Problem 30
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. $$ \begin{array}{l}{\text { Parabolic cylinder The parabolic cylinder surface } \mathbf{r}(x, y)=} \\ {x \mathbf{i}+y \mathbf{j}-x^{2} \mathbf{k},-\infty < x <\infty,-\infty < y < \infty, \text { at the point }} \\ {P_{0}(1,2,-1) \text { corresponding to }(x, y)=(1,2)}\end{array} $$
Step-by-Step Solution
Verified Answer
The tangent plane equation is \(2x + z = 1\); the surface is \(z = -x^2\).
1Step 1: Calculate Partial Derivatives
First, we need to find the partial derivatives \( \mathbf{r}_x \) and \( \mathbf{r}_y \) of the vector function \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k} \). This gives us: \( \mathbf{r}_x(x, y) = \frac{\partial}{\partial x}(x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k}) = \mathbf{i} - 2x \mathbf{k} \) and \( \mathbf{r}_y(x, y) = \frac{\partial}{\partial y}(x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k}) = \mathbf{j} \).
2Step 2: Evaluate at Given Point
Substitute \((x_0, y_0) = (1, 2)\) into \( \mathbf{r}_x \) and \( \mathbf{r}_y \) to find their values at the point \( P_0 \). We get \( \mathbf{r}_x(1, 2) = \mathbf{i} - 2 \cdot 1 \mathbf{k} = \mathbf{i} - 2 \mathbf{k} \) and \( \mathbf{r}_y(1, 2) = \mathbf{j} \).
3Step 3: Compute Cross Product
Calculate the cross product \( \mathbf{r}_x \times \mathbf{r}_y \). Simplifying, \( (\mathbf{i} - 2 \mathbf{k}) \times \mathbf{j} = \mathbf{i} \times \mathbf{j} - 2 \mathbf{k} \times \mathbf{j} = \mathbf{k} + 2 \mathbf{i} \). The resulting normal vector is \( 2 \mathbf{i} + \mathbf{k} \).
4Step 4: Write Equation of Tangent Plane
The equation of the tangent plane at \( P_0 (1, 2, -1) \) can be written as \( 2(x - 1) + 0(y - 2) + 1(z + 1) = 0 \). Simplifying, we get \( 2x + z = 1 \).
5Step 5: Cartesian Equation for Surface
The Cartesian equation for the surface from \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k} \) is \( z = -x^2 \). This describes a parabolic cylinder extending along the y-axis.
6Step 6: Sketch the Surface and Tangent Plane
Sketch the parabolic cylinder using the equation \( z = -x^2 \) and sketch the tangent plane given by \( 2x + z = 1 \) to illustrate that it touches the surface at the point (1, 2, -1).
Key Concepts
Parametrized SurfacePartial DerivativesCross ProductNormal VectorCartesian Equation
Parametrized Surface
A parametrized surface is a mapping that sets two parameters, usually denoted as \( u \) and \( v \), to describe a surface in three-dimensional space. Think of it as a way to represent surfaces where each point is determined by a pair of values. This is particularly helpful because it allows us to define complex surfaces easily.
In the given exercise, we have the surface described by the vector function \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k} \). Here, \( x \) and \( y \) act as the parameters, and the surface equation forms a parabolic cylinder.
Understanding parametrized surfaces is crucial because they lay the groundwork for more complex concepts like tangent planes and normal vectors.
In the given exercise, we have the surface described by the vector function \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k} \). Here, \( x \) and \( y \) act as the parameters, and the surface equation forms a parabolic cylinder.
Understanding parametrized surfaces is crucial because they lay the groundwork for more complex concepts like tangent planes and normal vectors.
Partial Derivatives
Partial derivatives are used when you need to calculate the rate of change of a function with respect to one variable, while holding the other variables constant. It’s like peeling layers of an onion to see how one ingredient affects the whole blend.
In this exercise, we calculate the partial derivatives of \( \mathbf{r}(x, y) \) with respect to \( x \) and \( y \):
These derivatives give us the tangent vectors at any given point. They are essential for understanding how changes in \( x \) or \( y \) direction affect the surface.
In this exercise, we calculate the partial derivatives of \( \mathbf{r}(x, y) \) with respect to \( x \) and \( y \):
- \( \mathbf{r}_x(x, y) = \frac{\partial}{\partial x}(x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k}) = \mathbf{i} - 2x \mathbf{k} \)
- \( \mathbf{r}_y(x, y) = \frac{\partial}{\partial y}(x \mathbf{i} + y \mathbf{j} - x^2 \mathbf{k}) = \mathbf{j} \)
These derivatives give us the tangent vectors at any given point. They are essential for understanding how changes in \( x \) or \( y \) direction affect the surface.
Cross Product
The cross product is a vector operation that outputs a vector perpendicular to two input vectors. This is key in finding normal vectors, which are essential for determining tangent planes.
For our exercise, we compute the cross product of \( \mathbf{r}_x \) and \( \mathbf{r}_y \), the partial derivatives:
The cross product provides us with the normal vector \( 2 \mathbf{i} + \mathbf{k} \), which is crucial for the next steps, such as writing the equation of the tangent plane.
For our exercise, we compute the cross product of \( \mathbf{r}_x \) and \( \mathbf{r}_y \), the partial derivatives:
- \( (\mathbf{i} - 2 \mathbf{k}) \times \mathbf{j} = \mathbf{i} \times \mathbf{j} - 2 \mathbf{k} \times \mathbf{j} = \mathbf{k} + 2 \mathbf{i} \)
The cross product provides us with the normal vector \( 2 \mathbf{i} + \mathbf{k} \), which is crucial for the next steps, such as writing the equation of the tangent plane.
Normal Vector
A normal vector is a vector that is perpendicular to a surface at a given point. It is essential in defining planes, especially when you need to find a tangent plane to a surface.
In this exercise, the normal vector is derived from the cross product \( \mathbf{r}_x \times \mathbf{r}_y \). We determined that the normal vector is \( 2 \mathbf{i} + \mathbf{k} \).
This normal vector aids in writing the equation of the tangent plane, capturing the direction that is perpendicular to the surface at point \( P_0 (1, 2, -1) \). Without it, we wouldn't be able to ensure that the plane is tangent to the surface at the specific point needed.
In this exercise, the normal vector is derived from the cross product \( \mathbf{r}_x \times \mathbf{r}_y \). We determined that the normal vector is \( 2 \mathbf{i} + \mathbf{k} \).
This normal vector aids in writing the equation of the tangent plane, capturing the direction that is perpendicular to the surface at point \( P_0 (1, 2, -1) \). Without it, we wouldn't be able to ensure that the plane is tangent to the surface at the specific point needed.
Cartesian Equation
A Cartesian equation of a surface or plane expresses one coordinate in terms of the others, typically in the form of \( x, y, \) and \( z \). It’s a straightforward way to represent geometric shapes algebraically.
For the parabolic cylinder surface in the exercise, its Cartesian equation is \( z = -x^2 \). This describes how the surface extends through space, showing us the parabolic profile along the \( z \)-axis as \( x \) changes.
Furthermore, once we have the normal vector, we can write the tangent plane's equation. In this case, the tangent plane at point \( P_0 (1, 2, -1) \) is given by \( 2x + z = 1 \).
This equation succinctly describes the plane in the \( xyz \) coordinate space, which just touches the surface without intersecting it at multiple points.
For the parabolic cylinder surface in the exercise, its Cartesian equation is \( z = -x^2 \). This describes how the surface extends through space, showing us the parabolic profile along the \( z \)-axis as \( x \) changes.
Furthermore, once we have the normal vector, we can write the tangent plane's equation. In this case, the tangent plane at point \( P_0 (1, 2, -1) \) is given by \( 2x + z = 1 \).
This equation succinctly describes the plane in the \( xyz \) coordinate space, which just touches the surface without intersecting it at multiple points.
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