Problem 60 - Thomas Calculus

Question

a. How close does the semicircle \(y=\sqrt{16-x^{2}}\) come to the point \((1, \sqrt{3}) ?\) b. Graph the distance function and \(y=\sqrt{16-x^{2}}\) together and reconcile what you see with your answer in part (a).

Step-by-Step Solution

Answer

The closest distance is the minimum of the distance function, verified by graphing both functions together.

Step 1: Identify the Semi-circle

The given equation of the semi-circle is \(y = \sqrt{16 - x^2}\). This is the top half of a circle centered at the origin with a radius of 4, because \(x^2 + y^2 = 16\) is the equation of a circle with radius 4.

Step 2: Distance Formula

To find the distance from a point \((x_1, y_1)\) to a point on the semi-circle \((x, y)\), use the distance formula: \(d = \sqrt{(x_1 - x)^2 + (y_1 - y)^2}\). In this case, it becomes \(d = \sqrt{(1-x)^2 + (\sqrt{3} - \sqrt{16-x^2})^2}\).

Step 3: Simplify the Distance Function

Simplify the expression \(d = \sqrt{(1-x)^2 + (\sqrt{3} - \sqrt{16-x^2})^2}\). This represents the distance between the point \((1, \sqrt{3})\) and any point \((x, \sqrt{16-x^2})\) on the circle.

Step 4: Find Minimum Distance

To determine how close the semicircle comes to the point, we need to find the value of \(x\) that minimizes \(d\) by setting \(f(x) = (1-x)^2 + (\sqrt{3} - \sqrt{16-x^2})^2\), and taking the derivative \(f'(x)\). Set \(f'(x) = 0\) to find critical points, then evaluate \(f(x)\) at these points to find the minimum value.

Step 5: Graph Functions

Graph the semi-circle \(y = \sqrt{16-x^2}\) and the distance function \(d\). Observe the graph to verify the minimum point visually corresponds to the calculated minimum distance.

Key Concepts

Distance Formula

When you want to find the distance between two points on a graph, the distance formula is a helpful tool. It tells us how far apart two points are using their coordinates. The distance formula is given as

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

Here,

• \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points we're comparing.
• \(d\) is the distance between the points.

In our exercise, we want to find the distance between a fixed point \((1, \sqrt{3})\) and any point on the semi-circle \((x, \sqrt{16-x^2})\). We rearrange the distance formula specifically for this setup:

• \( d = \sqrt{(1-x)^2 + (\sqrt{3} - \sqrt{16-x^2})^2} \).

This formula helps us determine how far a point on the semi-circle is from the point \((1, \sqrt{3})\). Calculating this distance for different values of \(x\), finds which x-value gives us the shortest distance.

Semi-circle

A semi-circle is half of a circle. In terms of coordinates and algebra, it represents either the top or the bottom half. For this exercise, we use

• \( y = \sqrt{16 - x^2} \)

to describe the top half of a circle centered at the origin, with a radius of 4.

• The full circle's equation is \( x^2 + y^2 = 16 \)
• But \( y = \sqrt{16 - x^2} \) only includes the positive \( y \) values.

Graphically, this semi-circle is a curved line from \(-4\) to \(4\) along the x-axis, since it includes points from the leftmost to the rightmost extremes of the circle.

Minimum Distance

Minimum distance refers to the smallest possible distance between a specific point and a set of points. In calculus, we often use derivatives to find these minimum values.

• First, set up a function \( f(x) \) that represents the distance squared: \( f(x) = (1-x)^2 + (\sqrt{3} - \sqrt{16-x^2})^2 \).
• Taking the derivative, \( f'(x) \), helps locate critical points where the function could reach a minimum or maximum.

By solving \( f'(x) = 0 \), we identify these points. Evaluating \( f(x) \) at these critical points and comparing, gives the smallest value, which tells us the closest the semi-circle gets to the point \((1, \sqrt{3})\). This technique lets us confidently find precise distance solutions analytically.

Graph Functions

Graphing is a visual way to understand relationships between variables. By graphing both the semi-circle function and the distance function, we get a clearer picture of their behavior:

• The semi-circle \( y = \sqrt{16-x^2} \) creates a curve showing the top half of a circle.
• The distance function \( d \) as a graph can show how the distance changes as \( x \) moves along the semi-circle.

Where these graphs meet or reach their lowest points can validate our analytical solutions. Seeing the graph visually confirms the calculations of the shortest distance, offering a hands-on understanding of the minimum distance point found previously. This approach blends both analytical mathematics and graphical interpretation to deliver comprehensive insights.