Problem 36 - 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 \(t=t_{0}\) . $$ \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(\sin 2 t) \mathbf{k}, \quad t_{0}=\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The parametric equations are \(x = -s\), \(y = 1\), \(z = -2s\).

Step 1: Find the Velocity Vector

First, find the velocity vector \( \mathbf{v}(t) \) by differentiating \( \mathbf{r}(t) \) with respect to \( t \), which gives the components for \( \mathbf{i}, \mathbf{j}, \text{and} \mathbf{k} \):\[\mathbf{v}(t) = \frac{d}{dt}[(\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}] = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (2 \cos 2t) \mathbf{k}\]

Step 2: Evaluate the Velocity Vector at t=t0

Substitute \( t = \frac{\pi}{2} \) into the velocity vector \( \mathbf{v}(t) \):\[\mathbf{v}\left(\frac{\pi}{2}\right) = (-\sin \frac{\pi}{2}) \mathbf{i} + (\cos \frac{\pi}{2}) \mathbf{j} + (2 \cos \pi) \mathbf{k} = -\mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} = -\mathbf{i} - 2 \mathbf{k}\]

Step 3: Find the Point on the Curve at t=t0

Calculate \( \mathbf{r}\left(\frac{\pi}{2}\right) \) to find the point on the curve:\[\mathbf{r}\left(\frac{\pi}{2}\right) = (\cos \frac{\pi}{2}) \mathbf{i} + (\sin \frac{\pi}{2}) \mathbf{j} + (\sin \pi) \mathbf{k} = 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} = \mathbf{j}\]

Step 4: Write the Parametric Equations for the Tangent Line

The parametric equations for the tangent line use the point from Step 3 and the velocity vector from Step 2. The general formula is \((x, y, z) = (x_0, y_0, z_0) + t(v_x, v_y, v_z)\), which translates to:\[(x, y, z) = (0, 1, 0) + s(-1, 0, -2)\]Breaking this down into parametric equations:- \(x(s) = -s\)- \(y(s) = 1\)- \(z(s) = -2s\)

Key Concepts

Velocity Vector

Let's dive into the concept of velocity vectors, especially in the context of parametric equations. A velocity vector describes the direction and speed of a moving object along a path defined by parametric equations. In this exercise, the curve is given by \[ \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(\sin 2t) \mathbf{k} \]
To find the velocity vector \( \mathbf{v}(t) \), we need to differentiate the given curve \( \mathbf{r}(t) \) with respect to \( t \). This is a critical part of studying motion along curves, as it tells us how the curve progresses over time.

In this instance, differentiating results in the velocity vector:\[ \mathbf{v}(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (2 \cos 2t) \mathbf{k} \]At \( t = \frac{\pi}{2} \), substituting back into the velocity vector provides us with the specific vector \( \mathbf{v}\left(\frac{\pi}{2}\right) = -\mathbf{i} - 2 \mathbf{k} \). This means that at \( t = \frac{\pi}{2} \), the curve is moving in the negative \( \mathbf{i} \) and \( \mathbf{k} \) direction with no change in the \( \mathbf{j} \) component.

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it nearby, effectively "hugging" the curve at that point. For a smooth curve described by parametric equations, such as our given curve \( \mathbf{r}(t) \), the tangent line at a specific parameter \( t=t_0 \) passes through the point on the curve at \( t_0 \) and is parallel to the velocity vector at \( t_0 \).

In the case of this exercise, once the velocity vector \( \mathbf{v}\left(\frac{\pi}{2}\right) \) is determined, we pair this with the point on the curve \( \mathbf{r}\left(\frac{\pi}{2}\right) \) to form the equation of the tangent line.

The parametric equations for the tangent line become:

• \( x(s) = -s \)
• \( y(s) = 1 \)
• \( z(s) = -2s \)

This tells us that at \( t = \frac{\pi}{2} \), the tangent line dips in the \( x \) and \( z \) directions while maintaining a constant \( y \). It epitomizes the tangency by occupying the same space as the curve just at this point.

Differentiation

Differentiation is a powerful mathematical tool used to compute the rate at which one quantity changes with respect to another. In the context of curves and parametric equations, differentiation helps us derive the velocity vector from the position vector. This is crucial because velocity vectors indicate not only the path's direction at any point but also how fast it's changing.

When differentiating the curve \( \mathbf{r}(t) \) in this exercise:\[ \frac{d}{dt}[(\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}] = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (2 \cos 2t) \mathbf{k} \]
we arrive at \( \mathbf{v}(t) \), the crucial velocity vector. This process of differentiation converts our understanding from a static position at \( t \) to a dynamic interpretation of movement. It captures the essence of how the curve naturally unfolds with time.

Curves

Curves in mathematics define elegant, flowing paths that can be described using functions or parametric equations. These curves can exist in various dimensions, and their shapes can range from simple lines to complex spirals.

In this exercise, the curve is given by the function\[ \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2 t) \mathbf{k} \]This vector function outputs a point in a 3D space for every value of \( t \), effectively painting a path. At any given \( t \), the position vector \( \mathbf{r}(t) \) defines the exact point on this curve.

Understanding these aspects of curves is essential for many applications, from engineering to physics. Each point on a curve connects seamlessly to the next, leading to practical uses such as modeling trajectories or creating animations. The smooth nature of curves represented by a parameter like \( t \) allows for sophisticated analysis and can embed complex information, including changes over time.