find normal unit vector

Question

ksaimouli · Answer

$$<t,2t,t^2> t=1$$

ksaimouli · Answer

@ganeshie8 I tried this couldn't get. Plz correct me if I am wrong

ksaimouli · Answer

$$r'(t)=<1,2,2t> and |r(t)|= \sqrt{4t^2+5}$$

ksaimouli · Answer

and then $$ <1,2,2t> \frac{ 1 }{ \sqrt{4t^2+5} }$$

ksaimouli · Answer

now unit normal

ksaimouli · Answer

$$<\frac{ -4t }{ \sqrt{4t^2+5}(4t^2+5) }, \frac{ -8t }{ \sqrt{4t^2+5}4t^2+5 },\frac{ 2\sqrt{4t^2+5 }8t }{ sqrt{(4t^2+5)} 4t^2+5}$$

ganeshie8 · Answer

ganeshie8 · Answer

check the last component..

ksaimouli · Answer

substituting t=1 $$<\frac{ -4 }{ 27 },\frac{ -8 }{ 27 },\frac{ 48 }{ 27 }$$

ganeshie8 · Answer

check the last component... 
wolfram gives 10/27

ksaimouli · Answer

hmm how did they get that $$\frac{ 2\sqrt{4t^2+5}-2t \frac{ 1 }{ 2}(4t^2+5)^{(-1/2)} }{4t^2+5  }$$

ganeshie8 · Answer

also its not unit vector yet, u need to divide by its magnitude

ksaimouli · Answer

can I find magnitude for <-4/27,-8/27,10/27>

ksaimouli · Answer

instead of the equation

ganeshie8 · Answer

absolutely !

ganeshie8 · Answer

wat u have is a normal vector

ganeshie8 · Answer

to get unit vector in that direction,
simply divide by its mag

ganeshie8 · Answer

we dont need to mess wid ||T'|| again

ksaimouli · Answer

|dw:1395972549287:dw|

ksaimouli · Answer

yup I got this but didnt understand how they got the derivative for last component

ganeshie8 · Answer

ganeshie8 · Answer

ksaimouli · Answer

yup got that if i simplify