Problem 11
Question
Find an equation of the plane tangent to the following surfaces at the given points. $$x y+x z+y z-12=0 ;(2,2,2) \text { and }(2,0,6)$$
Step-by-Step Solution
Verified Answer
Question: Find the equations of the tangent planes to the surface xy + xz + yz = 12 at the points (2,2,2) and (2,0,6).
Answer: The equations of the tangent planes are:
1. 4(x-2) + 4(y-2) + 4(z-2) = 0
2. 6(x-2) + 8y + 2(z-6) = 0
1Step 1: Calculate gradient of the surface at the given points
First, let's rewrite the equation in terms of a single function:
$$
F(x,y,z) = xy + xz + yz - 12 =0
$$
Now, let's find the gradient (\(∇F\)), which consists of partial derivatives with respect to \(x\), \(y\), and \(z\).
$$
∇F = \begin{bmatrix}\frac{\partial F}{\partial x} \\ \frac{\partial F}{\partial y} \\ \frac{\partial F}{\partial z}\end{bmatrix} = \begin{bmatrix} y + z \\ x + z \\ x + y \end{bmatrix}
$$
Using the given points, we can find the gradient at each of those points:
$$
∇F(2,2,2) = \begin{bmatrix} 4 \\ 4 \\ 4 \end{bmatrix}
$$
$$
∇F(2,0,6) = \begin{bmatrix} 6 \\ 8 \\ 2 \end{bmatrix}
$$
2Step 2: Find the equation of the tangent planes using the point-normal form
The point-normal form of a plane equation is given by \((\)normal vector\()⋅(\)vector to a point on the plane\() = 0\). For our case, the equation becomes
$$
a(x-x_0) + b(y-y_0) + c(z-z_0) = 0
$$
where \((a, b, c)\) is the normal vector of the plane and \((x_0, y_0, z_0)\) is the point on the plane.
For the first point \((2,2,2)\), the normal vector is \((4, 4, 4)\), and the equation of the tangent plane becomes
$$
4(x-2) + 4(y-2) + 4(z-2) = 0
$$
For the second point \((2,0,6)\), the normal vector is \((6, 8, 2)\), and the equation of the tangent plane becomes
$$
6(x-2) + 8(y-0) + 2(z-6) = 0
$$
So the two tangent planes are:
$$
4(x-2) + 4(y-2) + 4(z-2) = 0
$$
and
$$
6(x-2) + 8y + 2(z-6) = 0
$$
Key Concepts
Gradient VectorPartial DerivativesPoint-Normal Form
Gradient Vector
The gradient vector is a fundamental concept in multivariable calculus, especially when working with functions of several variables. At its core, the gradient vector \(abla F\) at a point on a surface is like a multi-dimensional "slope". It points in the direction of the greatest rate of increase of the function. This makes it highly valuable in locating tangent planes.
For a function like \(F(x, y, z) = xy + xz + yz - 12\), the gradient vector is found by calculating the partial derivatives of \(F\) with respect to each variable: \(x\), \(y\), and \(z\). These derivatives tell us how \(F\) changes as we tweak these variables a little bit.
Given the specific function, the gradient vector is \(abla F = \begin{bmatrix} y + z \ x + z \ x + y \end{bmatrix}\). Evaluating the gradient at certain points, like \((2, 2, 2)\) or \((2, 0, 6)\), gives us vectors which are essential for constructing tangent planes.
For a function like \(F(x, y, z) = xy + xz + yz - 12\), the gradient vector is found by calculating the partial derivatives of \(F\) with respect to each variable: \(x\), \(y\), and \(z\). These derivatives tell us how \(F\) changes as we tweak these variables a little bit.
- \(\frac{\partial F}{\partial x}\) indicates how \(F\) changes as \(x\) changes (holding \(y\) and \(z\) constant).
- \(\frac{\partial F}{\partial y}\) shows the change in \(F\) with respect to \(y\).
- \(\frac{\partial F}{\partial z}\) describes the change in \(F\) when \(z\) changes.
Given the specific function, the gradient vector is \(abla F = \begin{bmatrix} y + z \ x + z \ x + y \end{bmatrix}\). Evaluating the gradient at certain points, like \((2, 2, 2)\) or \((2, 0, 6)\), gives us vectors which are essential for constructing tangent planes.
Partial Derivatives
Partial derivatives form the backbone of understanding changes in multivariable functions. When dealing with a function of several variables, a partial derivative measures how the function varies as one specific variable changes, while keeping the others fixed.
In the equation \(F(x, y, z) = xy + xz + yz - 12\), each partial derivative corresponds to the effect of isolating and changing a single variable. For example:
In the equation \(F(x, y, z) = xy + xz + yz - 12\), each partial derivative corresponds to the effect of isolating and changing a single variable. For example:
- \(\frac{\partial F}{\partial x} = y + z\) tells how \(F\) changes when only \(x\) is tweaked.
- \(\frac{\partial F}{\partial y} = x + z\) represents \(F\)'s response to changes solely in \(y\).
- \(\frac{\partial F}{\partial z} = x + y\) depicts the variation of \(F\) with respect to \(z\).
Point-Normal Form
The point-normal form is a straightforward method to describe the equation of a plane using a point on the surface and a normal vector. A normal vector to a plane is perpendicular to every line lying on the plane. In our context, the gradient vector at a point serves as this normal vector.
For a tangent plane to a surface at a specific point \((x_0, y_0, z_0)\), the equation is represented as:\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]Here, \((a, b, c)\) are the coordinates of the gradient vector (normal vector), and \((x_0, y_0, z_0)\) is a point through which the plane passes.
For a tangent plane to a surface at a specific point \((x_0, y_0, z_0)\), the equation is represented as:\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]Here, \((a, b, c)\) are the coordinates of the gradient vector (normal vector), and \((x_0, y_0, z_0)\) is a point through which the plane passes.
- For the point \((2, 2, 2)\) with the normal vector \((4, 4, 4)\), the tangent plane is \(4(x-2) + 4(y-2) + 4(z-2) = 0\).
- For the point \((2, 0, 6)\) with the normal vector \((6, 8, 2)\), it is \(6(x-2) + 8y + 2(z-6) = 0\).
Other exercises in this chapter
Problem 11
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