Answer :
To find the equation of the tangent plane to the parametric surface defined by [tex]\( x = u + v \)[/tex], [tex]\( y = 7u^2 \)[/tex], and [tex]\( z = u - v \)[/tex] at the point [tex]\((2,7,0)\)[/tex], we will follow a systematic approach.
### Step 1: Establish the Relationship Between Parametric Variables and Coordinates
We start by identifying the parametric values [tex]\(u\)[/tex] and [tex]\(v\)[/tex] that correspond to the point [tex]\((2, 7, 0)\)[/tex] on the surface.
Given:
- [tex]\(x = u + v\)[/tex]
- [tex]\(y = 7u^2\)[/tex]
- [tex]\(z = u - v\)[/tex]
We need to find [tex]\(u\)[/tex] and [tex]\(v\)[/tex] such that:
- [tex]\(2 = u + v\)[/tex]
- [tex]\(7 = 7u^2\)[/tex]
- [tex]\(0 = u - v\)[/tex]
From [tex]\(y = 7u^2\)[/tex], we solve for [tex]\(u\)[/tex]:
[tex]\[ y = 7 \Rightarrow 7u^2 = 7 \Rightarrow u^2 = 1 \Rightarrow u = 1 \text{ or } -1. \][/tex]
Next, consider the coordinates [tex]\( (2, 0) = (u + v, u - v) \)[/tex].
1. If [tex]\(u = 1\)[/tex]:
[tex]\[ x = 2 \Rightarrow 1 + v = 2 \Rightarrow v = 1 \][/tex]
[tex]\[ z = 0 \Rightarrow 1 - v = 0 \Rightarrow v = 1 \][/tex]
2. If [tex]\(u = -1\)[/tex]:
[tex]\[x = 2 \Rightarrow -1 + v = 2 \Rightarrow v = 3\][/tex]
[tex]\[z = 0 \Rightarrow -1 - v = 0 \Rightarrow v = -1\][/tex]
We discard [tex]\(u = -1\)[/tex] since it does not give the given [tex]\(z = 0\)[/tex]. Thus, we use [tex]\(u = 1\)[/tex] and [tex]\(v = 1\)[/tex].
### Step 2: Compute the Partial Derivatives of Parametric Equations
Consider the partial derivatives of [tex]\(x, y,\)[/tex] and [tex]\(z\)[/tex] with respect to [tex]\(u\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ \frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 1 \][/tex]
[tex]\[ \frac{\partial y}{\partial u} = 14u, \quad \frac{\partial y}{\partial v} = 0 \][/tex]
[tex]\[ \frac{\partial z}{\partial u} = 1, \quad \frac{\partial z}{\partial v} = -1 \][/tex]
### Step 3: Find the Normal Vector to the Tangent Plane
The normal vector [tex]\(\mathbf{N}\)[/tex] to the tangent plane is given by the cross product of the gradient vectors [tex]\(\frac{\partial \mathbf{R}}{\partial u}\)[/tex] and [tex]\(\frac{\partial \mathbf{R}}{\partial v}\)[/tex]:
[tex]\[ \mathbf{R}_u = \begin{pmatrix} 1 \\ 14u \\ 1 \end{pmatrix}, \quad \mathbf{R}_v = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \][/tex]
Calculate the cross product:
[tex]\[ \mathbf{N} = \mathbf{R}_u \times \mathbf{R}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 14u & 1 \\ 1 & 0 & -1 \end{vmatrix} = \mathbf{i} (14u \cdot (-1) - 1 \cdot 0) - \mathbf{j} (1 \cdot (-1) - 1 \cdot 1) + \mathbf{k} (1 \cdot 0 - 1 \cdot 14u) \][/tex]
[tex]\[ = \mathbf{i} (-14u) - \mathbf{j} (-2) + \mathbf{k} (-14u) \][/tex]
[tex]\[ = \begin{pmatrix} -14u \\ 2 \\ -14u \end{pmatrix} \][/tex]
Now, substitute [tex]\(u = 1\)[/tex] into the normal vector [tex]\(\mathbf{N}\)[/tex]:
[tex]\[ \mathbf{N} = \begin{pmatrix} -14(1) \\ 2 \\ -14(1) \end{pmatrix} = \begin{pmatrix} -14 \\ 2 \\ -14 \end{pmatrix} \][/tex]
### Step 4: Formulate the Equation of the Tangent Plane
The equation of the tangent plane at a given point [tex]\((x_0, y_0, z_0)\)[/tex] using the normal vector [tex]\((A, B, C)\)[/tex] can be written as:
[tex]\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \][/tex]
Substitute the point coordinates [tex]\((x_0, y_0, z_0) = (2, 7, 0)\)[/tex] and the normal vector components [tex]\(A = -14\)[/tex], [tex]\(B = 2\)[/tex], [tex]\(C = -14\)[/tex]:
[tex]\[ -14(x - 2) + 2(y - 7) - 14(z - 0) = 0 \][/tex]
Expand and simplify:
[tex]\[ -14x + 28 + 2y - 14 - 14z = 0 \][/tex]
[tex]\[ -14x + 2y - 14z + 14 = 0 \Rightarrow 14x - 2y + 14z = 14 \][/tex]
Divide through by 14:
[tex]\[ x - \frac{1}{7} y + z = 1 \][/tex]
Thus, the equation of the tangent plane to the parametric surface at the point [tex]\((2, 7, 0)\)[/tex] is:
[tex]\[ x - \frac{1}{7} y + z = 1 \][/tex]
### Step 1: Establish the Relationship Between Parametric Variables and Coordinates
We start by identifying the parametric values [tex]\(u\)[/tex] and [tex]\(v\)[/tex] that correspond to the point [tex]\((2, 7, 0)\)[/tex] on the surface.
Given:
- [tex]\(x = u + v\)[/tex]
- [tex]\(y = 7u^2\)[/tex]
- [tex]\(z = u - v\)[/tex]
We need to find [tex]\(u\)[/tex] and [tex]\(v\)[/tex] such that:
- [tex]\(2 = u + v\)[/tex]
- [tex]\(7 = 7u^2\)[/tex]
- [tex]\(0 = u - v\)[/tex]
From [tex]\(y = 7u^2\)[/tex], we solve for [tex]\(u\)[/tex]:
[tex]\[ y = 7 \Rightarrow 7u^2 = 7 \Rightarrow u^2 = 1 \Rightarrow u = 1 \text{ or } -1. \][/tex]
Next, consider the coordinates [tex]\( (2, 0) = (u + v, u - v) \)[/tex].
1. If [tex]\(u = 1\)[/tex]:
[tex]\[ x = 2 \Rightarrow 1 + v = 2 \Rightarrow v = 1 \][/tex]
[tex]\[ z = 0 \Rightarrow 1 - v = 0 \Rightarrow v = 1 \][/tex]
2. If [tex]\(u = -1\)[/tex]:
[tex]\[x = 2 \Rightarrow -1 + v = 2 \Rightarrow v = 3\][/tex]
[tex]\[z = 0 \Rightarrow -1 - v = 0 \Rightarrow v = -1\][/tex]
We discard [tex]\(u = -1\)[/tex] since it does not give the given [tex]\(z = 0\)[/tex]. Thus, we use [tex]\(u = 1\)[/tex] and [tex]\(v = 1\)[/tex].
### Step 2: Compute the Partial Derivatives of Parametric Equations
Consider the partial derivatives of [tex]\(x, y,\)[/tex] and [tex]\(z\)[/tex] with respect to [tex]\(u\)[/tex] and [tex]\(v\)[/tex]:
[tex]\[ \frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 1 \][/tex]
[tex]\[ \frac{\partial y}{\partial u} = 14u, \quad \frac{\partial y}{\partial v} = 0 \][/tex]
[tex]\[ \frac{\partial z}{\partial u} = 1, \quad \frac{\partial z}{\partial v} = -1 \][/tex]
### Step 3: Find the Normal Vector to the Tangent Plane
The normal vector [tex]\(\mathbf{N}\)[/tex] to the tangent plane is given by the cross product of the gradient vectors [tex]\(\frac{\partial \mathbf{R}}{\partial u}\)[/tex] and [tex]\(\frac{\partial \mathbf{R}}{\partial v}\)[/tex]:
[tex]\[ \mathbf{R}_u = \begin{pmatrix} 1 \\ 14u \\ 1 \end{pmatrix}, \quad \mathbf{R}_v = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \][/tex]
Calculate the cross product:
[tex]\[ \mathbf{N} = \mathbf{R}_u \times \mathbf{R}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 14u & 1 \\ 1 & 0 & -1 \end{vmatrix} = \mathbf{i} (14u \cdot (-1) - 1 \cdot 0) - \mathbf{j} (1 \cdot (-1) - 1 \cdot 1) + \mathbf{k} (1 \cdot 0 - 1 \cdot 14u) \][/tex]
[tex]\[ = \mathbf{i} (-14u) - \mathbf{j} (-2) + \mathbf{k} (-14u) \][/tex]
[tex]\[ = \begin{pmatrix} -14u \\ 2 \\ -14u \end{pmatrix} \][/tex]
Now, substitute [tex]\(u = 1\)[/tex] into the normal vector [tex]\(\mathbf{N}\)[/tex]:
[tex]\[ \mathbf{N} = \begin{pmatrix} -14(1) \\ 2 \\ -14(1) \end{pmatrix} = \begin{pmatrix} -14 \\ 2 \\ -14 \end{pmatrix} \][/tex]
### Step 4: Formulate the Equation of the Tangent Plane
The equation of the tangent plane at a given point [tex]\((x_0, y_0, z_0)\)[/tex] using the normal vector [tex]\((A, B, C)\)[/tex] can be written as:
[tex]\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \][/tex]
Substitute the point coordinates [tex]\((x_0, y_0, z_0) = (2, 7, 0)\)[/tex] and the normal vector components [tex]\(A = -14\)[/tex], [tex]\(B = 2\)[/tex], [tex]\(C = -14\)[/tex]:
[tex]\[ -14(x - 2) + 2(y - 7) - 14(z - 0) = 0 \][/tex]
Expand and simplify:
[tex]\[ -14x + 28 + 2y - 14 - 14z = 0 \][/tex]
[tex]\[ -14x + 2y - 14z + 14 = 0 \Rightarrow 14x - 2y + 14z = 14 \][/tex]
Divide through by 14:
[tex]\[ x - \frac{1}{7} y + z = 1 \][/tex]
Thus, the equation of the tangent plane to the parametric surface at the point [tex]\((2, 7, 0)\)[/tex] is:
[tex]\[ x - \frac{1}{7} y + z = 1 \][/tex]
