Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x = t,   y = e−2t,   z = 3t − t3,   (0, 1, 0)

We are given the parametric equations of a curve:

\[
x(t) = t, \quad y(t) = e^{-2t}, \quad z(t) = 3t - t^3
\]

and we need to find the parametric equations for the tangent line at the point \((0, 1, 0)\).

### Step 1: Verify the parameter \(t\) at the specified point
The point \((x, y, z) = (0, 1, 0)\) corresponds to some value of \(t\). We can determine \(t\) by substituting the point into the parametric equations:

- From \(x(t) = t\), we know that \(t = 0\).
- Substituting \(t = 0\) into \(y(t) = e^{-2t}\), we get \(y(0) = e^{-2(0)} = 1\), which matches the \(y\)-coordinate of the point.
- Substituting \(t = 0\) into \(z(t) = 3t - t^3\), we get \(z(0) = 3(0) - (0)^3 = 0\), which matches the \(z\)-coordinate of the point.

Thus, \(t = 0\) is the correct parameter value.

### Step 2: Find the tangent vector
The tangent vector to the curve at any point is given by the derivative of the position vector \(\mathbf{r}(t) = (x(t), y(t), z(t))\) with respect to \(t\). Let's compute the derivatives of \(x(t)\), \(y(t)\), and \(z(t)\):

\[
\frac{dx}{dt} = \frac{d}{dt}(t) = 1
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(e^{-2t}) = -2e^{-2t}
\]
\[
\frac{dz}{dt} = \frac{d}{dt}(3t - t^3) = 3 - 3t^2
\]

Now, evaluate these derivatives at \(t = 0\):

\[
\frac{dx}{dt}\bigg|_{t=0} = 1
\]
\[
\frac{dy}{dt}\bigg|_{t=0} = -2e^{0} = -2
\]
\[
\frac{dz}{dt}\bigg|_{t=0} = 3 - 3(0)^2 = 3
\]

Thus, the tangent vector at \(t = 0\) is:

\[
\mathbf{v} = \left(1, -2, 3\right)
\]

### Step 3: Write the parametric equations for the tangent line
The parametric equations for the tangent line can be written as:

\[
x = x_0 + v_x t, \quad y = y_0 + v_y t, \quad z = z_0 + v_z t
\]

where \((x_0, y_0, z_0)\) is the point on the curve and \((v_x, v_y, v_z)\) is the tangent vector. For our case:

- The point is \((0, 1, 0)\),
- The tangent vector is \((1, -2, 3)\).

Thus, the parametric equations of the tangent line are:

\[
x = 0 + 1t = t
\]
\[
y = 1 - 2t
\]
\[
z = 0 + 3t = 3t
\]

### Final Answer
The parametric equations for the tangent line are:

\[
x = t, \quad y = 1 - 2t, \quad z = 3t
\]

Would you like any further details or clarifications on this process?

Here are 5 related questions:
1. How do you find the tangent line to a curve in space at a specific point?
2. What is the role of the tangent vector in parametric curves?
3. How do parametric equations differ from regular equations in describing curves?
4. Can a curve have more than one tangent line at a given point?
5. How can you find the curvature of a parametric curve?

**Tip:** The derivative of a parametric curve provides the direction of the tangent line at any point, while the curve itself provides the position!

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = t, y = e−2t, z = 3t − t^3, (0, 1, 0)

Find the parametric equations for the tangent line to the curve given by x = t, y = e^(-2t), z = 3t - t^3 at the point (0, 1, 0). This solution covers parametric equations, tangent vectors, and derivative calculations, suitable for high school students in Grades 10-12.

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