer 22

nah, ln(sec+tan) i think

Answer 23

lol, im integrating

Answer 24

yah you had me confused for a second.

Answer 25

so basically need the derivitive of 1/sqrt(1+t^2) and derivitive of t/sqrt(1+t^2)

Answer 26

taking the derivative ....

\[T' = 0~ i - \frac{4t}{(1+t^2)^{3/2}~}~ j +(...)~k\]
yeah

Answer 27

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

Answer 28

ok thank you