Problem 26 - 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 \(23-26\) , 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 for the tangent line are: \( x = -t, y = 1, z = -2t \).

Step 1: Compute the Point on the Curve

First, evaluate the curve \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k} \) at \( t_0 = \frac{\pi}{2} \).\Substituting \( t = \frac{\pi}{2} \) into \( \mathbf{r}(t) \), we get:\[\mathbf{r}\left(\frac{\pi}{2}\right) = \left( \cos\left(\frac{\pi}{2}\right) \mathbf{i} + \sin\left(\frac{\pi}{2}\right) \mathbf{j} + \sin\left( 2 \times \frac{\pi}{2}\right) \mathbf{k} \right) = 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} = \mathbf{j}\]

Step 2: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \).Find \( \mathbf{v}(t) = \frac{d}{dt}\left((\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (\sin 2t) \mathbf{k}\right) \).This gives:\[\mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + 2\cos(2t) \mathbf{k}\]Evaluate this at \( t_0 = \frac{\pi}{2} \):\[\mathbf{v}\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) \mathbf{i} + \cos\left(\frac{\pi}{2}\right) \mathbf{j} + 2\cos\left(\pi\right) \mathbf{k} = -\mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} = -\mathbf{i} - 2\mathbf{k}\]

Step 3: Write the Parametric Equations

Now that we have the point and the direction vector, we can write the parametric equations of the tangent line. The equations are:\[x(t) = 0 - t = -t, \]\[y(t) = 1,\]\[z(t) = 0 - 2t = -2t\]This gives the parametric equations of the tangent line:\\[x = -t, \y = 1, \z = -2t\]

Key Concepts

Parametric Equations

Parametric equations are used to represent the coordinates of points that make up a geometric shape, like a line or a curve, with the use of one or more variables. In our context, we deal with curves that are defined by three parametric equations related to a parameter, usually denoted as \( t \). Each equation describes how a coordinate point changes as the parameter changes.
For example, a curve in space can be expressed as \( \mathbf{r}(t) = (x(t), y(t), z(t)) \). As \( t \) varies, we sweep through different points on the curve. This way, parametric equations provide a powerful tool for defining more complex shapes than typical Cartesian coordinates.

• Advantages: They offer great flexibility, especially when dealing with curves that loop or twist.
• Utilization: Commonly used in physics and engineering to represent trajectories.

Velocity Vector

The velocity vector is a fundamental concept when analyzing motion along curves. It is essentially the derivative of the position vector with respect to time. In calculus terms, if the position of an object is given by \( \mathbf{r}(t) \), then its velocity vector \( \mathbf{v}(t) \) is \( \frac{d\mathbf{r}}{dt} \).
To physically interpret it, think of the velocity vector as showing the direction and speed at which a point moves along a curve. In our problem, we compute \( \mathbf{v}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + 2\cos(2t) \mathbf{k} \). Evaluating this vector at \( t_0 = \frac{\pi}{2} \) gives the directional component of the tangent line.

• Function: Determines the line's direction at the curve.
• Use in Tangents: Tangent lines at a point on a curve are parallel to this velocity vector.

Smooth Curve

A smooth curve, in the context of parametric equations, refers to a curve that is continuously differentiable and doesn’t have sharp corners or cusps. This means the derivatives exist and are continuous for all points on the curve.
Smoothness is important when defining tangent lines because it ensures that a unique tangent can be drawn at every point of the curve. For the given parametric equations, if the velocity vector is non-zero for all \( t \), the curve is typically smooth.

• Properties: Continuous derivatives ensure a well-defined tangent and curvature at every point.
• Applications: Critical for ensuring stability in engineering designs and computer graphics.

Calculus

Calculus is the mathematical study of continuous change, and it is vital for understanding the intricacies of curves and motion as described by parametric equations. Fundamental techniques of calculus, such as differentiation and integration, are essential tools.
When it comes to tangent lines, differentiation plays a critical role. By differentiating the position vector \( \mathbf{r}(t) \), we obtain the velocity vector which then helps in defining the tangent line.
Calculus not only helps in understanding how objects move but also provides insights into changes over time, rates of change, and spatial dynamics.

• Key Concepts: Derivatives for identifying tangent directions, integrals for calculating areas under curves.
• Relevance: Essential in fields ranging from physics to economics where modeling change is crucial.