Problem 25 - 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 \(t=t_{0}\). $$ \mathbf{r}(t)=\ln t \mathbf{i}+\frac{t-1}{t+2} \mathbf{j}+t \ln t \mathbf{k}, \quad t_{0}=1 $$

Step-by-Step Solution

Answer

The parametric equations for the tangent line are \(x = s\), \(y = \frac{1}{3}s\), \(z = s\).

Step 1: Find the position vector at t=t₀

Substitute the given parameter value into the position vector to find the point through which the tangent line passes. Calculate:\[ f(1) = \ln(1) = 0 \]\[ g(1) = \frac{1-1}{1+2} = 0 \]\[ h(1) = 1 \cdot \ln(1) = 0 \]Therefore, the position vector at \(t = 1\) is \( (0, 0, 0) \).

Step 2: Determine the velocity vector

To find the velocity vector, differentiate each component of the position vector with respect to \(t\):\[ \frac{d}{dt}[\ln t] = \frac{1}{t} \]\[ \frac{d}{dt}\left[\frac{t-1}{t+2}\right] = \frac{(t+2) \cdot 1 - (t-1) \cdot 1}{(t+2)^2} = \frac{3}{(t+2)^2} \]\[ \frac{d}{dt}[t \ln t] = \ln t + 1 \]Substitute \(t = 1\) into the derivatives to find the velocity vector:\[ v(1) = \left(1, \frac{3}{9}, 1\right) = (1, \frac{1}{3}, 1) \]

Step 3: Write the parametric equations of the tangent line

The tangent line can be expressed as a vector equation derived from the position vector and velocity vector:\[ (x, y, z) = (f(t_0), g(t_0), h(t_0)) + s \cdot (v_1, v_2, v_3) \]Substitute the found values:\[ (x, y, z) = (0, 0, 0) + s \cdot (1, \frac{1}{3}, 1) \]This gives the parametric equations:\[ x = s \]\[ y = \frac{1}{3}s \]\[ z = s \]

Key Concepts

Position Vector

In the world of vectors and curves, a position vector is a fundamental concept. It represents the position of a point in space relative to an origin.

• The position vector is usually denoted as \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \), where \(i, j, k\) are unit vectors along the x, y, and z axes.
• In this specific exercise, to find the position vector at a given parameter \( t_0 \), you substitute \( t_0 \) into the functions \( f(t), g(t), h(t) \).

Here, at \( t = 1 \), we calculated the position vector as \( (0, 0, 0) \). This means that the point at which the tangent line passes is at the origin for this particular curve and parameter value. It's crucial to understand the role of the position vector as it serves as the starting point for defining the tangent line.

Velocity Vector

A velocity vector indicates the direction and speed at which an object's position is changing. It is pivotal when discussing curves as it helps us determine the tangent's direction at any given point.

• To find the velocity vector, you need to differentiate each component of the position vector \( \mathbf{r}(t) \).
• The resultant vector \( \mathbf{v}(t) = \left( \frac{1}{t}, \frac{3}{(t+2)^2}, \ln t + 1 \right) \) gives the velocity at a parameter \( t \).

For \( t = 1 \), the velocity vector calculated is \( (1, \frac{1}{3}, 1) \). This tells us that right at the point \( (0, 0, 0) \), the curve is moving in the direction described by this vector. Understanding velocity vectors is crucial for correctly plotting the path of an object and determining tangent lines.

Differentiation

Differentiation is the process of finding how a function changes at any point. It is crucial for calculating velocity vectors from position vectors.

• Every component of the vector function \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \) is differentiated with respect to \( t \) to find the velocity vector.
• This involves applying derivative rules, for example, the derivative of \(\ln t\) is \(\frac{1}{t}\).

In the exercise's context, differentiating each component, namely \(\ln t\), \(\frac{t-1}{t+2}\), and \(t \ln t\), gives us the individual components of the velocity vector. The ability to differentiate correctly affects the accuracy of calculating how curves behave over time. This forms the backbone of analyzing motion and related phenomena.

Smooth Curve

A smooth curve is a continuous curve without sharp angles or discontinuities, making it differentiable at all points.

• Such curves are essential for discussing tangent lines because differentiability ensures that a tangent exists at each point.
• In practical applications, a smooth curve like \( \mathbf{r}(t) = \ln t \mathbf{i} + \frac{t-1}{t+2} \mathbf{j} + t \ln t \mathbf{k} \) allows for efficient analysis and computations.

In our exercise, identifying \( \mathbf{r}(t) \) as a smooth curve allows us to confidently find derivatives to obtain velocity vectors and thus the tangent lines. Understanding what makes a curve smooth underpins much of the work in calculus relating to motion and changes. It allows for precise calculations and predictions of a path's behavior.