Problem 27

Question

Find points on the surface $x^{2}+4 x+y^{2}+z^{2}-2 z=11$ at which the tangent plane is horizontal.

Step-by-Step Solution

Verified

Answer

The points are $(-2, 0, 5)$ and $(-2, 0, -3)$.

1Step 1: Understand the problem

We need to find the points on the given surface where the tangent plane is horizontal. This means that the normal vector to the tangent plane has to be vertical and thus only has a component in the z-direction.

2Step 2: Rewrite implicit equation

The given surface equation is $x^{2} + 4x + y^{2} + z^{2} - 2z = 11$. A horizontal tangent plane has a normal with zero x and y components, implying the partial derivatives with respect to $x$ and $y$ both equal zero.

3Step 3: Find partial derivatives

Determine the partial derivatives of the function $f(x, y, z) = x^{2} + 4x + y^{2} + z^{2} - 2z$:$f_x = 2x + 4$, $f_y = 2y$, and $f_z = 2z - 2$.

4Step 4: Set the x and y plane partial derivatives to zero

For the tangent plane to be horizontal, $f_x = 0$ and $f_y = 0$. Thus:1. $2x + 4 = 0$ implies $x = -2$.2. $2y = 0$ implies $y = 0$.

5Step 5: Determine z value for points

Substitute $x = -2$ and $y = 0$ back into the original surface equation to find $z$:$(-2)^{2} + 4(-2) + 0^{2} + z^{2} - 2z = 11$ which simplifies to:$4 - 8 + z^{2} - 2z = 11$, therefore $z^{2} - 2z - 15 = 0$.

6Step 6: Solve quadratic equation for z

Solve the quadratic equation $z^{2} - 2z - 15 = 0$ using the quadratic formula:$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1, b = -2, c = -15$. This gives roots:$z = 5$ and $z = -3$.

7Step 7: Find points on the surface

Use the values for $z$ to determine the points:- For $z = 5$, the point is $(-2, 0, 5)$.- For $z = -3$, the point is $(-2, 0, -3)$.

Key Concepts

Partial DerivativesQuadratic EquationHorizontal TangentNormal Vector

Partial Derivatives

Understanding partial derivatives is key when analyzing functions of multiple variables, like the surface given in the problem. A partial derivative represents how a function changes with respect to one variable, while keeping others constant. In the context of this problem, the surface is defined by the equation:

$ f(x, y, z) = x^2 + 4x + y^2 + z^2 - 2z $

To find out where the tangent plane becomes horizontal, we calculate the partial derivatives of this function with respect to each variable $x$, $y$, and $z$. Here's how these are derived:

Partial derivative with respect to $x$, denoted as $f_x$, is $2x + 4$.
Partial derivative with respect to $y$, denoted as $f_y$, is $2y$.
Partial derivative with respect to $z$, denoted as $f_z$, is $2z - 2$.

The conditions for a horizontal tangent plane involve setting $f_x = 0$ and $f_y = 0$, indicating no slope in those directions.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable. It takes the form:
\[ ax^2 + bx + c = 0 \] where $a, b,$ and $c$ are constants. In this problem, we derive such an equation from the surface equation to solve for $z$.
Once we substitute values of $x$ and $y$ found from setting partial derivatives to zero, the equation simplifies:

$z^2 - 2z - 15 = 0$

This is a quadratic equation, solved using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in $a = 1$, $b = -2$, and $c = -15$, the solution yields the roots:

$z = 5$
$z = -3$

These roots represent the possible $z$-values at which the tangent plane is horizontal.

Horizontal Tangent

A horizontal tangent plane implies that the plane is perfectly flat with respect to the horizontal axes $x$ and $y$. For a surface, this occurs when the partial derivatives of the surface equation with respect to $x$ and $y$ are zero, meaning there is no change in height as you move along these directions.
In our problem, this requirement leads to:

$f_x = 0$ simplifying to $2x + 4 = 0$, or $x = -2$
$f_y = 0$ simplifying to $2y = 0$, or $y = 0$

By finding these points, we can solve the surface equation to find corresponding $z$-values that make the tangent plane horizontal. The solutions, $(-2, 0, 5)$ and $(-2, 0, -3)$, confirm such points on the surface.

Normal Vector

A normal vector is a vector that is perpendicular to the tangent plane of a surface at a given point. For a plane described by the equation $f(x, y, z) = 0$, the normal vector is formed by its gradient:
$ abla f = (f_x, f_y, f_z) $ In this case, since we want the tangent plane to be horizontal, the normal vector should only point in the vertical ($z$) direction. Hence, we set $f_x = 0$ and $f_y = 0$, ensuring the normal vector only depends on $f_z$.
This adjustment results in the normal vector:

$(0, 0, 2z - 2)$

A normal vector of this form signifies that our tangent plane is indeed horizontal, as it aligns vertically. With the calculated values of $z$, the normal vectors verify the essential property of the tangent plane when the slopes in $x$ and $y$ directions are zero.

Problem 27

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