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Chapter 4: Problem 72

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

Short Answer

Expert verified
The semicircle comes closest to the point \((1, \sqrt{3})\) at the shortest distance found from the minimal value of the distance function. Graph both functions to see the minimum distance visually.

Step by step solution

01

Understand the Semicircle Equation

The equation \( y = \sqrt{16 - x^2} \) represents the upper half, or the semicircle, of a circle centered at the origin with a radius of 4. This is because \( x^2 + y^2 = 16 \) is the general equation of a circle centered at the origin with radius 4.
02

Distance Formula Setup

To find how close the semicircle comes to the point \((1, \sqrt{3})\), we use the distance formula \( D = \sqrt{(x - 1)^2 + (y - \sqrt{3})^2} \). Substitute \( y = \sqrt{16 - x^2} \) into this expression to get a distance function in terms of \( x \).
03

Set the Distance Function

The function for the distance between a point on the semicircle and \((1, \sqrt{3})\) is \[ D(x) = \sqrt{(x - 1)^2 + (\sqrt{16 - x^2} - \sqrt{3})^2} \].
04

Minimize the Distance Function

To find the minimum distance, take the derivative of \( D(x) \) with respect to \( x \), set it to zero, and solve for \( x \). This will give the point on the semicircle that is closest to \((1, \sqrt{3})\).
05

Find Critical Points

Simplify and differentiate \( D(x) \), and determine the critical points by solving \( \frac{dD}{dx} = 0 \). Check these points to find the minimum distance.
06

Evaluate the Distance

Substitute the critical points found back into \( D(x) \) to determine the numerical value of the minimum distance. This will give the shortest distance from the point \((1, \sqrt{3})\) to the semicircle.
07

Graphing the Functions

Graph the semicircle \( y = \sqrt{16 - x^2} \) and the distance function. Observe the behavior of the distance as \( x \) varies, confirming the minimum point found in the earlier steps.
08

Reconcile Observation with Analytical Findings

Ensure the graphical representation aligns with the analytical solution, confirming that the minimum distance corresponds to the closest approach on the graph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Semicircle Equation
The equation of the semicircle, given as \( y = \sqrt{16 - x^2} \), represents the upper half of a circle. This circle is centered at the origin \((0, 0)\) and has a radius of 4. We derive this semicircle equation from the general circle equation \( x^2 + y^2 = 16 \), which explains the circular shape.
  • The full circle has a diameter of 8, since the radius is 4.
  • The semicircle extends from \(x = -4\) to \(x = 4\) along the x-axis.
  • Only positive y-values are considered, creating the 'top' half of the circle as the semicircle.
This understanding is crucial for dealing with the distance problem and further graphical analysis.
Minimizing Distance with Calculus
To determine how closely the semicircle approaches the point \((1, \sqrt{3})\), we use the distance formula. The key is to express this distance as \(D(x) = \sqrt{(x - 1)^2 + (\sqrt{16 - x^2} - \sqrt{3})^2}\). This distance formula helps measure the gap between a point \((x, y)\) on the semicircle and \((1, \sqrt{3})\).
  • First, substitute \( y = \sqrt{16 - x^2} \) into the distance formula to express distance as a function of \( x \).
  • The challenge is to find the \( x \) that minimizes this distance.
For optimization, apply calculus by differentiating \(D(x)\) with respect to \(x\), and find where the derivative equals zero. This indicates a potential minimum. This step is crucial as it allows us to find the optimal \(x\) where the semicircle is closest to \((1, \sqrt{3})\).
Using Derivatives to Find Minimum Distance
The derivative of a function shows how the function value changes as its input changes. For our distance function \(D(x)\), taking its derivative helps us identify critical points where \(D(x)\)'s slope is zero, indicating potential minimum distances.
  • Differentiate \(D(x)\) with respect to \(x\). This step can be algebraically complex but critical for precise analysis.
  • Set \( \frac{dD}{dx} = 0 \) to solve for \(x\), finding critical points that suggest minimum or maximum values.
  • Test each critical point to confirm if they are truly minimizing the distance by evaluating \(D(x)\) at these points.
Through solving \( \frac{dD}{dx} = 0 \), you find the \(x\)-coordinate, which can then be used to find \(y\) using the semicircle equation. This analysis ultimately gives the exact coordinates on the semicircle closest to \((1, \sqrt{3})\).
Graphing Functions for Visual Insight
Graphing both the semicircle equation \( y = \sqrt{16 - x^2} \) and the distance function \(D(x)\) provides a graphical perspective on the problem.
  • The semicircle graph represents all possible points that could be the closest to \((1, \sqrt{3})\).
  • Plotting \(D(x)\) alongside helps visualize how the distance varies across different points on the semicircle.
  • When these functions intersect at the minimum point, it visually confirms the calculations for minimum distance.
A graph effectively supports the analytical solution, allowing students to see how the minimum distance found mathematically aligns with the closest approach on the graph.

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Most popular questions from this chapter

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