We are asked to find the minimum distance from the point $P = (2, 0, 0)$ to the cone defined by the equation:

$z^2 = x^2 + y^2$

Step 1: Understanding the Geometry

The equation $z^2 = x^2 + y^2$ describes a cone that opens along the $z$-axis. The cone's apex is at the origin $(0, 0, 0)$, and its surface consists of points where the distance from the $z$-axis is equal to the $z$-coordinate in absolute value.

Step 2: Distance Formula

The distance from a point $P(x_1, y_1, z_1)$ to a point $Q(x_2, y_2, z_2)$ in three-dimensional space is given by the formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

For our case, the point $P = (2, 0, 0)$ and any point on the cone $Q(x, y, z)$ satisfies the equation $z^2 = x^2 + y^2$. So, we need to minimize the distance from $(2, 0, 0)$ to any point on the cone.

Step 3: Setting up the Optimization Problem

We can set up the distance function $D$ from the point $(2, 0, 0)$ to any point on the cone $(x, y, z)$:

$D = \sqrt{(x - 2)^2 + y^2 + z^2}$

Given the equation of the cone $z^2 = x^2 + y^2$, we substitute $z^2$ with $x^2 + y^2$ into the distance formula:

$D = \sqrt{(x - 2)^2 + y^2 + (x^2 + y^2)}$

Now, simplifying the expression:

$D = \sqrt{(x - 2)^2 + y^2 + x^2 + y^2} = \sqrt{(x - 2)^2 + 2y^2 + x^2}$

Step 4: Minimizing the Distance Function

To minimize $D$, it's easiest to minimize $D^2$ (which is the squared distance):

$D^2 = (x - 2)^2 + 2y^2 + x^2$

Expanding this expression:

$D^2 = (x^2 - 4x + 4) + 2y^2 + x^2$ $D^2 = 2x^2 - 4x + 4 + 2y^2$

We now minimize $D^2$ with respect to $x$ and $y$. First, take partial derivatives of $D^2$ with respect to $x$ and $y$, and set them equal to zero.

Partial Derivative with respect to $x$:

$\frac{\partial D^2}{\partial x} = 4x - 4 = 0$ $4x - 4 = 0 \implies x = 1$

Partial Derivative with respect to $y$:

$\frac{\partial D^2}{\partial y} = 4y = 0$ $y = 0$

Step 5: Substituting into the Cone Equation

Now, substitute $x = 1$ and $y = 0$ into the cone equation $z^2 = x^2 + y^2$:

$z^2 = 1^2 + 0^2 = 1$ $z = \pm 1$

Step 6: Finding the Minimum Distance

Thus, the closest points on the cone to $P(2, 0, 0)$ are $(1, 0, 1)$ and $(1, 0, -1)$.

Now, we compute the distance from $P = (2, 0, 0)$ to these points.

For $Q(1, 0, 1)$:

$D = \sqrt{(1 - 2)^2 + (0 - 0)^2 + (1 - 0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$

For $Q(1, 0, -1)$:

$D = \sqrt{(1 - 2)^2 + (0 - 0)^2 + (-1 - 0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$

Final Answer

Thus, the minimum distance from the point $(2, 0, 0)$ to the cone $z^2 = x^2 + y^2$ is $\sqrt{2}$.

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Would you like any further details or clarifications on this solution?

Here are 5 related questions to explore:

1. How does the distance from a point to a cone change if the point is off-center along the z-axis?
2. What is the method for finding the shortest distance from a point to a surface in general?
3. How can we apply the concept of gradients to find the shortest distance to a geometric surface?
4. What happens to the minimum distance if the cone's axis is rotated around the z-axis?
5. How do the equations of a cone change in three-dimensional space if it's not aligned with the z-axis?

Tip: When minimizing the distance function, using symmetry (like the fact that the cone is symmetric about the z-axis) can sometimes make solving the problem much easier.