Problem 30 - 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.) $$ \begin{array}{l}{\text { Circle: }(x-4)^{2}+y^{2}=4,(0,10)} \\ {\text { Minimize } d^{2}=x^{2}+(y-10)^{2}}\end{array} $$

Step-by-Step Solution

Answer

The minimum distance from the point (0,10) to the circle is at the point (4,2) on the circle.

Step 1: Write out the given Circle equation and the equation for distance (squared)

The given equation for the circle is \((x-4)^{2}+y^{2}=4\). Now substitute this into the equation for the square of the distance from a point on the circle to the given point (0,10), getting \[d^{2}=x^{2}+(y-10)^{2}\] which will become \[d^{2}=(x^{2}+(y-10)^{2})=(x^{2}+(16-x^{2})-10)^{2}\] by rearranging the equation of the circle to express y in terms of x.

Step 2: Simplify and Differentiate

The equation from step 1 simplifies to \[d^{2}=(16-x^{2})^2\]. Now we differentiate that equation with respect to x to find the minimum value. Let's call the derivative \(d^{'2}\). This gives us \[d^{'2}=2*(16-x^{2})*-2x\].

Step 3: Find the x value which gives the minimum distance

Set the derivative \(d^{'2}\) equal to zero and solve for x, so \[2*(16-x^{2})*-2x=0\]. By solving that equation we get \(x=4\).

Step 4: Find the y value corresponding to \(x=4\)

Substitute \(x=4\) into the circle equation to get the y value for the point on the circle closest to the given point. This gives \(y=sqrt(4)\), which simplifies to \(y=2\).

Key Concepts

Understanding the Distance Formula

The distance formula helps us calculate how far apart two points are located in a space. In this problem, our task is to find the minimum distance between a circle and a specific point, which is (0, 10).
The distance formula for two points

• (x extsubscript{1}, y extsubscript{1})
• (x extsubscript{2}, y extsubscript{2})

can be written as:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
In our case, the first point (x, y) is on the circle, while the second point is (0, 10). However, instead of minimizing the distance \(d\), we're minimizing \(d^2\), which eliminates the square root for simplicity in calculation.
This approach reduces computational complexity without affecting the location of the minimum distance.

The Role of Derivatives in Optimization

Derivatives are incredibly valuable in finding the points where a function reaches its extrema, such as a maximum or minimum. In this problem, our main objective is to find the point on the circle that is closest to a given point.
We aim to minimize \(d^2\). By taking the derivative, \(d^{'2}\), with respect to \(x\), we can identify these critical points.
The derivative tells us how \(d^2\) varies as \(x\) changes. The mathematical expression is simplified to

• \(d^{'2} = 2(16 - x^2)(-2x)\)

Setting the derivative to zero helps us find the value of \(x\) that achieves the minimum distance.

Differentiation: Solving for Critical Points

Differentiation is the process of finding the derivative of a function. In our context, it allows us to identify minimum points in \(d^2\).
The equation after differentiating is \[d^{'2} = 2(16 - x^2)(-2x)\].
To find the critical points, we set the derivative equal to zero. This equation simplifies to give \(x = 4\).
This means the x-coordinate of the point on the circle that is closest to (0, 10) is 4.
Differentiation helps us find these "turning points" where the rate of change is zero and where curves tend to flatten, revealing spots of minimum or maximum.

Applying Geometry to the Problem

Solving the distance problem with an understanding of geometry provides a complete view of what's occurring visually. We have a circle represented by

• \((x-4)^2 + y^2 = 4\)

and a point

• (0, 10)

we need to find the shortest line connecting these.
Once we find out \(x=4\), we substitute this back into the circle's equation, \((x-4)^2 + y^2 = 4\), to find the y-coordinate.
Calculating gives us \(y = 2\).
So, the closest point on the circle to the point (0, 10) is (4, 2).
This geometry problem taps into coordinate geometry, helping in visualizing the shortest path between a curve and a point.