To find the parametric equations for the tangent line to the curve at a specified point, we need to go through the following steps:

Given parametric equations for the curve:

$x(t) = t \cos(t), \quad y(t) = t^4, \quad z(t) = t \sin(t)$

Point specified: $(- \pi, \pi^4, 0)$

The point corresponds to $t = -\pi$, which means we need to find the parametric equation of the tangent line at $t = -\pi$.

Step 1: Find the derivative of each component

The derivatives of $x(t)$, $y(t)$, and $z(t)$ will give us the velocity vector at any point on the curve. This velocity vector will provide the direction of the tangent line.

• Derivative of $x(t)$: $\frac{dx}{dt} = \frac{d}{dt} \left(t \cos(t)\right) = \cos(t) - t \sin(t)$

• Derivative of $y(t)$: $\frac{dy}{dt} = \frac{d}{dt} \left(t^4\right) = 4t^3$

• Derivative of $z(t)$: $\frac{dz}{dt} = \frac{d}{dt} \left(t \sin(t)\right) = \sin(t) + t \cos(t)$

Step 2: Evaluate the derivatives at $t = -\pi$

Substitute $t = -\pi$ into the derivatives to get the components of the tangent vector at the point.

• For $x'(t)$ at $t = -\pi$: $x'(-\pi) = \cos(-\pi) - (-\pi) \sin(-\pi) = -1 + \pi(0) = -1$

• For $y'(t)$ at $t = -\pi$: $y'(-\pi) = 4(-\pi)^3 = 4(-\pi^3) = -4\pi^3$

• For $z'(t)$ at $t = -\pi$: $z'(-\pi) = \sin(-\pi) + (-\pi) \cos(-\pi) = 0 - \pi(-1) = \pi$

So, the tangent vector at $t = -\pi$ is $\langle -1, -4\pi^3, \pi \rangle$.

Step 3: Write the parametric equations of the tangent line

The parametric equations for the tangent line can be written as: $x(t) = x_0 + t \cdot x'(-\pi)$ $y(t) = y_0 + t \cdot y'(-\pi)$ $z(t) = z_0 + t \cdot z'(-\pi)$ where $(x_0, y_0, z_0)$ is the point $(- \pi, \pi^4, 0)$ and $\langle x'(-\pi), y'(-\pi), z'(-\pi) \rangle = \langle -1, -4\pi^3, \pi \rangle$.

So the parametric equations of the tangent line are:

• $x(t) = -\pi + t(-1) = -\pi - t$
• $y(t) = \pi^4 + t(-4\pi^3)$
• $z(t) = 0 + t(\pi) = t\pi$

Step 4: Graph the curve and the tangent line

Using a graphing tool, plot both the curve defined by:

• $x = t \cos(t)$
• $y = t^4$
• $z = t \sin(t)$

and the tangent line defined by:

• $x(t) = -\pi - t$
• $y(t) = \pi^4 - 4\pi^3 t$
• $z(t) = \pi t$

The graph should show how the tangent line touches the curve exactly at the point $(- \pi, \pi^4, 0)$.

Additional Questions:

1. What does the tangent vector tell us about the local behavior of the curve at the point?
2. How would the tangent line equation change if we chose a different point on the curve?
3. Can you find the normal vector to the curve at this point?
4. How do we interpret the derivatives in terms of the curve's motion?
5. How does the curve behave as $t$ approaches infinity?

Tip: When graphing 3D curves and tangent lines, it’s helpful to use parametric plotting tools to get an accurate picture of the relationship between the curve and its tangent line.