Find parametric equations for the tangent line at the point
\(\displaystyle \((\cos(\frac{5 \pi}{6}) ,\sin(\frac{5 \pi}{6}) ,\frac{5 \pi}{6}) )\)\)
on the curve \(\displaystyle \(x=\cos t,\ y=\sin t, \ z=t\)\)
This is what I've done so far:
r(t) = <cos(t) , sin(t), t>
r'(t) = <-sin(t) , cos(t), 1>
So tangent line at r(t): L(t) = r(t) + tr'(t)
so since my point (x, y, z) = \(\displaystyle \((\cos(\frac{5 \pi}{6}) ,\sin(\frac{5 \pi}{6}) ,\frac{5 \pi}{6}) )\)\)
wouldn't the parametric equations for the tangent at the above point be:
x(t) = r(x) + tr'(x) (for x = cos(5(pi)/6))
y(t) = r(y) + tr'(y) (for y = sin(5(pi)/6))
z(t) = r(z) + tr'(z)
Thanks for any help
EDIT: Solved:
t = 5(pi)/6 for r(t) satisfies the given point
thus:
x(t) = r(t) + tr'(t)
x(t) = cos(5(pi)/6) - (1/2)t
and we do the same for y(t) and z(t)
\(\displaystyle \((\cos(\frac{5 \pi}{6}) ,\sin(\frac{5 \pi}{6}) ,\frac{5 \pi}{6}) )\)\)
on the curve \(\displaystyle \(x=\cos t,\ y=\sin t, \ z=t\)\)
This is what I've done so far:
r(t) = <cos(t) , sin(t), t>
r'(t) = <-sin(t) , cos(t), 1>
So tangent line at r(t): L(t) = r(t) + tr'(t)
so since my point (x, y, z) = \(\displaystyle \((\cos(\frac{5 \pi}{6}) ,\sin(\frac{5 \pi}{6}) ,\frac{5 \pi}{6}) )\)\)
wouldn't the parametric equations for the tangent at the above point be:
x(t) = r(x) + tr'(x) (for x = cos(5(pi)/6))
y(t) = r(y) + tr'(y) (for y = sin(5(pi)/6))
z(t) = r(z) + tr'(z)
Thanks for any help
EDIT: Solved:
t = 5(pi)/6 for r(t) satisfies the given point
thus:
x(t) = r(t) + tr'(t)
x(t) = cos(5(pi)/6) - (1/2)t
and we do the same for y(t) and z(t)