Problem 22 - 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)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(\sin 2 t) \mathbf{k}, \quad t_{0}=\frac{\pi}{2} \end{equation}

Step-by-Step Solution

Answer

The tangent line is given by \(x = -t, y = 1, z = -2t\).

Step 1: Evaluate the Curve at t_0

Find the position vector of the curve at \(t = t_0\), which is \(t_0 = \frac{\pi}{2}\). Plug this value into \(\mathbf{r}(t)\):\[\mathbf{r}\left(\frac{\pi}{2}\right) = \left(\cos\frac{\pi}{2}\right) \mathbf{i} + \left(\sin\frac{\pi}{2}\right) \mathbf{j} + \left(\sin 2\times\frac{\pi}{2}\right) \mathbf{k}\]This simplifies to:\[\mathbf{r}\left(\frac{\pi}{2}\right) = 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} = \mathbf{j}\]

Step 2: Find the Velocity Vector

Calculate the derivative of the curve \(\mathbf{r}(t)\) with respect to \(t\), which gives the velocity vector \(\mathbf{v}(t)\):\[\mathbf{v}(t) = \frac{d}{dt}\left((\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}\right)\]Solving, we find:\[\mathbf{v}(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (2 \cos 2t) \mathbf{k}\]Evaluate this at \(t_0 = \frac{\pi}{2}\):\[\mathbf{v}\left(\frac{\pi}{2}\right) = (-\sin\frac{\pi}{2}) \mathbf{i} + (\cos\frac{\pi}{2}) \mathbf{j} + (2 \cos \pi) \mathbf{k}\]Which simplifies to:\[\mathbf{v}\left(\frac{\pi}{2}\right) = -\mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} = -\mathbf{i} - 2 \mathbf{k}\]

Step 3: Write Parametric Equations for the Tangent Line

Using the point \(\mathbf{r}(\frac{\pi}{2}) = \mathbf{j}\) and the velocity vector \(\mathbf{v}(\frac{\pi}{2}) = -\mathbf{i} - 2\mathbf{k}\), the parametric equations of the tangent line at \(t_0\) are given by:\[x = 0 - t\y = 1\z = -2t\]This provides the complete set of parametric equations for the tangent line:\[x(t) = - t, \quad y(t) = 1, \quad z(t) = -2t\]

Key Concepts

Parametric Equations

Parametric equations describe a curve or geometrical object in space through equations that express the coordinates as continuous functions of one or more parameters. This technique is very useful to represent curves that are not easily described using simple functions. In the given problem, the curve is defined by the position vector \[ \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}, \]which involves trigonometric functions of a parameter \( t \). Here, each coordinate \((x, y, z)\) is represented as a function of \( t \):

• \( x(t) = \cos t \)
• \( y(t) = \sin t \)
• \( z(t) = \sin 2t \)

At a specific point along the curve, such as when \( t = \frac{\pi}{2} \), these parametric equations allow us to easily determine the corresponding coordinates of the point: \( (0, 1, 0) \). This is because at \( \frac{\pi}{2} \), \( \cos \frac{\pi}{2} = 0 \), \( \sin \frac{\pi}{2} = 1 \), and \( \sin \pi = 0 \). Parametric equations provide a powerful way to describe complex curves and enable visualization of paths and objects in space.

Velocity Vector

The velocity vector is the derivative of the position vector with respect to time. It indicates the rate of change of the position vector, giving insight into both the speed and direction of motion at any given point. In this problem, the curve's velocity vector is derived from the position vector:\[ \mathbf{v}(t) = \frac{d}{dt}\big((\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}\big). \]This results in \[ \mathbf{v}(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (2 \cos 2t) \mathbf{k}. \]When evaluating this at \( t = \frac{\pi}{2} \), you find:

• \( \mathbf{v}\left(\frac{\pi}{2}\right) = -\mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} \)

This expresses that, at \( t = \frac{\pi}{2} \), the direction of motion is in the negative \( x \)-axis and \( z \)-axis at rates of \(-1\) and \(-2\) respectively. Therefore, the curve is not changing in the \( y \)-direction at this specific point. The velocity vector effectively conveys how the curve progresses through space.

Differentiation

Differentiation is a fundamental concept in calculus that measures the rate at which a function changes at any point. It helps us find tangents, velocities, and much more, by analyzing how small changes in the input impact the output. In the context of curves described by parametric equations, differentiation allows us to derive the velocity vector. Given \[ \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}, \]calculating \( \frac{d}{dt} \mathbf{r}(t) \) gives:

• \( \frac{d}{dt}(\cos t) = -\sin t \)
• \( \frac{d}{dt}(\sin t) = \cos t \)
• \( \frac{d}{dt}(\sin 2t) = 2\cos 2t \)

Through differentiation, we can efficiently derive that \[ \mathbf{v}(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (2 \cos 2t) \mathbf{k}. \]Differentiation not only helps in finding the velocity vector but also plays a crucial role in defining tangent lines to curves at particular points. It is a versatile mathematical tool essential for understanding changes in functions, whether they relate to motion, slopes, or other types of rates of change.