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OpenStudy (anonymous):

A curve C is given parametrically. Find parametric equations for the tangent line to C at P. x=2t^3-1 y=-5t^2+3 z=8t+2; P(1, -2, 10) what does that even mean? to find parametric equations for the tangent line lol

OpenStudy (phi):

Here's a description of parametric equations, and why you might use them http://tutorial.math.lamar.edu/Classes/CalcII/ParametricEqn.aspx But as you can see, the curves x,y,z points are all a function of t rather than y= f(x,z) for example. To do this problem, find the derivative of each component wrt to t. (dx/dt, dy/dt, dz/dt) Next, notice that the given point P (1,-2,10) corresponds to t=1 (plug 1 into the definition of x,y,z). So in the derivative, plug in t=1. You should get <6,-10,8> Think of this a direction (a vector pointing in that direction). You can move up and down in that direction by scaling it (multiplying by a number): <6,-10,8>*t (here t is the scalar) If we start at position (x,y,z)= (1,-2,10) and move along the vector <6,-10,8>*t we have a line. So the answer is <1+6t,-2-10t,10+8t> Or, being explicit about x,y,z: x= 1+6t y= -2-10t z= 10+8t See 3D graph of the curve and the tangent line

OpenStudy (anonymous):

how am i supposed to know that it corresponds to 1? should i be guessing random numbers?

OpenStudy (phi):

I just took the def of x=2t^3-1 and set it = to 1 (the x value of point P)

OpenStudy (phi):

But I notice z= 8t+2 = 10 is the easiest to solve.

OpenStudy (anonymous):

ohh okay

OpenStudy (anonymous):

another question then :D, if you don't mind? A curve C is given parametrically. Find two unit tangent vectors to C at point P. x=e^2t, y=e^-t and z=t^2 +4, p(1 1 4)

OpenStudy (phi):

You try it.

OpenStudy (anonymous):

i'm not quite sure how to find unit tangent vectors? i'm assuming first i find r' so i get.. x'=2e^2t y=-e^-2 z=2t, if i'm doing the same thing i was doing previously, i would plug y to get e^-1 which is 1/e? .. i think i'm stuck lol i tried plugging in all the numbers to each equation and i'm not sure where to go form here

OpenStudy (phi):

x'=2e^2t y=-e^-t z=2t I hope you see that P(1,1,4) means t=0 so evaluate the derivatives at t=0. That gives you the change in x as you change t, etc so your "slope vector" is <x' , y', z'> evaluated at t=0 then do as in the previous problem. scale this vector by t, add in the starting point P

OpenStudy (phi):

Btw, I did not normalize the vectors, but that does sound like a good idea!

OpenStudy (anonymous):

wait.. how did you get that?, that t=0?

OpenStudy (phi):

pick z for example z=t^2 +4, at P (1,1,4), z is 4 so 4= t^2+4, and t=0 (t=0 should work for x, and y also)

OpenStudy (phi):

Because any point on the curve is a function of t.

OpenStudy (anonymous):

ohh i did that! but i thought that if its 0 i was doing something wrong lol kk let me keep going

OpenStudy (anonymous):

so r' evaluated at 0 is x=2, y =1, z=0, is that correct?

OpenStudy (anonymous):

so, the parametrics would be x=1+2t y=1+t and z=4

OpenStudy (phi):

isn't y' = -1

OpenStudy (anonymous):

OH! yeah you're right, it is xD

OpenStudy (anonymous):

so then how do i find two unit tangent vectors?

OpenStudy (phi):

Good question. But from a graph, or looking at the definition of the curve, it has a minimum at z=4. So any line in the z=4 plane will work. if your first set is x= 1+2t y= 1-t z=4 then another set might be x= 1+t y= 1+2t z=4 (These should be orthogonal - not that this was required- )

OpenStudy (anonymous):

@phi Thank you so much for the fantastic lesson :)

OpenStudy (anonymous):

I'm thinking there's will be the set ± T

OpenStudy (anonymous):

@itzmashy Thanks for asking this question!

OpenStudy (anonymous):

according to this site, http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/arc/arc.html, a unit tangent vector is equal to R'(t)/||r'(t)|| so if i have those parametrics (x' y' z') how do i form r' from them? is it just (1+2t}i +(1-t)j + (4)k so if i'm follow that equation is it, (1+2t)+(1-t)+(4)? /over magnitude of r'.. but how do i find magnitude of r'? and how do you know what the curve looks like? =\

OpenStudy (anonymous):

magnitute = sqrt of sum all the square coordinators

OpenStudy (anonymous):

yeah so if you're given r(t) then how do you know what coordinates are?

OpenStudy (anonymous):

is it like <2 1 0> when at r(0)?

OpenStudy (anonymous):

r = < x, y, z > T = r' <x', y', z' > Slope T = r' ( Po)

OpenStudy (anonymous):

Yes!

OpenStudy (anonymous):

so if i'm trying to find two unit vectors, i should use the T formula for two points of r? like r(0) and r(1)?

OpenStudy (anonymous):

It's just a set of ±T/ ||T||

OpenStudy (anonymous):

and i find T by r'/||r'|| right?

OpenStudy (anonymous):

Yes!

OpenStudy (anonymous):

@phi I'm not expert at 3 dim. Am I correct?

OpenStudy (anonymous):

so i have r'(0) = <2,1,0> does that mean that T = <2,0,0>/\[3\sqrt{?}\]

OpenStudy (anonymous):

oops thats not waht i wana put lol i mean <2,1,0> / \[\sqrt{3}\]

OpenStudy (phi):

Here is how you can plot these: http://www.wolframalpha.com/input/?i=parametric+plot+%28e%5E%282t%29%2C+e%5E-t%2C+t%5E2%2B4%29 It's not necessary to normalize the tangent vector (unless a question is asking you to). But if you have a vector <2,-1,0>, its norm is sqrt(2^2+ (-1)^2 +0) = sqrt(5) T by r'/||r'|| right? Yes ±T/ ||T|| no, I think if you reverse direction (the minus sign) you get the same line

OpenStudy (anonymous):

what does it mean to normalize the tangent vector?

OpenStudy (anonymous):

so the only way you plotted was through the graphing calculator right? you didn't do it by hand?

OpenStudy (anonymous):

@phi, you're my hero.

OpenStudy (phi):

the tangent vector is the vector that is parallel to the point. Its the <dx/dt, dy/dt dz/dt>

OpenStudy (anonymous):

is there a difference between that and a tangent unit vector?

OpenStudy (phi):

unit vector has unit length, which you get by normalizing by the vector's length

OpenStudy (anonymous):

I guess you should open another post!

OpenStudy (anonymous):

so the answer in the back of the book is \[\pm(1/\sqrt{5}) (2i-j)\] i'm so confused how they got this... if T = r'/||r|| then isn't it <2,1,0> / sq5? where did they get the 1 on top? and where did they get the other vector? the (2i-j)

OpenStudy (phi):

First, you still keep losing the minus sign on y. your vector is <2, -1, 0> to normalize divide each component by the length of the vector or multiply by 1/sqrt(5)

OpenStudy (anonymous):

so 2/sq5, -1/sq5, 0/sq5?

OpenStudy (phi):

And thinking about, it looks like ±T is correct. Apologies to chlor

OpenStudy (anonymous):

i guess i'm not sure exactly what normalizing means?

OpenStudy (phi):

yes, but you can factor out the 1/sqrt(5) and they use i,j,k (unit vectors) to denote the x,y,z dimensions.

OpenStudy (anonymous):

@phi Pls don't say that! I'm learning great deal from you right now :)

OpenStudy (anonymous):

so why don't they use the - OH THEY DON'T USE Z CUZ ITS 0 RIGHT?

OpenStudy (anonymous):

yep

OpenStudy (phi):

In 2D, you know how to find the length of a vector (pythagoras) so a vector <1,1> has length sqrt(2) if you divide by sqrt(2) so each component is 1/sqrt(2) and do pythagoras you will get a length of 1.

OpenStudy (anonymous):

so let me get this straight.. to do this problem.. you normalize the vector by dividing each component by sq5, which you can then factor out so you get 1/sq5<2, -1, 0>, but to put it in terms of the vector you just change it to 1/sq5 (2i -j) ?

OpenStudy (anonymous):

T = < 2, -1, 0 > -> Its magnitude = sqrt ( 2^2 _ -1^2 ) = sqrt 5

OpenStudy (phi):

yes, but you can factor out the 1/sqrt(5) and they use i,j,k (unit vectors) to denote the x,y,z dimensions.

OpenStudy (phi):

And thinking about, it looks like ±T is correct. Apologies to chlor

OpenStudy (anonymous):

Unit vector = vector/ magnitude

OpenStudy (anonymous):

Just happy that I can be some helpful here :)

OpenStudy (anonymous):

omg my internet died at the most inopportune moment lol anyway.. lemme run this by you guys... x=sint +2 y=cost z=t, p(2 1 0) x'=cost y'=-sint z'=1 x'(o)=1, y'(o)=0 z'(o)=1 <1,0,1> so then plusminus 1/sq2 (1i + 1k) ??!?!?!

OpenStudy (anonymous):

Yep, you got it :D

OpenStudy (anonymous):

Unit tangent line: < 1, 1 >/ √2

OpenStudy (anonymous):

yayy :D k i have another question.. \[\int\limits_{0}^{\pi/4}(sinti - costj +tantk)\] i know you integrate them separately so for example it'll be \[\int\limits_{0}^{\pi/4} sint = -cost| -\cos(\pi/4)+\cos(0)\] but i think i'm doing something wrong..cuz cos(pi/4) is 1 and cos(0) is 1 but the answer says 1-1/sq2i

OpenStudy (anonymous):

cospi/4 = √2/2

OpenStudy (anonymous):

how do you know that? =\ what did i do wrong? *glares at calculator*

OpenStudy (anonymous):

Just memorize it!

OpenStudy (anonymous):

but, but.. if its sq2/2, then why is the answer 1-1/sq2? o.O

OpenStudy (anonymous):

Are you asking to integral all three?

OpenStudy (anonymous):

well, yes but, for simplicities sake i'm just focusing on i right now

OpenStudy (anonymous):

Do you know that √2/2 = 1/√2

OpenStudy (anonymous):

2 = √2 * √2

OpenStudy (anonymous):

how.. what.>?!?

OpenStudy (anonymous):

- ( 1/√2 -1) = 1 - 1/√2

OpenStudy (anonymous):

They just simplify it!

OpenStudy (anonymous):

I believe you need to grant your brain a break :P

OpenStudy (anonymous):

wait, i don't get how you got that sq2/2 = 1/sq2 o.o

OpenStudy (anonymous):

i can't, i have a test on this tomorrow, and this is only the first section.. :(

OpenStudy (anonymous):

because ( √2 )² = 2

OpenStudy (anonymous):

so √2 / 2 = √2 /( √2 )² = 1/√2

OpenStudy (anonymous):

Poor you :(

OpenStudy (anonymous):

i know :( but you've been so helpful, so thank you so much!!!

OpenStudy (anonymous):

i has another question? ^.^ find r(t) given the conditions: r''(t)=6ti+3j, r'(0)=4i-j+k, r(o)=5j i know i have to integrate r'' so i get \[r'(t)=(t^6/6) + (t^3/3) + c\] but i'm not exactly sure what to do to get c?

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