Questions: Find parametric equations for the tangent line at the point (cos(5/6 pi), sin(5/6 pi), 5/6 pi) on the curve x=cos t, y=sin t, z=t x(t)= y(t)= z(t)= (Your line should be parametrized so that it passes through the given point at t=0 ).

Solution Steps

To find the parametric equations for the tangent line at the given point on the curve, we need to:

1. Evaluate the point on the curve at the given parameter value.
2. Compute the derivative of the curve with respect to the parameter \( t \).
3. Use the point and the derivative to form the parametric equations of the tangent line.

Given the parametric equations of the curve: \[ x = \cos(t), \quad y = \sin(t), \quad z = t \]

Evaluate these at \( t = \frac{5\pi}{6} \): \[ x\left(\frac{5\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \] \[ y\left(\frac{5\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \] \[ z\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} \]

So, the point on the curve is: \[ \left( -\frac{\sqrt{3}}{2}, \frac{1}{2}, \frac{5\pi}{6} \right) \]

The derivatives of the parametric equations with respect to \( t \) are: \[ \frac{dx}{dt} = -\sin(t) \] \[ \frac{dy}{dt} = \cos(t) \] \[ \frac{dz}{dt} = 1 \]

Evaluate these derivatives at \( t = \frac{5\pi}{6} \): \[ \frac{dx}{dt}\bigg|_{t = \frac{5\pi}{6}} = -\sin\left(\frac{5\pi}{6}\right) = -\frac{1}{2} \] \[ \frac{dy}{dt}\bigg|_{t = \frac{5\pi}{6}} = \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \] \[ \frac{dz}{dt}\bigg|_{t = \frac{5\pi}{6}} = 1 \]

So, the tangent vector at \( t = \frac{5\pi}{6} \) is: \[ \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2}, 1 \right) \]

Using the point and the tangent vector, the parametric equations of the tangent line are: \[ x(t) = -\frac{\sqrt{3}}{2} + \left( -\frac{1}{2} \right) t \] \[ y(t) = \frac{1}{2} + \left( -\frac{\sqrt{3}}{2} \right) t \] \[ z(t) = \frac{5\pi}{6} + t \]

Final Answer