Problem 143 - Calculus Volume 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

The osculating plane equation is \(4x + 8z - 2\pi = 0\).

Step 1: Find the Point on the Curve

To find the point on the curve at the given parameter value \( t = \pi/4 \), substitute \( t = \pi/4 \) into the curve equation \( \mathbf{r}(t) = \cos(2t)\mathbf{i} + \sin(2t)\mathbf{j} + t\mathbf{k} \). This gives:\[\mathbf{r}(\pi/4) = \cos(\pi/2)\mathbf{i} + \sin(\pi/2)\mathbf{j} + \frac{\pi}{4}\mathbf{k} = 0\mathbf{i} + 1\mathbf{j} + \frac{\pi}{4}\mathbf{k}.\]So, the point is \( (0, 1, \frac{\pi}{4}) \).

Step 2: Compute the Tangent Vector

The tangent vector is obtained by differentiating \( \mathbf{r}(t) \) with respect to \( t \). We have:\[\mathbf{r}'(t) = \frac{d}{dt}(\cos(2t)\mathbf{i} + \sin(2t)\mathbf{j} + t\mathbf{k}) = -2\sin(2t)\mathbf{i} + 2\cos(2t)\mathbf{j} + \mathbf{k}.\]Substituting \( t = \pi/4 \), we get:\[\mathbf{r}'(\pi/4) = -2\sin(\pi/2)\mathbf{i} + 2\cos(\pi/2)\mathbf{j} + \mathbf{k} = -2\mathbf{i} + 0\mathbf{j} + \mathbf{k}.\]

Step 3: Compute the Principal Normal Vector

The principal normal vector is found by computing the derivative of the tangent vector \( \mathbf{r}''(t) \) and then evaluating it at \( t = \pi/4 \). First, compute:\[\mathbf{r}''(t) = \frac{d}{dt}(-2\sin(2t)\mathbf{i} + 2\cos(2t)\mathbf{j} + \mathbf{k}) = -4\cos(2t)\mathbf{i} - 4\sin(2t)\mathbf{j}.\]At \( t = \pi/4 \), this becomes:\[\mathbf{r}''(\pi/4) = -4\cos(\pi/2)\mathbf{i} - 4\sin(\pi/2)\mathbf{j} = 0\mathbf{i} - 4\mathbf{j}.\]

Step 4: Find the Normal Vector to the Osculating Plane

The normal vector to the osculating plane is the cross product of the tangent vector \( \mathbf{r}'(\pi/4) \) and the principal normal vector \( \mathbf{r}''(\pi/4) \). Compute:\[\mathbf{n} = \mathbf{r}'(\pi/4) \times \mathbf{r}''(\pi/4) = (-2\mathbf{i} + \mathbf{k}) \times (-4\mathbf{j}) = i(-4 \cdot 0 - 1 \cdot -4) - j(-2 \cdot 0 - 0 \cdot -4) + k(-2 \cdot -4 - 0 \cdot 0).\]Simplifying gives:\[\mathbf{n} = 4\mathbf{i} + 8\mathbf{k}.\]

Step 5: Write the Equation of the Osculating Plane

The equation of the osculating plane can be expressed as:\[4(x - 0) + 0(y - 1) + 8\left(z - \frac{\pi}{4}\right) = 0.\]Simplifying, the equation is:\[4x + 8z - 2\pi = 0.\]

Key Concepts

Tangent Vector

When studying curves, the tangent vector is crucial for understanding the direction of a curve at any given point. Imagine holding a rope at any point along its path; the tangent vector would indicate the direction the rope is naturally pointing at that spot.

To find the tangent vector for a curve represented by a vector function \( \mathbf{r}(t) \), differentiate the function with respect to \( t \). In this case, the tangent vector \( \mathbf{r}'(t) \) at \( t = \pi/4 \) becomes \(-2\mathbf{i} + \mathbf{k} \), aligning with the instantaneous direction of the curve.

This vector provides valuable information about the curve's behavior and how it moves through space.

Principal Normal Vector

The principal normal vector helps us understand how a curve is bending at a specific point. It is perpendicular to the tangent vector and indicates the direction towards the center of curvature.

• To find this vector, further differentiate the tangent vector with respect to \( t \).
• For our curve, the principal normal vector at \( t = \pi/4 \) is \(0\mathbf{i} - 4\mathbf{j} \).

These calculations reveal how sharply the curve is turning at that point. The principal normal vector is essential in defining the osculating or the best-fit plane around the curve.

Cross Product

The cross product is a handy operation in vector algebra, particularly when you want to find a vector perpendicular to two given vectors in three-dimensional space.

For the osculating plane exercise, we used the cross product to determine the normal vector. It involved the tangent vector \((-2\mathbf{i} + \mathbf{k})\) and the principal normal vector \((0\mathbf{i} - 4\mathbf{j})\).

Computing the cross product provides \(4\mathbf{i} + 8\mathbf{k} \), a vector orthogonal to both the curves directions, which serves as the normal vector to the osculating plane. This product is pivotal in establishing the plane's orientation in space.

Equation of a Plane

The equation of a plane in three-dimensional space can be derived once a point on the plane and a normal vector to it are known.

• The normal vector, \(4\mathbf{i} + 8\mathbf{k} \), obtained from the cross product, is vital.
• The point on the osculating plane at \(t=\pi/4\) is \((0, 1, \pi/4)\).

Using these, we form the standard equation of the plane: \[4(x - 0) + 8(z - \frac{\pi}{4}) = 0\].

Simplifying yields \(4x + 8z - 2\pi = 0\), capturing the exact osculating plane at \(t = \pi/4\). This equation defines the plane that closely hugs the curve at that specific point.