Question

Find parametric equations for the tangent line at the point (cos(3pi/6),sin(3pi/6), (3pi/6))
on the curve x=cost, y= sint, z=t
x(t)=?
y(t)=?
z(t)=?

Answer 1

just derive each component on its own metir

Answer 2

merit even

Answer 3

the derivative of a vector defined function is the tangent to the curve ... like normal

Answer 4

its when you derive the components of the surface equations that you get a normal

Answer 5

x=cost; x' = -sin(t)

y= sint; y' = cos(t)

z = t ; z' = 1

Answer 6

now just fill in your given point and youve got the slope of the tangent at least

Answer 7

i take it the given t is 3pi/6 ....

Answer 8

establish the values of the given point to create the parametric line equations

Answer 9

so x=-sin(3pi/6)
y= cos(3pi/6)
z=1

Answer 10

that would be the vector tangent at that point yes

Answer 11

to be lazy about it i would express it like this:

x =cos(3pi/6) -sin(3pi/6)t
y = sin(3pi/6) +cos(3pi/6)t
z = 3pi/6 + t

Answer 12

3pi/6 ... really? whats wrong with pi/2 lol; 90 degress

Answer 13

x = t
y = 1
z = pi/2 + t

maybe if i simplified it right

Answer 14

you agree?

Answer 15

agree