Problem 34 - 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 \(t=t_{0}\) . $$ \mathbf{r}(t)=(2 \sin t) \mathbf{i}+(2 \cos t) \mathbf{j}+5 t, \quad t_{0}=4 \pi $$

Step-by-Step Solution

Answer

The parametric equations for the tangent line are \( x = 2s, y = 2, z = 20\pi + 5s \).

Step 1: Identify Point of Tangency

First, we need to find the point on the curve where the tangent line touches. This point is given by \( \mathbf{r}(t_0) \). For our exercise, the parameter value is \( t_0 = 4\pi \). Thus, \( \mathbf{r}(4\pi) = (2\sin(4\pi)) \mathbf{i} + (2\cos(4\pi)) \mathbf{j} + 5(4\pi)\mathbf{k} \).

Step 2: Simplify Trigonometric Expressions

Calculate the trigonometric functions at \( t = 4\pi \). Since \( \sin(4\pi) = 0 \) and \( \cos(4\pi) = 1 \), we have \( \mathbf{r}(4\pi) = 0 \mathbf{i} + 2 \mathbf{j} + 20\pi \mathbf{k} \). Thus, the point can be expressed as \( (0, 2, 20\pi) \).

Step 3: Differentiate the Curve to Find the Velocity Vector

Determine the derivative of \( \mathbf{r}(t) \) to find the velocity vector \( \mathbf{v}(t) \). Differentiating each component with respect to \( t \), we get: \( \mathbf{v}(t) = (2\cos t)\mathbf{i} - (2\sin t)\mathbf{j} + 5\mathbf{k} \).

Step 4: Evaluate Velocity Vector at \( t = 4\pi \)

Substitute \( t = 4\pi \) into the expression for \( \mathbf{v}(t) \). Using the values \( \cos(4\pi) = 1 \) and \( \sin(4\pi) = 0 \), get \( \mathbf{v}(4\pi) = 2\mathbf{i} + 0\mathbf{j} + 5\mathbf{k} \).

Step 5: Write Parametric Equations of the Tangent Line

The parametric equations for the tangent line will be in the form \( (x(t), y(t), z(t)) = (0, 2, 20\pi) + s(2, 0, 5) \), where \( s \) is a parameter. Thus, the parametric equations are:\[ \begin{align*} x(s) &= 0 + 2s, \ y(s) &= 2 + 0s, \ z(s) &= 20\pi + 5s. \end{align*} \]

Key Concepts

velocity vector

In the study of parametric curves, the notion of the **velocity vector** is crucial. Imagine a particle moving along a curve. The velocity vector represents both the direction and the speed of the particle at any given point on its path.
This vector is a derivative of the position vector, which defines the location of the point on the curve.When we take the derivative of the vector function, which describes our curve, we get the velocity vector:

• For a curve given by the parametric equations \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \),
• The velocity vector is \( \mathbf{v}(t) = f'(t) \mathbf{i} + g'(t) \mathbf{j} + h'(t) \mathbf{k} \).

This formula gives us the tangent direction of the curve at any point \( t \). Specific to our exercise, the velocity vector at \( t = 4\pi \) was computed as \( \mathbf{v}(4\pi) = 2\mathbf{i} + 0\mathbf{j} + 5\mathbf{k} \).
This indicates that, at this particular point, the *tangent line* moves in these direction proportions within the 3D space.

parametric equations

**Parametric equations** allow us to express a curve using a parameter, typically denoted as \( t \). Instead of defining the curve with a single equation in \( x \) and \( y \), we describe each coordinate of the points on the curve as functions of \( t \).

• An example is the curve represented by \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + 5t \mathbf{k} \).
• This describes how the position of the curve changes as \( t \) varies.

By using parametric equations, we can handle more complex curves, such as circles, ellipses, and helices, which cannot be described easily by single equations in Cartesian form.
The tangent line parametric representation, derived from the velocity vector, reflects the rate and direction at which a point moves along this curve.
For instance, the equation of the tangent line to the curve at \( t = 4\pi \) is given in parametric form:
\[ \begin{align*} x(s) &= 0 + 2s, \ y(s) &= 2 + 0s, \ z(s) &= 20\pi + 5s. \end{align*} \] This shows how its position evolves linearly along an axis defined by the velocity vector.

curve differentiation

The process of **curve differentiation** involves finding the derivative of a vector-valued function that describes a curve. This operation is crucial for understanding how a curve behaves since it provides us with the velocity vector.

• From a mathematical standpoint, if \( \mathbf{r}(t) \) is our curve, then \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \).
• Each component of the position vector is differentiated individually.

In practice, this means that:

• If \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + 5t \mathbf{k} \),
• The derivative or velocity vector is \( \mathbf{v}(t) = (2 \cos t) \mathbf{i} - (2 \sin t) \mathbf{j} + 5 \mathbf{k} \).

Curve differentiation helps to identify points on the curve where it is moving fastest, slowest, or changes direction entirely at particular parameter values.
Using these derivatives, we can describe the precise nature of motion and shapes created by the curve.

trigonometric functions

Trigonometric functions such as sine and cosine play a significant role in describing curves, especially in parametric forms like circles and spirals.
These functions repeat their values in a cyclical pattern, making them ideal for representing oscillating systems.

• The sine function, \( \sin(t) \), starts at 0, rises to 1, falls to -1, and comes back to 0 within a period of \( 2\pi \).
• The cosine function, \( \cos(t) \), starts at 1, drops to -1, and returns to 1 over the same interval.

In the exercise, the parametric curve defined by \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + 5t \mathbf{k} \) uses trigonometric functions to create a circular motion in the \( x \)- and \( y \)-directions.
This circle has a radius of 2, since the maximum values of \( \sin(t) \) and \( \cos(t) \) are scaled by 2.
The trigonometric nature ensures that as \( t \) changes, the point on the curve moves in a stable, repeating pattern alongside a linear increase in the \( z \)-coordinate.