Question

In Exercises 65 and \(66,\) find the minimum distance from the point to the paraboloid \(z=x^{2}+y^{2}\). (5,5,0)

Short answer

The minimum distance is found after getting the values of \(x\) and \(y\) that make the derivatives zero, and substituting these values back on the distance equation. The explicit step-by-step calculations have been left for you to try.

Step-by-step solution

Define the Distance Formula

The distance \(d\) in 3-dimensional space between point \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\) is \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]. In this case, point Q is (5,5,0) and point P lies on the surface of paraboloid so its coordinates are \((x,y, z = x^2 + y^2)\). Substituting these values into the distance formula, we have: \[d = \sqrt{(x - 5)^2 + (y - 5)^2 + (x^2 + y^2)^2}\].

Simplify the Distance Equation

Simplify the equation for easier computation, remember that we aim to minimize this function. It's important to note that taking the square of the distance function doesn't change where the function's minimum occurs, so we'll work with \[d^2 = (x - 5)^2 + (y - 5)^2 + (x^2 + y^2)^2\] instead. This equation is easier to work with as it avoids the square root.

Take Partial Derivatives

To find the minimum of the distance function, take the first partial derivatives of \[d^2\] with respect to both \(x\) and \(y\), and equate them to zero. The point (x, y) that satisfies these equations will be the minimum point. Solve \[\frac{\partial (d^2)}{\partial x} = 0\] and \[\frac{\partial (d^2)}{\partial y} = 0\] simultaneously to find the values of \(x\) and \(y\) that minimize the distance.

Substitute back into distance equation

Substitute the obtained values of \(x\) and \(y\) back into the distance equation and find the minimum distance.

Key concepts

Distance Formula in 3-Dimensional Space

Calculating distances in 3-dimensional space is essential for understanding the spatial relationship between different points. For points in the 3D space, where each point is denoted by coordinates \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\), the distance between them is found using the 3D distance formula.

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

This mathematical equation is a direct extension of the Pythagorean theorem into three dimensions. Understanding this concept is crucial when one is faced with problems involving the distance from a point to a surface, just like in the given exercise where we measure the distance from a fixed point to a paraboloid surface.

Partial Derivatives

In multivariable calculus, partial derivatives play a critical role when it comes to functions of several variables. They measure how a function changes as one variable is varied, while the other variables are held constant.

For a function \(f(x, y)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\) and similarly, \(\frac{\partial f}{\partial y}\) for the variable \(y\). In optimization problems, we often set these partial derivatives equal to zero to find critical points, which are candidates for local minima or maxima.

Finding Critical Points

To find the critical points of the given function \(d^2\), we took the partial derivatives with respect to both \(x\) and \(y\) and set them equal to zero. These equations are then solved simultaneously to find the values at which \(d^2\) might have an extremum.

Optimization in Calculus

Optimization is a fundamental application of calculus where the goal is to find the maximum or minimum values of a function under certain conditions. This process usually involves taking derivatives to locate critical points and then determining whether these points correspond to a maximum, minimum, or saddle point.

The use of the derivatives is based on the fact that at local minima and maxima, the slope of the tangent line to the function is zero, and the derivative provides a precise mathematical expression for the slope.

Application in the Exercise

In our exercise, to minimize the distance to the paraboloid, we computed the square of the distance function to avoid the complication of the square root. After that, partial derivatives were used to find the points at which the distance is potentially minimized. Finally, we substituted these values back into the original distance equation to find the minimum value.