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!