Problem 20 - Thomas Calculus

Question

As mentioned in the text, the tangent line to a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) at \(t=t_{0}\) is the line that passes through the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0}\) Find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . \begin{equation} \mathbf{r}(t)=t^{2} \mathbf{i}+(2 t-1) \mathbf{j}+t^{3} \mathbf{k}, \quad t_{0}=2 \end{equation}

Step-by-Step Solution

Answer

The tangent line is given by parametric equations: \(x = 4 + 4t, y = 3 + 2t, z = 8 + 12t\).

Step 1: Find the Point on the Curve

The first step is to find the point on the curve where the tangent line is to be drawn. We plug in the given value of \( t_0 = 2 \) into the curve equation. The curve \( \mathbf{r}(t) = t^2 \mathbf{i} + (2t - 1) \mathbf{j} + t^3 \mathbf{k} \). Let's evaluate this at \( t_0 = 2 \):\[\mathbf{r}(2) = 2^2 \mathbf{i} + (2(2) - 1) \mathbf{j} + 2^3 \mathbf{k} = 4 \mathbf{i} + 3 \mathbf{j} + 8 \mathbf{k}.\]So, the point on the curve is \( (4, 3, 8) \).

Step 2: Compute the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is given by the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Differentiating each component, we have\[ \mathbf{v}(t) = \frac{d}{dt}(t^2) \mathbf{i} + \frac{d}{dt}(2t - 1) \mathbf{j} + \frac{d}{dt}(t^3) \mathbf{k} = 2t \mathbf{i} + 2 \mathbf{j} + 3t^2 \mathbf{k}. \]Evaluate this at \( t_0 = 2 \):\[ \mathbf{v}(2) = 2(2) \mathbf{i} + 2 \mathbf{j} + 3(2)^2 \mathbf{k} = 4 \mathbf{i} + 2 \mathbf{j} + 12 \mathbf{k}. \]Thus, the velocity vector is \( \mathbf{v}(2) = 4 \mathbf{i} + 2 \mathbf{j} + 12 \mathbf{k} \).

Step 3: Write the Parametric Equations for the Tangent Line

A tangent line can be expressed in parametric form using a point \((x_0, y_0, z_0)\) and a direction vector \((a, b, c)\). The parametric equations are:\[(x, y, z) = (x_0, y_0, z_0) + t(a, b, c).\]Here, \( x_0 = 4 \), \( y_0 = 3 \), \( z_0 = 8 \), and the direction vector \((a, b, c)\) is the same as the velocity vector at \( t_0 = 2 \), which is \((4, 2, 12)\). Thus, the parametric equations for the tangent line are:\[x = 4 + 4t, \quad y = 3 + 2t, \quad z = 8 + 12t.\]

Key Concepts

Parametric Equations

In mathematics, parametric equations are a great tool to describe curves using parameters. Instead of expressing variables like x, y, and z independently, we describe each one in terms of one or more parameters, such as the parameter t used here. This approach allows for more flexibility, especially when dealing with curves and surfaces in three-dimensional space.

For instance, consider the curve \( \mathbf{r}(t) = t^2 \mathbf{i} + (2t - 1) \mathbf{j} + t^3 \mathbf{k} \). Here, t is the parameter that drives both the direction and position as t changes. Each component (x, y, z) is expressed as a function of t:

• x = f(t) = t^2
• y = g(t) = 2t - 1
• z = h(t) = t^3

Parametric equations are crucial for tackling problems involving curves, especially when we need to find particular properties like tangent lines at specific points.

Velocity Vector

The velocity vector is a pivotal concept in the study of motion and curves described in space. It's derived from the curve's parametric equation by taking its derivative with respect to the parameter t. Essentially, it represents the direction and rate of change of the position as the parameter varies.

In our example, the velocity vector \( \mathbf{v}(t) \) is found by differentiating each component of \( \mathbf{r}(t) \) with respect to t:

• \( \frac{d}{dt}(t^2) = 2t \)
• \( \frac{d}{dt}(2t - 1) = 2 \)
• \( \frac{d}{dt}(t^3) = 3t^2 \)

This gives the velocity vector: \( \mathbf{v}(t) = 2t \mathbf{i} + 2 \mathbf{j} + 3t^2 \mathbf{k} \).

Evaluated at \( t_0 = 2 \), the velocity vector becomes \( \mathbf{v}(2) = 4 \mathbf{i} + 2 \mathbf{j} + 12 \mathbf{k} \). This vector not only determines the direction of the tangent line at that point but also provides insights into the motion along the curve.

Derivative

Derivatives are foundational in calculus, measuring how a quantity changes as another quantity changes. In the context of parametric curves, the derivative with respect to the parameter t gives us critical information such as velocity and acceleration.

When we compute the derivative of the parametric equation, \( \mathbf{r}(t) = t^2 \mathbf{i} + (2t - 1) \mathbf{j} + t^3 \mathbf{k} \), we are actually determining how the position changes as t varies. This results in what we call the velocity vector:

• x-component: 2t (from differentiating \( t^2 \))
• y-component: 2 (from differentiating \( 2t - 1 \))
• z-component: 3t^2 (from differentiating \( t^3 \))

So, the derivative gives us the velocity vector: \( \, \mathbf{v}(t) = 2t \mathbf{i} + 2 \mathbf{j} + 3t^2 \mathbf{k} \).

This derivative is extremely useful because it not only helps us find the velocity but also lays the groundwork for understanding acceleration and changes in motion along the curve.