Problem 35 - 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 \(33-36,\) find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . $$ \mathbf{r}(t)=(a \sin t) \mathbf{i}+(a \cos t) \mathbf{j}+b t \mathbf{k}, \quad t_{0}=2 \pi $$

Step-by-Step Solution

Answer

The parametric equations of the tangent line are: \( x = ta, \ y = a, \ z = 2\pi b + tb \).

Step 1: Determine the curve's point at t_0

First, identify the point on the curve at the given parameter value \( t_0 = 2\pi \). Substitute \( t_0 \) into the vector-valued function \( \mathbf{r}(t) = (a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + bt \mathbf{k} \):\[ \mathbf{r}(2\pi) = (a \sin 2\pi) \mathbf{i} + (a \cos 2\pi) \mathbf{j} + b(2\pi) \mathbf{k} \]Since \( \sin(2\pi) = 0 \) and \( \cos(2\pi) = 1 \), the point is:\[ \mathbf{r}(2\pi) = 0 \mathbf{i} + a \mathbf{j} + 2\pi b \mathbf{k} \] Thus, the point on the curve at \( t_0 = 2\pi \) is \( (0, a, 2\pi b) \).

Step 2: Find the velocity vector at t_0

The tangent line is parallel to the velocity vector of the curve at \( t_0 \). To find this, calculate the derivative of \( \mathbf{r}(t) \), which gives the velocity vector \( \mathbf{v}(t) = \mathbf{r}'(t) \):\[ \mathbf{r}'(t) = (a \cos t) \mathbf{i} - (a \sin t) \mathbf{j} + b \mathbf{k} \] Then, substitute \( t_0 = 2\pi \) into \( \mathbf{r}'(t) \):\[ \mathbf{v}(2\pi) = (a \cos(2\pi)) \mathbf{i} - (a \sin(2\pi)) \mathbf{j} + b \mathbf{k} \]\[ \mathbf{v}(2\pi)= a\mathbf{i} + 0 \mathbf{j} + b \mathbf{k} \] Thus, the velocity vector at \( t_0 = 2\pi \) is \( (a, 0, b) \).

Step 3: Write the parametric equations of the tangent line

The tangent line is determined by the point \((0, a, 2\pi b)\) and is parallel to the vector \((a, 0, b)\). The parametric equations of a line can be written as:\[ x = x_0 + tv_x \]\[ y = y_0 + tv_y \]\[ z = z_0 + tv_z \]Substituting \((x_0, y_0, z_0) = (0, a, 2\pi b)\) and \((v_x, v_y, v_z) = (a, 0, b)\), the parametric equations become:\[ x = 0 + ta = ta \]\[ y = a + t \cdot 0 = a \]\[ z = 2\pi b + tb \] Therefore, the parametric equations for the tangent line are:\[ x = ta, \quad y = a, \quad z = 2\pi b + tb \].

Key Concepts

Parametric Equations

In the context of curves, parametric equations are a way of defining the path of a point on a plane. Instead of using just one variable as we do in standard function equations like \( y = f(x) \), parametric equations use parameters, often time \( t \), to define components separately. For example, the vector-valued function \( \mathbf{r}(t) = (a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + bt \mathbf{k} \) gives us the parametric form of each component of a point in a 3D space:

• \( x = a \sin t \)
• \( y = a \cos t \)
• \( z = bt \)

Each of these functions depends on \( t \), the parameter, which you can think of as a time variable. The path traced out by \( (x, y, z) \) as \( t \) changes is the curve in space. By manipulating \( t \), you get different points along this curve. Parametric equations are particularly useful for describing complex curves and three-dimensional paths, like the one given in our example.

Velocity Vector

The velocity vector of a curve at a given point helps us understand how the curve is changing at that point. If we imagine the curve representing the path of a moving object, the velocity vector indicates the speed and direction of that movement. For our vector function \( \mathbf{r}(t) = (a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + bt \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is found by taking the derivative of each component with respect to \( t \).

The derivative, \( \mathbf{r}'(t) \), is:

• \( \frac{d}{dt}(a \sin t) \to a \cos t \)
• \( \frac{d}{dt}(a \cos t) \to -a \sin t \)
• \( \frac{d}{dt}(bt) \to b \)

Thus, the velocity vector is \( \mathbf{v}(t) = (a \cos t) \mathbf{i} - (a \sin t) \mathbf{j} + b \mathbf{k} \).

When you substitute \( t = 2\pi \) into this expression, there's a point on the curve where \( \cos(2\pi) = 1 \) and \( \sin(2\pi) = 0 \). The velocity vector becomes \( (a, 0, b) \), which tells us how fast and in which direction the curve is moving precisely at \( t = 2\pi \).

Vector-Valued Function

A vector-valued function lets us describe curves in multidimensional space. This extends the idea of a standard function y = f(x) to vector outputs. Instead of giving a single value as the output, vector-valued functions give a vector, which can have multiple dimensions.

Consider \( \mathbf{r}(t) = (a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + bt \mathbf{k} \). This function maps \( t \) to a vector in three-dimensional space:

• The \( \mathbf{i} \) component is influenced by \( \sin t \)
• The \( \mathbf{j} \) component by \( \cos t \)
• The \( \mathbf{k} \) component grows linearly in \( t \)

The power of vector-valued functions lies in their ability to represent complex motions, like spirals or helixes, with simple and compact equations. They're essential in physics, computer graphics, and engineering to simulate movement and changes over time in multiple dimensions.

Derivative of Vector Functions

Derivatives aren't just about slopes for lines or curves in a plane. With vector functions, derivatives tell us much more about motion along those curves. The derivative of a vector-valued function \( \mathbf{r}'(t) \) gives the tangent vector to the curve at any point. It tells us which way the curve points as \( t \) changes.

When you differentiate each component of \( \mathbf{r}(t) = (a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + bt \mathbf{k} \), you differentiate with respect to \( t \), treating each basis vector independently:

• The i-component is \( \frac{d}{dt}(a \sin t) = a \cos t \)
• The j-component is \( \frac{d}{dt}(a \cos t) = -a \sin t \)
• The k-component is simply \( \frac{d}{dt}(bt) = b \)

This vector derivative \( \mathbf{r}'(t) = (a \cos t) \mathbf{i} - (a \sin t) \mathbf{j} + b \mathbf{k} \) becomes the velocity vector. It conveys not just direction but also the rate of change (or speed) along the curve, key for anything involving motion analysis in space.