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 \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) 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 \(P_{0} .\) In Exercises \(27-30,\) 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. $$ \begin{array}{l}{\text { Circular cylinder } \text { The circular cylinder } \mathbf{r}(\theta, z)=(3 \sin 2 \theta) \mathbf{i}+} \\ {\left(6 \sin ^{2} \theta\right) \mathbf{j}+z \mathbf{k}, 0 \leq \theta \leq \pi, \text { at the point } P_{0}(3 \sqrt{3} / 2,9 / 2,0)} \\ {\text { corresponding to }(\theta, z)=(\pi / 3,0) \text { (See Example } 3 . )}\end{array} $$

Step-by-Step Solution

Answer

Tangent plane equation: \(3\sqrt{3}x + 3y = 27\). Surface equation: \(x^2 + 3y = 36\).

Step 1: Parametrize the Surface

The parametrized surface given is \( \mathbf{r}(\theta, z) = (3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k} \). We want to find the tangent plane at the point \( P_0(3\sqrt{3}/2, 9/2, 0) \).

Step 2: Calculate Partial Derivatives

Find the partial derivatives of \( \mathbf{r}(\theta, z) \) with respect to \( \theta \) and \( z \). The derivative with respect to \( \theta \) is: \[ \mathbf{r}_\theta = \frac{\partial}{\partial\theta} (3 \sin 2\theta) \mathbf{i} + \frac{\partial}{\partial\theta} (6 \sin^2 \theta) \mathbf{j} + 0 \mathbf{k} \] \[ = 6 \cos 2\theta \mathbf{i} + 12 \sin \theta \cos \theta \mathbf{j} \] The derivative with respect to \( z \) is: \[ \mathbf{r}_z = \frac{\partial}{\partial z} (3 \sin 2\theta) \mathbf{i} + \frac{\partial}{\partial z} (6 \sin^2 \theta) \mathbf{j} + 1 \mathbf{k} = \mathbf{k} \]

Step 3: Evaluate Derivatives at Given Point

Evaluate \( \mathbf{r}_\theta \) and \( \mathbf{r}_z \) at \( (\theta, z) = (\pi/3, 0) \). This gives: \[ \mathbf{r}_\theta (\pi/3, 0) = 6 \cos(2\pi/3) \mathbf{i} + 12 \sin(\pi/3)\cos(\pi/3) \mathbf{j} \] \[ = -3 \mathbf{i} + 3\sqrt{3} \mathbf{j} \] and \[ \mathbf{r}_z (\pi/3, 0) = \mathbf{k} \]

Step 4: Calculate the Cross Product

Find the cross product \( \mathbf{r}_\theta(\pi/3, 0) \times \mathbf{r}_z(\pi/3, 0) \): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 3\sqrt{3} & 0 \ 0 & 0 & 1 \end{vmatrix} = \mathbf{i}(3\sqrt{3}) - \mathbf{j}(-3) + \mathbf{k}(0) = 3\sqrt{3} \mathbf{i} + 3 \mathbf{j} \]

Step 5: Formulate Equation of Tangent Plane

The equation of the tangent plane is given by the normal vector \( (3\sqrt{3}, 3, 0) \) and point \( (3\sqrt{3}/2, 9/2, 0) \). The equation is: \[ 3\sqrt{3}(x - 3\sqrt{3}/2) + 3(y - 9/2) + 0(z - 0) = 0 \] Simplifying gives: \[ 3\sqrt{3}x + 3y = 27 \]

Step 6: Convert to Cartesian Coordinates

The circular cylinder in Cartesian form can be expressed as follows. Set \( x = 3 \sin 2\theta \) and \( y = 6 \sin^2 \theta \). Therefore, \( x^2 = 9 \sin^2 2\theta = 9(1 - \cos^2 2\theta) \) and \( y = 3(1 - \cos 2\theta) \). Combining the identities, \( y = 3 - 3\cos 2\theta \), leads to: \[ 36 - 3y = x^2 \] or \[ x^2 + 3y = 36 \]

Step 7: Sketching the Surface and Tangent Plane

The Cartesian equation of the surface \( x^2 + 3y = 36 \) describes a circular cylinder. The tangent plane equation is \(3\sqrt{3}x + 3y = 27\). You can sketch this by first plotting the curve of the cylinder on the xy-plane and then showing how the plane intersects perpendicularly with it, touching at exactly one point, \( (3\sqrt{3}/2, 9/2) \).

Key Concepts

Parametrized Surfaces

Parametrized surfaces are a way to describe two-dimensional surfaces in three-dimensional space, using parameters. These parameters are usually denoted as variables, like \( u \) and \( v \) or in the exercise \( \theta \) and \( z \). In the formula \( \mathbf{r}(u, v) = f(u, v) \mathbf{i} + g(u, v) \mathbf{j} + h(u, v) \mathbf{k} \), each coordinate is expressed as a function of these parameters.
This way, the surface becomes a continuous map over a region of the \( uv \)-plane, helping us handle more complex surfaces easily. Parametrized surfaces make it simpler to calculate things such as areas, normal vectors, and curvature.

• Allows easy description of surfaces that aren't easily expressed as a single equation in Cartesian coordinates.
• Helps visualize the surface in three-dimensional space by relating the parameter space to the physical space.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions with multiple variables. They measure the rate of change of a function when only one variable is allowed to change while others are held constant.
In the context of parametrized surfaces, partial derivatives of the position vector \( \mathbf{r}(u, v) \) with respect to the parameters give us tangent vectors. These indicate the direction of the surface's slope concerning a particular parameter.
For the given surface \( \mathbf{r}(\theta, z) = (3 \sin 2\theta) \mathbf{i} + (6 \sin^2 \theta) \mathbf{j} + z \mathbf{k} \), the partial derivative with respect to \( \theta \), \( \mathbf{r}_\theta \), represents how the surface bends or twists when \( \theta \) changes, while \( \mathbf{r}_z \) shows the change along the axis defined by incrementing \( z \).

• Essential for finding tangent planes as they provide the slopes along each parameter axis.
• Used to compute the cross product leading to the normal vector of the plane.

Cross Product

The cross product is a mathematical operation used to find a vector perpendicular to two other vectors in three-dimensional space. It's essential in calculus when we need to determine a plane's orientation.
Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), their cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector perpendicular to both. It's vital for computing the normal vector to a surface, which is crucial for defining tangent planes.
In the exercise problem, the cross product of the partial derivatives \( \mathbf{r}_\theta \) and \( \mathbf{r}_z \) gives a normal vector \( 3\sqrt{3} \mathbf{i} + 3 \mathbf{j} \), which is perpendicular to the surface at the point in question. This ensures that the tangent plane is correctly oriented.

• Provides a vector perpendicular to the tangent plane, crucial for its equation formation.
• Helps understanding surfaces' spatial properties in physics and engineering.

Cylinder in Cartesian Coordinates

Describing a cylinder in Cartesian coordinates starts by understanding its axis and radius. Cylinders are surfaces created by moving a circle along a line, and they have distinct equations in different coordinate forms.
The exercise deals with a circular cylinder, meaning its cross-sections are circles. By converting the parametrized form \( \mathbf{r}(\theta, z) \) to a Cartesian form, we derive an equation \( x^2 + 3y = 36 \). This represents the cylinder without needing the \( \theta \) and \( z \) parameters.
In this form, we can see that the variable \( x \) is related to \( \theta \) through trigonometric functions, and \( y \) is connected via calculations involving these functions. Removing \( \theta \) gives us a simple algebraic expression that defines the actual geometry in ordinary space.

• Transforms complex parametrized descriptions into simpler, visualizable equations.
• Allows for integration, differentiation, and graphical representation without parameter-specific complexity.