Problem 21 - 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)=\ln t \mathbf{i}+\frac{t-1}{t+2} \mathbf{j}+t \ln t \mathbf{k}, \quad t_{0}=1 \end{equation}

Step-by-Step Solution

Answer

The parametric equations are \(x(t) = t\), \(y(t) = \frac{1}{3}t\), and \(z(t) = t\).

Step 1: Evaluate the position vector at t=t_0

Firstly, we need to find the position vector at the specific parameter value \(t_0 = 1\). This is done by substituting \(t_0 = 1\) into \(\mathbf{r}(t)\). So, we have:\( \mathbf{r}(1) = \ln(1) \mathbf{i} + \frac{1-1}{1+2} \mathbf{j} + 1 \ln(1) \mathbf{k} \).Since \(\ln(1) = 0\), the point is \((0, 0, 0)\).

Step 2: Find the velocity vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\). Compute the derivatives of each component:1. For \(\ln t\), the derivative is \(\frac{1}{t}\).2. For \(\frac{t-1}{t+2}\), use the quotient rule: \(\frac{(t+2) - (t-1)}{(t+2)^2} = \frac{3}{(t+2)^2}\).3. For \(t \ln t\), use the product rule: \(\ln t + 1\).So, \(\mathbf{v}(t) = \frac{1}{t} \mathbf{i} + \frac{3}{(t+2)^2} \mathbf{j} + (\ln t + 1) \mathbf{k}\).

Step 3: Evaluate the velocity vector at t=t_0

Now substitute \(t = 1\) into \(\mathbf{v}(t)\) to find \(\mathbf{v}(1)\):\( \mathbf{v}(1) = \frac{1}{1} \mathbf{i} + \frac{3}{3^2} \mathbf{j} + (\ln 1 + 1) \mathbf{k} \).This simplifies to \(\mathbf{v}(1) = \mathbf{i} + \frac{1}{3} \mathbf{j} + \mathbf{k}\).

Step 4: Write the parametric equations of the tangent line

The parametric equations for the tangent line are based on the form: \(x(t) = x_0 + a t\), \(y(t) = y_0 + b t\), \(z(t) = z_0 + c t\), where \((x_0, y_0, z_0)\) is the point found in Step 1 and \((a, b, c)\) are the components of the velocity vector from Step 3. Thus, we have:- \(x(t) = 0 + 1t = t\)- \(y(t) = 0 + \frac{1}{3}t = \frac{1}{3}t\)- \(z(t) = 0 + t = t\)The parametric equations of the tangent line are:\[ x(t) = t, \quad y(t) = \frac{1}{3}t, \quad z(t) = t. \]

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and covers operations on vectors such as differentiation and integration. In problems like finding a tangent line to a curve, vector calculus is essential because it provides tools to handle multi-dimensional fields and derivatives.

In the context of this exercise, vector calculus is used to derive the velocity vector, which is the derivative of the position vector. This derivative indicates the direction of change of the position vector at any time \( t \). By calculating it, we reveal important insights about how an object moves in space.

The steps involve taking derivatives of each component function of the position vector, as seen in the solution where the derivative of \( \mathbf{r}(t) \) is calculated component-wise.

• The derivative of the \( \ln(t) \) function leads to \( \frac{1}{t} \).
• For the \( \frac{t-1}{t+2} \) function, a quotient rule is required, which is a standard technique in calculus.
• The \( t \ln(t) \) function needs the product rule to find its derivative.

Understanding these basics opens the door to efficiently apply vector calculus in complex dynamics and physical systems.

Parametric Equations

Parametric equations help describe a curve in terms of a parameter—in this case, \( t \). They express each coordinate of a point on the curve as functions of the parameter.

In this particular exercise, the curve is defined parametrically by the position vector \( \mathbf{r}(t) = \ln t \mathbf{i} + \frac{t-1}{t+2} \mathbf{j} + t \ln t \mathbf{k} \). With parametric equations, the path followed by a moving point is described without directly referring to its form in \( x, y, \) or \( z \).

By substituting the value \( t_0 = 1 \) into the position vector, we find a specific point on the curve. Then the tangent line, which touches the curve precisely at this point, can also be parametric. This produces easy-to-follow descriptions of the line's direction and position over time, which are derived using the parameter \( t \).

• For the tangent line: \( x(t) = 0 + t \)
• \( y(t) = 0 + \frac{1}{3}t \)
• \( z(t) = 0 + t \).

These parametric equations beautifully simplify the representation of the tangent line in three-dimensional space.

Velocity Vector

The concept of the velocity vector is crucial when discussing movement along curves. In this problem, the velocity vector \( \mathbf{v}(t) \) is derived by differentiating the position vector \( \mathbf{r}(t) \), a key step in understanding motion.

The velocity vector offers a powerful insight: it denotes the rate and direction at which a point moves along the curve. It's essentially the first derivative of the position vector regarding time \( t \). For instance, in this exercise, the velocity vector was found to be \( \mathbf{i} + \frac{1}{3} \mathbf{j} + \mathbf{k} \) when evaluated at the specific parameter \( t_0 = 1 \).

Here's why the velocity vector matters:

• It provides the direction along which the tangent line is aligned, crucial for plotting or analyzing the motion of particles.
• The magnitude of the velocity vector tells how fast the point is moving at any given moment on the curve.

Therefore, understanding the velocity vector not only helps plot tangent lines but also models the dynamic behavior of systems in physical and mathematical simulations.