Problem 24 - 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 \(t_{0}\) . In Exercises \(23-26\) , find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\). $$ \mathbf{r}(t)=t^{2} \mathbf{i}+(2 t-1) \mathbf{j}+t^{3} \mathbf{k}, \quad t_{0}=2 $$

Step-by-Step Solution

Answer

The parametric equations for the tangent line are \(x = 4 + 4t\), \(y = 3 + 2t\), and \(z = 8 + 12t\).

Step 1: Identify the position vector

We start with the given position vector \(\mathbf{r}(t) = t^2 \, \mathbf{i} + (2t-1) \, \mathbf{j} + t^3 \, \mathbf{k}\). Our task is to find the tangent line at \(t_0 = 2\) by utilizing the point and the velocity vector at this parameter value.

Step 2: Calculate the point on the curve

Substitute \(t_0 = 2\) into \(\mathbf{r}(t)\) to find the point on the curve. \(\mathbf{r}(2) = 2^2 \, \mathbf{i} + (2(2) - 1) \, \mathbf{j} + (2^3) \, \mathbf{k} = 4 \, \mathbf{i} + 3 \, \mathbf{j} + 8 \, \mathbf{k}\). Thus, the point on the curve is \((4, 3, 8)\).

Step 3: Calculate the velocity vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\). Calculate \(\mathbf{v}(t) = \frac{d}{dt}(t^2) \, \mathbf{i} + \frac{d}{dt}(2t-1) \, \mathbf{j} + \frac{d}{dt}(t^3) \, \mathbf{k}\). Thus, \(\mathbf{v}(t) = 2t \, \mathbf{i} + 2 \, \mathbf{j} + 3t^2 \, \mathbf{k}\).

Step 4: Evaluate the velocity vector at the given parameter

Substitute \(t_0 = 2\) into the velocity vector equation: \(\mathbf{v}(2) = 2(2) \, \mathbf{i} + 2 \, \mathbf{j} + 3(2)^2 \, \mathbf{k} = 4 \, \mathbf{i} + 2 \, \mathbf{j} + 12 \, \mathbf{k}\). So the velocity vector is \((4, 2, 12)\).

Step 5: Write the parametric equations for the tangent line

The line tangent to the curve at \(t = 2\) can be expressed with the parametric equations: \(x = 4 + 4t\), \(y = 3 + 2t\), \(z = 8 + 12t\) by using the point and the direction provided by the velocity vector. Thus, the final tangent line equations are based on moving along the velocity vector from the point \((4, 3, 8)\).

Key Concepts

Parametric Equations

Parametric equations use one or more parameters to describe a set of related variables. In the case of a curve in three-dimensional space, parametric equations describe the x, y, and z coordinates as functions of a parameter, often denoted as \( t \). This parameter makes it easier to express and manipulate geometric figures. Instead of working with lengthy algebraic equations, parametric equations

• Provide a convenient way to describe paths, curves, and motions.
• Allow the representation of curves that cannot be expressed easily with a single function in terms of x and y.
• Simplify the expression of points and vectors in space.

In our example, the position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + (2t-1) \mathbf{j} + t^3 \mathbf{k} \) defines a curve where each point depends on parameter \( t \). To find the parametric equations of a tangent line to this curve at a specific point, you derive these equations from both the position and velocity vectors.

Velocity Vector

The velocity vector is crucial to understanding the movement of an object along a path described by parametric equations. It is essentially the derivative of the position vector with respect to the parameter, representing the rate of change of position. This means:

• The velocity vector points in the direction in which the position is changing fastest.
• It provides the direction and the speed of the object along the curve.

In the exercise, the velocity vector is derived by differentiating the position vector. Starting with: \(\mathbf{v}(t) = \frac{d}{dt}(t^2) \mathbf{i} + \frac{d}{dt}(2t-1) \mathbf{j} + \frac{d}{dt}(t^3) \mathbf{k}\), we calculate: \(\mathbf{v}(t) = 2t \mathbf{i} + 2 \mathbf{j} + 3t^2 \mathbf{k}\). The velocity vector at \( t_0 = 2 \) is \( \mathbf{v}(2) = 4 \mathbf{i} + 2 \mathbf{j} + 12 \mathbf{k} \), which is used to determine the direction of the tangent line at that point.

Position Vector

The position vector \( \mathbf{r}(t) \) describes the coordinates of a point on the curve as a function of the parameter \( t \). This vector gives a geometric position in space and is fundamental in understanding the curve's shape and behavior. In particular:

• It identifies a specific point by substituting a value for \( t \).
• It is used in calculating the velocity vector, which describes how the position changes over time.

In the step-by-step solution, the position vector is given as \( \mathbf{r}(t) = t^2 \mathbf{i} + (2t-1) \mathbf{j} + t^3 \mathbf{k} \). By substituting \( t_0 = 2 \) in the position vector, we determine that the point on the curve is \( (4, 3, 8) \). This is crucial for deriving the parametric equations of the tangent line. The position vector thus sets the anchor point through which the tangent line passes.