To find the parametric equations of the tangent line to the given parametric curve at the specified point, we follow these steps:

Step 1: Given Parametric Equations

$x = 1 + 6t, \quad y = t^5 - t, \quad z = t^5 + t$ and the given point: (7, 0, 2).

Step 2: Solve for $t$ at (7,0,2)

We solve for $t$ by using the $x$-coordinate equation: $7 = 1 + 6t$ $6t = 6$ $t = 1.$

Step 3: Compute the Tangent Vector

To find the tangent vector, we differentiate each component with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt} (1 + 6t) = 6$

$\frac{dy}{dt} = \frac{d}{dt} (t^5 - t) = 5t^4 - 1$

$\frac{dz}{dt} = \frac{d}{dt} (t^5 + t) = 5t^4 + 1$

Evaluate at $t = 1$:

$\frac{dx}{dt} = 6, \quad \frac{dy}{dt} = 5(1)^4 - 1 = 4, \quad \frac{dz}{dt} = 5(1)^4 + 1 = 6.$

So, the tangent vector is $\mathbf{v} = (6, 4, 6)$.

Step 4: Write the Parametric Equations for the Tangent Line

The parametric equations for the tangent line passing through $(7,0,2)$ in the direction of $\mathbf{v} = (6,4,6)$ are:

$x = 7 + 6t$ $y = 0 + 4t$ $z = 2 + 6t$

These equations describe the tangent line to the curve at the point $(7,0,2)$.

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Related Questions:

1. How do you determine the tangent vector at a given point for parametric curves?
2. What does a tangent line represent in 3D space?
3. How can you find the normal plane to a parametric curve?
4. How do you determine if a point lies on a given parametric curve?
5. What is the geometric interpretation of the derivative in parametric equations?

Tip:

When finding the tangent line, always verify that the given point satisfies the original parametric equations before proceeding with differentiation.