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

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

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

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

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

ohh okay

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)

You try it.

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

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

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

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

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)

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

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

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

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

isn't y' = -1

OH! yeah you're right, it is xD

so then how do i find two unit tangent vectors?

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- )

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

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

@itzmashy Thanks for asking this question!

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? =\

magnitute = sqrt of sum all the square coordinators

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

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

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

Yes!

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)?

It's just a set of ±T/ ||T||

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

Yes!

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

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

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

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

what does it mean to normalize the tangent vector?

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

@phi, you're my hero.

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

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

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

I guess you should open another post!

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)

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)

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

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

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

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.

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

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

yep

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.

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) ?

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

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.

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

Unit vector = vector/ magnitude

Just happy that I can be some helpful here :)

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) ??!?!?!

Yep, you got it :D

Unit tangent line: < 1, 1 >/ √2

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

cospi/4 = √2/2

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

Just memorize it!

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

Are you asking to integral all three?

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

Do you know that √2/2 = 1/√2

2 = √2 * √2

how.. what.>?!?

- ( 1/√2 -1) = 1 - 1/√2

They just simplify it!

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

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

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

because ( √2 )² = 2

so √2 / 2 = √2 /( √2 )² = 1/√2

Poor you :(

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

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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