Question

Find the minimum distance from the curve or surface to the given point. (Hint: Start by minimizing the square of the distance.) Cone: \(z=\sqrt{x^{2}+y^{2}},(4,0,0)\) Minimize \(d^{2}=(x-4)^{2}+y^{2}+z^{2}\)

Short answer

The minimum distance is at coordinates (2,0,2).

Step-by-step solution

Express the Distance in Terms of x, y, and z

The distance between two points (a,b,c) and (x,y,z) in space is given by \(d=\sqrt{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}\). Given point is (4,0,0) so, \(d^{2}=(x-4)^{2}+y^{2}+z^{2}\)

Substitute z with Miximum Value

Given that the cone is specified by the equation \(z=\sqrt{x^{2}+y^{2}}\). So, replacing z in the distance equation, we have \(d^{2}=(x-4)^{2}+y^{2}+x^{2}+y^{2}\)

Simplify the equation

Combining the like terms, we get \(d^{2}=2(x^{2}+y^{2})+(x-4)^{2}\)

Take Partial Derivative of d^2 With Respect to x and y

To find the minimum value of \(d^{2}\), we take the derivative and set it equal to 0. The partial derivative with respect to x is \(4x+(x-4)=0\) and with respect to y is \(4y=0\)

Solve for x and y

Solving the above equations we get the minimum point at x = 2 and y = 0

Substitute x and y into Cone Equation for Z

Substitute x = 2 and y = 0 into the Cone equation to get z = 2

Key concepts

Distance Formula in 3D Space

Understanding the distance formula in 3D space is fundamental when solving geometry problems involving the shortest distance. The general idea is to find the length of the segment connecting two points in space. The formula derived from the Pythagorean theorem is \( d = \sqrt{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}} \), where \( (x, y, z) \) and \( (a, b, c) \) are coordinates of the two points.

To illustrate, consider the exercise's requirement of finding the minimum distance between a cone, described by \( z=\sqrt{x^{2}+y^{2}} \), and the point \( (4,0,0) \). But why minimize the square of the distance instead of the distance itself? It's simpler because squaring the distance eliminates the square root, avoiding a complicated derivative and making the optimization problem more manageable.

Partial Derivatives Optimization

When dealing with functions of several variables, optimization problems often require the use of partial derivatives. The concept of partial derivatives involves holding all variables except one constant and finding the derivative concerning the variable of interest. This is crucial in identifying points where the function may have a local maximum or minimum.

For the cone exercise, to find the minimum distance to the point \( (4,0,0) \) we must express the square of the distance, \( d^{2} \), as a function of the variables x and y. Then, by taking partial derivatives with respect to x and y, we set them to zero to find the critical points. This process is known as setting up the system of equations that will lead us to the optimum value—ultimately, the minimum distance.

Substitution in Equations

Substitution is a valuable technique for solving equations, especially when working with multiple variables. It involves replacing one variable with another expression that equals that variable, which can simplify the problem significantly.

In this optimization problem, we substitute the z variable with \( \sqrt{x^{2}+y^{2}} \) from the cone's equation before we take the derivatives. This kind of substitution allows us to write the distance square, \( d^{2} \), entirely in terms of x and y, making it possible to find the minimum distance on the cone's surface with respect to a fixed point. Once we have determined the values of x and y that minimize \( d^{2} \), we substitute back into the original cone equation to find the corresponding z-value. Such systematic substitution is a cornerstone technique in solving complex algebraic and calculus problems.