Question

Curve C:

r(t) = (cos(t), sin(t), t-cos(t))

Tangent line C through P = r(pi/2)?

Answer 1

r'(t) = (-sin(t) , cos(t) , 1+sin(t) )

Answer 2

Now you just need a position vector

r(pi/2) = (0,1,pi/2)

Answer 3

1+sin(pi/2) = 2 right?

but now you have the equation r(t) = (0,1,pi/2) + t(-1,0,0) and then?

Answer 4

but what is the the step after you find r(t) ...

Answer 5

r'(pi/2) = (-1,0,2)

Answer 6

\[\gamma (t) = (0,1, \frac{\pi}{2}) + t (-1,0,2) \]\[t \in \mathbb{R}\]

Answer 7

called it gamma, because technically the curve was called r , so yeah.

Answer 8

oke that make sense.. but than you will have 3 tangent lines?

one for the x, y en z?

Answer 9

The equation a line is always

\[x(t) = a + bt\]
Where "x" is a component vector, "a" is the position vector and "b" is the direction vector, which you "span" along. As it is called in linear algebra. t is the free parameter, which you vary through the real numbers.

Answer 10

No, you have ONE line, in 3 dimensional space.

Answer 11

Picture a supporting cable on a tower (assuming it is not loose). That is a line in three dimensional space.

Answer 12

and similarly it works for as many dimensions as you like. The general equation of a line will always be the same.

Answer 13

The only difference will be the number of components in the position and direction vectors.

Answer 14

could you type in the final answer for me? (a)
so i can understand it :)

Answer 15

teh vector functions derives componentwise into a tangent vector

Answer 16

if you want to know the line of the tangent, you simply attach the tangent vector to a point

Answer 17

elec did a good derivative for you; not you just need to know its value at pi/2

Answer 18

r'(t) = (-sin(t) , cos(t) , 1+sin(t) )

r'(pi/2):

Tx = -sin(pi/2
Ty = cos(pi/2)
Tz = 1+sin(pi/2)

Answer 19

what is that then, <0,1,1> ?

Answer 20

so we nee to attach that to the point at the vector functions values of pi/2

Answer 21

i got my tangent points wrong .. confused my values

Answer 22

where did you found <0,1,1> ?

Answer 23

Tx=0
Ty=0
Tz=2

right?

Answer 24

oops Tx = 1

Answer 25

.... ok, the only way I see for you to discover this is for you to get your hands dirty :) so lets start from the beginning and see where it goes awry at

How do we derive r(t) = (cos(t), sin(t), t-cos(t))

Answer 26

my calculating of sin and cos was wrong because I confused the 2

Answer 27

oke.. de derivative of r(t) is:

r'(t) = ( -sin(t), cos(t), 1+ sin(t) )

Answer 28

those are good, now what do they represent to us?

Answer 29

its the i dont know the word in english >.< but its the slope?

Answer 30

slope is fine, but slope for a vector is just its component parts:

r'(t) = Tangent vector <x(t),y(t),z(t)>

Answer 31

so lets calculate correctly the component parts of our tangent vector

Answer 32

owh jah ofcource the tangent vector =.=

Answer 33

oke i have to fill in the pi/2 for t

right?

Answer 34

correct

Answer 35

oke lets do that

Answer 36

r'(pi/2) = ( -sin(pi/2), cos(pi/2), 1+ sin(pi/2) )

r'(pi/2) = ( -1, 0, 2 )

Answer 37

I get, hopefully :)

Tx = -sin(pi/2 = -1
Ty = cos(pi/2) = 0
Tz = 1+sin(pi/2) = 1+1 = 2

T<-1,0,2>

yay we match!

Answer 38

hahah score ^^

Answer 39

now that we have a vector all we need is a point to attach it to and a scalar to strectch it to infinity by and we get a line

Answer 40

out point is just the original vector functions at pi/2

Answer 41

*our point

Answer 42

oke so:

r(t) = (0,1,pi/2)

Answer 43

thats a good point; now lets attach our vector to it :) ill use "s" for a scalar

x = 0 + s*Tx
y = 1 + s*Ty
z = pi/2 + s*Tz

Answer 44

T<-1,0,2>

x = -s
y = 1 + 0s
z = pi/2 + 2s


x = -s
y = 1
z = pi/2 + 2s

is simplified if i see it right

Answer 45

oke... and what is the next step.. or is this the answer?

Answer 46

this is what we needed, we defined a line by a point and any scaled version of its vector

Answer 47

there are 2 forms that I know of, this one is the parametric; the other is similar and seldom used tho

Answer 48

ahhh oke...

the assignment said find the tangent line. so i thought there i just 1 line... stupid me

Answer 49

:) oh its one "line" its just defined by its seperate components

Answer 50

there is really no good way to define a line in 3 coords other than this

Answer 51

ahhh oke :) thank you honey <3

i have another question could you watch if i do it right?

Answer 52

if need be, since tis line acts nicely in one plane; we could define it like normal, we would just have to amend the equation by saying y = 1 so that it resides in the y=1 plane

Answer 53

sure

Answer 54

r(t) = (1 - t^2, 2t)

point p= (0,2)

Answer 55

i also have to find the tangent line, but the t is not defined

Answer 56

i know that

r'(t) = ( -2t, 2 )

Answer 57

the t is defined, you just have to unbury it from the given point

Answer 58

(0,2) + t(-2t,2)

is this step correct?

Answer 59

not quite, :)

Answer 60

:(

Answer 61

youve got it correct up to r'(t), but the tangent vector needs to be defined at (0,2) so we have to discover the secret to "t"

Answer 62

tell me what x equals in the given r(t)

Answer 63

x = -2t
y = 2

point x = 0 en y = 2

so
x = -2t
0 = -2t
t=0?

Answer 64

oops i took the derivative

Answer 65

r(t) = ( x(t) , y(t) )
r(t) = (1 - t^2, 2t )
r(t) = ( 0 , 2 ) ; when t = ?

Answer 66

x= 1 - t^2
y= 2t

x = 0, y=2

0 = 1 - t^2
2 = 2t

t = 1

Answer 67

:)

Answer 68

very good :) now we can define our tangent vector with t=1

Answer 69

T=<-2,2>

i dont know what to do now :S

Answer 70

i know the point that is (0,2)

the tangent vector is <-2,2>

Answer 71

oh but you do :) scale that vector and add it to our point like you did before. But this time we can sepearte it out so it looks more conventional

Answer 72

(0,2) + s<-2,2> is cute and all, but not quite standard format

Answer 73

nikvIst can u help?

Answer 74

x = 0 -s2
y = 2 + s2?

Answer 75

yes, but lets clean it up some :)

x = -2s
y = 2+2s

this is parametric, we could define it like normal since we started out in 2 planes

Answer 76

x = -2s; s = -x/2

y = 2 + 2s; y = 2 + 2(-x/2)

y = 2-x

Answer 77

:O owhh i see... thats the tangent line!

Answer 78

it is :)

Answer 79

you're really a hero :)

Answer 80

thnx ;)

Answer 81

oke.. one question could you help me with the gradient derivative?

Answer 82

sure

Answer 83

i almost have the answer of it, but the last step

Answer 84

gradient f(2,2) = 7i-2j

Answer 85

now i have to calculate the direction of the gradient

Answer 86

lets start with the problem from the beginning so I can see where we are going ... or is this the beginnning?

Answer 87

\[\sqrt{7^{2} + 2^{2}} = \sqrt{53}\]

Answer 88

direction has to deal with trig, it is the tan-1(y,x)

Answer 89

owhh oke..


i have to find the maximum direction at point (2,2)

Answer 90

maximum direction; would that mean magnitude as well as direction?

Answer 91

i already calculated the gradient of f and that is (7i-2j)

Answer 92

the gradient is the normal vector and we can regard if as simply:

<7,-2>

Answer 93

correct :)