Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

ohh okay

Answer 6

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)

Answer 7

You try it.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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)

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

isn't y' = -1

Answer 18

OH! yeah you're right, it is xD

Answer 19

so then how do i find two unit tangent vectors?

Answer 20

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- )

Answer 21

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

Answer 22

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

Answer 23

@itzmashy Thanks for asking this question!

Answer 24

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? =\

Answer 25

magnitute = sqrt of sum all the square coordinators

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

Yes!

Answer 30

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)?

Answer 31

It's just a set of ±T/ ||T||

Answer 32

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

Answer 33

Yes!

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

what does it mean to normalize the tangent vector?

Answer 39

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

Answer 40

@phi, you're my hero.

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

I guess you should open another post!

Answer 45

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)

Answer 46

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)

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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.

Answer 51

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

Answer 52

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

Answer 53

yep

Answer 54

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.

Answer 55

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) ?

Answer 56

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

Answer 57

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.

Answer 58

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

Answer 59

Unit vector = vector/ magnitude

Answer 60

Just happy that I can be some helpful here :)

Answer 61

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) ??!?!?!

Answer 62

Yep, you got it :D

Answer 63

Unit tangent line: < 1, 1 >/ √2

Answer 64

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

Answer 65

cospi/4 = √2/2

Answer 66

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

Answer 67

Just memorize it!

Answer 68

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

Answer 69

Are you asking to integral all three?

Answer 70

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

Answer 71

Do you know that √2/2 = 1/√2

Answer 72

2 = √2 * √2

Answer 73

how.. what.>?!?

Answer 74

- ( 1/√2 -1) = 1 - 1/√2

Answer 75

They just simplify it!

Answer 76

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

Answer 77

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

Answer 78

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

Answer 79

because ( √2 )² = 2

Answer 80

so √2 / 2 = √2 /( √2 )² = 1/√2

Answer 81

Poor you :(

Answer 82

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

Answer 83

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?