Problem 29 - Thomas Calculus

Question

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, \boldsymbol{v})=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0}\) . Find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. Circular cylinder The circular cylinder \(\mathbf{r}(\theta, z)=(3 \sin 2 \theta) \mathbf{i}+\) \(\left(6 \sin ^{2} \theta\right) \mathbf{j}+z \mathbf{k}, 0 \leq \theta \leq \pi,\) at the point \(P_{0}(3 \sqrt{3} / 2,9 / 2,0)\) corresponding to \((\theta, z)=(\pi / 3,0)\) (See Example \(3 . )\)

Step-by-Step Solution

Answer

The tangent plane equation is \(x + y = 9\).

Step 1: Define the Parametrization and the Point

The surface is given by \[ \mathbf{r}(\theta, z) = (3 \sin 2\theta) \mathbf{i} + \left(6 \sin^2 \theta\right) \mathbf{j} + z \mathbf{k} \]The point \(P_0\) at which we want to find the tangent plane is given as \((3\sqrt{3}/2, 9/2, 0)\). The parameters corresponding to this point are \((\theta, z) = (\pi/3, 0)\).

Step 2: Calculate the Partial Derivatives

Calculate the partial derivatives of \(\mathbf{r}(\theta, z)\):\[ \mathbf{r}_{\theta} = \frac{d}{d\theta} \left[(3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k}\right] = (6 \cos 2\theta) \mathbf{i} + (12 \sin \theta \cos \theta) \mathbf{j} \]\[ \mathbf{r}_z = \frac{d}{dz} \left[(3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k}\right] = \mathbf{k} \]

Step 3: Evaluate the Partial Derivatives at the Point

Substitute \( (\theta, z) = (\pi/3, 0)\) to find the vectors \(\mathbf{r}_{\theta}\) and \(\mathbf{r}_z\):\[ \mathbf{r}_{\theta}(\pi/3, 0) = (6 \cos(2\pi/3)) \mathbf{i} + (12 \sin(\pi/3) \cos(\pi/3)) \mathbf{j} = -3 \mathbf{i} + 3 \mathbf{j} \]\[ \mathbf{r}_z(\pi/3, 0) = \mathbf{k} \]

Step 4: Compute the Normal Vector Using Cross Product

Find the cross product \(\mathbf{n} = \mathbf{r}_{\theta} \times \mathbf{r}_z\):\[ \mathbf{n} = \|-3 \mathbf{i} + 3 \mathbf{j} \| \times \| \mathbf{k} \|= \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 3 & 0 \ 0 & 0 & 1 \end{matrix} \right| \]Expanding the determinant gives:\[ \mathbf{n} = 3\mathbf{i} + 3\mathbf{j} \]

Step 5: Formulate the Equation of the Tangent Plane

The equation for the tangent plane through \(P_0(x_0, y_0, z_0)\) is given by \[ (x - x_0, y - y_0, z - z_0) \cdot \mathbf{n} = 0 \]Substitute \(P_0 = (3\sqrt{3}/2, 9/2, 0)\) and \(\mathbf{n} = 3\mathbf{i} + 3\mathbf{j}\) into the equation:\[ 3(x - 3\sqrt{3}/2) + 3(y - 9/2) = 0 \]This simplifies to:\[ x + y = 9 \]

Step 6: Derive the Cartesian Equation for the Surface

Using the given parametric form, express \(x\) and \(y\) in terms of \((\theta, z)\), and eliminate \(\theta\): \[ x = 3 \sin 2\theta \quad \text{and} \quad y = 6 \sin^2 \theta \]Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), convert: \[ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \sin \theta \sqrt{1 - \sin^2 \theta}\] Now, solve these to find a relationship between \(x\) and \(y\).

Step 7: Sketch the Surface and Tangent Plane

Sketch the surface by considering the range for \(\theta\) (\(0\) to \(\pi\)) and noting the repeating structure of sine functions. The plane \(x + y = 9\) can be drawn in 3D space intersecting the surface. It is planar and lies above the \(xy\)-plane.

Key Concepts

Parametrized Surface

Understanding a parametrized surface is crucial for dealing with surfaces in multivariable calculus. A parametrized surface is typically expressed as a vector function \( \mathbf{r}(u, v) \), where \( u \) and \( v \) are parameters that trace out points on the surface as they vary within a certain range. The function maps a region in the \( uv \)-plane to points in three-dimensional space.

For example, the surface given in the exercise is described by the vector function \( \mathbf{r}(\theta, z) = (3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k} \). Here, \( \theta \) and \( z \) act as parameters that span the surface as their values change.

Parametrized surfaces are powerful because they allow complex surfaces to be tackled using calculus tools, by breaking them down into manipulable components.

Partial Derivatives

Partial derivatives are a fundamental aspect of calculus applied to functions of multiple variables. They measure how a function changes as one parameter is varied while the others are held constant. In the context of parametrized surfaces, we calculate partial derivatives with respect to each parameter to understand the behavior of the surface.

In the given exercise, the partial derivatives \( \mathbf{r}_{\theta} \) and \( \mathbf{r}_z \) are calculated. For \( \mathbf{r}(\theta, z) \), we compute:

• \( \mathbf{r}_{\theta} = \frac{d}{d\theta}[(3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k}] = (6 \cos 2\theta) \mathbf{i} + (12 \sin \theta \cos \theta) \mathbf{j} \)
• \( \mathbf{r}_z = \frac{d}{dz}[(3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k}] = \mathbf{k} \)

These derivative vectors are essential in forming the tangent vectors of the surface, vital for subsequent calculations such as the cross product.

Cross Product

The cross product is a vector operation crucial for determining a normal vector to a surface. It involves taking two vectors in space and producing a third vector that is perpendicular to both. The magnitude of the cross product gives the area of the parallelogram spanned by the two input vectors.

In this exercise, the cross product \( \mathbf{n} = \mathbf{r}_{\theta} \times \mathbf{r}_z \) is used. This operation tells us the direction orthogonal to the surface at the given point. Calculating it involves working through this determinant:

\[ \mathbf{n} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 3 & 0 \ 0 & 0 & 1 \end{matrix} \right| \]
Upon expansion, the result is \( \mathbf{n} = 3\mathbf{i} + 3\mathbf{j} \). The cross product provides the necessary normal vector to define the equation of a tangent plane.

Normal Vector

A normal vector is essential for determining the tangent plane to a surface at a given point. It is the vector that is perpendicular to the plane, and it not only helps us ascertain the orientation of the plane but also plays a crucial role in formulating its equation.

In the exercise, the normal vector \( \mathbf{n} \) was found using the cross product of the tangent vectors \( \mathbf{r}_{\theta} \) and \( \mathbf{r}_z \). With \( \mathbf{n} = 3 \mathbf{i} + 3 \mathbf{j} \), we use this vector to define the tangent plane with the point \( P_0 \).

To form the plane's equation, use the dot product:

• The equation is \( \mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0 \), where \( \mathbf{r}_0 \) is the position vector of the point \( P_0 \).
• Substituting \( \mathbf{n} \) and \( P_0 = (3\sqrt{3}/2, 9/2, 0) \) gives us the equation of the tangent plane: \( x + y = 9 \).

This concise equation represents the geometry of the plane tangent to the surface at the specified point.