Question

Calculus 3 Question:
Write a parametric representation for the tangent line to the given curve at the given point in each of the following cases.
(a) (cos 4t, sin 4t, t) at the point t = pi/8
(b) (t, 2t, t^2) at the point (1, 2, 1)

Answer 1

anybody good with calculus 3 on openstudy?

Answer 2

nobody that im aware of ...

Answer 3

lol

Answer 4

tangent to a curve is derivative of the curve function

Answer 5

r = cos4t, sin4t, t

r' = cos'4t, sin'4t, t'

Answer 6

soo,

x = cos4t + n (cos'4t)
y = sin4t + n (sin'4t)
z = t + n (t')

all when t=pi/8

Answer 7

and since thats in my own notative flair, you might have questions

Answer 8

so first I get the derivative and then just plug in t?

Answer 9

yes

Answer 10

and since having a scalar named "t" would have been confusing; i renamed it "n"

Answer 11

so then I get r'(pi/8)=<4sin4(pi/8), -4cos4(pi/8), 1>

Answer 12

cos' = -sin

Answer 13

oh woops

Answer 14

okay, the book says the answer is. <0, 1, pi/8>

Answer 15

sin'=cos; so other than a few sign swaps; it looks fine

Answer 16

for point or vector?

Answer 17

your point ends in pi/8

Answer 18

your vector ends in 1

Answer 19

the "parametric representaion" is <0, 1, pi/8>

Answer 20

^the answer

Answer 21

r = cos4t, sin4t, t ; t=pi/8

r = 0,1, pi/8 is the point that we attach the tangent to

Answer 22

the equation for the tangent line requires the tangent vector at that point to define it

Answer 23

<0,1,pi/8> is not a parametrically defined tangent line; its simply a vector pointing to a point

Answer 24

waiiit. I made a mistake, i copied the answer wrong

Answer 25

answer: (0, 1, pi/8) + t(-4, 0, 1)

Answer 26

thats better

Answer 27

but it makes no sense to say t = pi/8 and then put a nonrelated "t" into teh equation as a scalar

Answer 28

AND ... that is not parametric form; that is vector form i believe

Answer 29

okay. if you do r'(pi/8) you get (-4, 0, pi/8)

Answer 30

http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx

Answer 31

t' = ?

Answer 32

you don't need t' i don't think

Answer 33

yes, you do, you need ALL the component derivatives

Answer 34

t' would just be zero anyway

Answer 35

t' = ?

Answer 36

t = pi/8
t' = 0

Answer 37

t = t

t' = t' = 1

Answer 38

after you get your derivative; then apply t=pi/8 to get a value

Answer 39

pi/8' = 0

t' not= pi/8'

Answer 40

okay I think I figured it out.
first simply plug pi/8 into the line equation and you get r=<cos 4t, sin 4t, t>
r(pi/8) = <0, 1, pi/8>

Answer 41

next, plus pi/8 into the derivative of the line
r'=<-4sin4t, 4cos4t, 1>
r'(pi/8) = -4, 0, 1

Answer 42

x = cos4t + n (cos'4t)
y = sin4t + n (sin'4t)
z = t + n (t')


x = cos4t + n (-4sin4t)
y = sin4t + n (cos4t)
z = t + n (1)

t = pi/8
.....................


x = 0 + n (-4)
y = 1 + n (0)
z = pi/8 + n (1)

Answer 43

put it into vector form

Answer 44

the question asks for parametric form; why would we need to put it into vector form?

Answer 45

get the norm of r'(pi/8) which is <-4, 0, 1>, then use X + Nt = L
(0, 1, pi/8) + t(-4, 0, 1)

Answer 46

get the norm?

Answer 47

the normal, if i read you right, is not the tangent ....

Answer 48

sorry not the norm a parallel vector