Multivariable Calculus - Tangent Line

[Larrytsai]

Homework Statement

Find parametric equations for the tangent line at the point
(cos (-5*pi/6), sin (-5*pi/6), -5*pi/6) on the curve
x(t) = cos t
y(t) = sin t
z(t) = t

(Your line should be parametrized so that it passes through the given point at t=0).

Im not really understanding the question that well, can anyone help me understand and solve this problem?

Homework Equations

The Attempt at a Solution

What I have so far is,

r'(t) = ( - sin t , cos t, 1)
so
x(t) = -sin t + cos(-5pi/6)
thats all I have so far and its wrong.

 

[mighty2000]

To find the parametric equations for the tangent line at the given point, we first need to find the slope of the tangent line at that point. This can be done by taking the derivative of each component of the given curve and evaluating it at t=-5pi/6. So, we have:

x'(t) = -sin t
y'(t) = cos t
z'(t) = 1

x'(-5pi/6) = -sin (-5pi/6) = -1/2
y'(-5pi/6) = cos (-5pi/6) = -√3/2
z'(-5pi/6) = 1

So, the slope of the tangent line at t=-5pi/6 is given by m = (-1/2, -√3/2, 1).

Next, we need to find a point on the tangent line. This can be done by simply plugging in t=0 into the given curve, since we want the tangent line to pass through the given point at t=0.

x(0) = cos 0 = 1
y(0) = sin 0 = 0
z(0) = 0

So, a point on the tangent line is (1, 0, 0).

Now, we can use the point-slope form of a line to find the parametric equations for the tangent line:

x(t) = 1 + (-1/2)t
y(t) = 0 + (-√3/2)t
z(t) = 0 + t

These equations can also be written in vector form as:

r(t) = (1, 0, 0) + t(-1/2, -√3/2, 1)

So, the parametric equations for the tangent line at the given point are:

x(t) = 1 - (1/2)t
y(t) = (-√3/2)t
z(t) = t