Problem 30 - College Algebra and Calculus: An Applied Approach

Question

Find the minimum distance from the curve or surface to the given point. (Hint: Start by minimizing the square of the distance.) Circle: \((x-4)^{2}+y^{2}=4,(0,10)\) Minimize \(d^{2}=x^{2}+(y-10)^{2}\)

Step-by-Step Solution

Answer

The minimum distance from the point (0,10) to the circle \((x-4)^{2}+y^{2}=4\) is calculated by finding the x and y points on the circle that minimize the square of the distance to the given point. This is done by taking the derivative of the distance formula and equating it to zero to find the critical points. The result will be the minimum distance.

Step 1: Write down the equation

Write down the equation of the square of the distance between point (0,10) and any point (x, y) on the surface of the circle. This equation is \(d^{2}=x^{2}+(y-10)^{2}\).

Step 2: Implement the constraint

The constraint here is the equation of the circle. Substitute \(y=\sqrt{4-(x-4)^{2}}\) (the equation of y in terms of x) into the equation obtained in step 1. This will eliminate one variable (y) and make the equation easier to handle.

Step 3: Simplify and Differentiate

Simplify the equation obtained in step 2. Then, find the derivative of the simplified equation with respect to x and set it equal to zero. This corresponds to finding the critical points (where the function may have a minimum or maximum) of the distance function.

Step 4: Solve for x and y

By setting the derivative equal to zero and solving for x, the x-coordinate point on the circle where the distance is minimized is found. Substitute this x value back into the equation of the circle to find the corresponding y point.

Step 5: Calculate the minimum distance

Substitute the x and y values obtained from step 4 into the square root of the distance formula obtained in step 1 to calculate the minimum distance.

Key Concepts

Distance Formula

The Distance Formula is a crucial tool in geometry that helps us calculate the distance between two points in a plane. The formula is derived from the Pythagorean theorem and is given by \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
In this problem, rather than working with the distance directly, we minimize the square of the distance, \(d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\). This is computationally simpler, eliminating the need for square roots, and still allows us to find where the distance is minimized.
This problem specifically uses the formula to find the minimum distance from a point to the circumference of a circle, which is slightly more complex as it involves additional constraints.

Critical Points

Critical points in calculus are where a function's derivative is zero or undefined. These points are potential locations where the function can have minimum, maximum, or saddle points.
In the problem at hand, once we've substituted and simplified the distance squared function in terms of a single variable, we need to find its derivative, set the derivative equal to zero, and solve for the variable. This step gives us the critical points.
By evaluating these critical points, we can determine which configurations result in the minimum possible distance between the point and the circle. This helps in efficiently solving optimization problems using calculus.

Constraint Substitution

Constraint Substitution is a method used to solve optimization problems with given constraints. By substituting constraints directly into an equation, we reduce the number of variables, simplifying the problem.
In this exercise, the constraint is represented by the equation of the circle \((x-4)^2 + y^2 = 4\). We rearrange this to express \(y\) in terms of \(x\), giving us \(y = \sqrt{4-(x-4)^2}\).
This allows us to substitute \(y\)'s expression into the distance equation, effectively reducing the problem to a function of one variable. Having fewer variables simplifies differentiation, making it easier to find the critical points.

Circle Equation

The Circle Equation is key in problems involving distances related to circles. The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the circle's center and \(r\) is its radius.
In this problem, the circle's equation is \((x-4)^2 + y^2 = 4\), revealing that the circle is centered at \((4,0)\) with a radius of 2.
This equation helps us understand the location and size of the circle, which is essential for tasks such as determining where a point on the circle minimizes or maximizes the distance to another point. Understanding the circle's properties gives us the foundation to apply the distance formula and substitute constraints effectively.