Problem 143 - Calculus 3

Question

Find the equation for the osculating plane at point \(t=\pi / 4\) on the curve \(\mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}+t\)

Step-by-Step Solution

Answer

Osculating plane equation: \(x + y + z = 1 + \pi/4\).

Step 1: Calculate the First Derivative

Differentiate the curve function \(\mathbf{r}(t) = \cos(2t)\mathbf{i} + \sin(2t)\mathbf{j} + t\mathbf{k}\) with respect to \(t\). The first derivative is \(\mathbf{r}'(t) = -2\sin(2t)\mathbf{i} + 2\cos(2t)\mathbf{j} + \mathbf{k}\).

Step 2: Calculate the Second Derivative

Differentiate the first derivative, \(\mathbf{r}'(t)\), to find the second derivative. \(\mathbf{r}''(t) = -4\cos(2t)\mathbf{i} - 4\sin(2t)\mathbf{j}\).

Step 3: Evaluate Derivatives at \(t = \pi/4\)

Evaluate \(\mathbf{r}'(\pi/4)\) and \(\mathbf{r}''(\pi/4)\). \(\mathbf{r}'(\pi/4) = -2\sin(\pi/2)\mathbf{i} + 2\cos(\pi/2)\mathbf{j} + \mathbf{k} = -2\mathbf{i} + \mathbf{k}\) and \(\mathbf{r}''(\pi/4) = -4\cos(\pi/2)\mathbf{i} - 4\sin(\pi/2)\mathbf{j} = -4\mathbf{j}\).

Step 4: Compute the Normal Vector to the Osculating Plane

Compute the cross product of \(\mathbf{r}'(\pi/4)\) and \(\mathbf{r}''(\pi/4)\) to find the normal vector. \(\mathbf{n} = \mathbf{r}'(\pi/4) \times \mathbf{r}''(\pi/4) = (-2\mathbf{i} + \mathbf{k}) \times (-4\mathbf{j}) = 8\mathbf{i} + 8\mathbf{j} + 8\mathbf{k}\).

Step 5: Find the Point on the Curve

Substitute \(t = \pi/4\) into \(\mathbf{r}(t)\) to find the point on the curve. \(\mathbf{r}(\pi/4) = \cos(\pi/2)\mathbf{i} + \sin(\pi/2)\mathbf{j} + (\pi/4)\mathbf{k} = \mathbf{j} + (\pi/4)\mathbf{k}\).

Step 6: Write the Equation of the Osculating Plane

The equation of the plane can be found using the normal vector \(\mathbf{n} = 8\mathbf{i} + 8\mathbf{j} + 8\mathbf{k}\) and the point \(\mathbf{r}(\pi/4)\). The equation is \(8(x - 0) + 8(y - 1) + 8(z - \pi/4) = 0\). Simplify to get \(x + y + z = 1 + \pi/4\).

Key Concepts

Parametric Curves

A parametric curve represents a set of points in space, determined by one or more parameters. In mathematics, these curves are defined by parametric equations where each spatial dimension is expressed as a function of one or more independent parameters.
For example, the curve given by \[\mathbf{r}(t) = \cos(2t)\mathbf{i} + \sin(2t)\mathbf{j} + t\mathbf{k}\] defines a path in 3D space parameterized by the variable \(t\). Here:

• \(\textbf{i}\), \(\textbf{j}\), and \(\textbf{k}\) are unit vectors in the x, y, and z directions respectively.
• The parameter \(t\) typically represents time.
• As \(t\) varies, the evaluated expressions for \(\cos(2t)\), \(\sin(2t)\), and \(t\) form a trajectory or path in space.

Understanding parametric curves is essential when analyzing the shape and behavior of complex functions, as they offer a precise way of describing motion and the geometry of paths.

Cross Product

The cross product is a fundamental operation between two vectors in three-dimensional space. It results in a third vector that is perpendicular to the plane containing the original vectors. This perpendicular vector is often critical in defining planes, as it serves as a normal vector.
Mathematically, if you have two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), their cross product \(\mathbf{a} \times \mathbf{b}\) is given by:\[\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}\]

• The direction of the cross product is determined by the right-hand rule.
• This operation is particularly useful in finding the normal vector to a plane, as seen when calculating the normal vector for the osculating plane.
• In the osculating plane problem, the cross product helps identify the normal vector needed to define the plane's equation.

By utilizing the cross product, we gain a tool for exploring orientations and defining geometric relationships in three-dimensional space.

Normal Vector

A normal vector is a vector that is perpendicular to a given surface or curve at a specific point. It plays a crucial role in defining geometric shapes and understanding orientations in space. In the context of curves, a normal vector provides insight into how the curve is oriented at each point along its path.
The normal vector to the osculating plane is found by computing the cross product of the first and second derivatives of the parametric curve. This is because:

• The first derivative \(\mathbf{r}'(t)\) represents the tangent vector to the curve.
• The second derivative \(\mathbf{r}''(t)\) provides a measure of how the curve is bending.
• The cross product \(\mathbf{r}'(t) \times \mathbf{r}''(t)\) gives a vector that is perpendicular to the plane made by \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\).

In the problem, the normal vector \(\mathbf{n} = 8\mathbf{i} + 8\mathbf{j} + 8\mathbf{k}\) defines the orientation of the osculating plane at \(t = \pi/4\). Having the normal vector allows us to write an equation for the plane, driving home its importance in understanding the three-dimensional behavior of curves.

Derivatives of Vector Functions

Derivatives of vector functions extend the concept of differentiation to functions that output vectors rather than numbers. This is essential in physics and engineering, where quantities such as position, velocity, and acceleration are expressed as vectors.
To find these derivatives, we differentiate each component of the vector individually with respect to the parameter. Given a parametric vector function:\[\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}\]The derivative \(\mathbf{r}'(t)\) is computed as:\[\mathbf{r}'(t) = \frac{df}{dt}\mathbf{i} + \frac{dg}{dt}\mathbf{j} + \frac{dh}{dt}\mathbf{k}\]These derivatives are foundational in:

• Assessing the velocity and acceleration of objects in motion.
• Defining tangents and norms to curves, crucial for understanding the behavior at specific points.
• Calculating further geometric attributes like curvatures and torsions.

In the context of the osculating plane, both the first and second derivatives of the curve function \(\mathbf{r}(t)\) help determine its immediate orientation and future path. This makes derivatives of vector functions a powerful tool in the study of parametric curves.