Problem 76 - Precalculus: Functions and Graphs

Question

Airplane maneuver An airplane is flying along the hyperbolic path illustrated in the figure. If an equation of the path is \(2 y^{2}-x^{2}=8,\) determine how close the airplane comes to a town located at \((3,0) .\) (Hint: Let \(S\) denote the square of the distance from a point \((x, y)\) on the path to \((3,0),\) and find the minimum value of \(S\).) PICTURE CANT COPY

Step-by-Step Solution

Answer

The closest distance the airplane comes to the town is \(\sqrt{7}\).

Step 1: Express the Distance

We need to find how close the airplane comes to the town located at \((3,0)\). Using the distance formula, the distance \(d\) from a point \((x, y)\) on the hyperbolic path to the town \((3,0)\) is given by \(d = \sqrt{(x - 3)^2 + y^2}\). To minimize the distance, we can instead minimize \(S = (x - 3)^2 + y^2\) since the square root is a monotonically increasing function.

Step 2: Use the Path Equation

Given the equation for the path is \(2y^2 - x^2 = 8\), we can express \(y^2\) in terms of \(x\) using the equation: \(y^2 = \frac{x^2 + 8}{2}\). This allows us to substitute \(y^2\) into the equation for \(S\).

Step 3: Substitute and Simplify

Substituting \(y^2 = \frac{x^2 + 8}{2}\) into \(S = (x - 3)^2 + y^2\), we get:\[S = (x - 3)^2 + \frac{x^2 + 8}{2}\]Simplify this to:\[S = (x - 3)^2 + \frac{x^2}{2} + 4\]

Step 4: Further Simplification

Expand \((x - 3)^2\) to get:\[S = (x^2 - 6x + 9) + \frac{x^2}{2} + 4\]Combining terms, this becomes:\[S = x^2 - 6x + 9 + \frac{x^2}{2} + 4\]or\[S = \frac{3x^2}{2} - 6x + 13\]

Step 5: Minimize the Function

To minimize \(S = \frac{3x^2}{2} - 6x + 13\), take the derivative \(\frac{dS}{dx}\), and set it to zero:\[\frac{dS}{dx} = 3x - 6 = 0\]Solve for \(x\):\[3x = 6\]\[x = 2\]

Step 6: Verify the Minimum

Check the second derivative \(\frac{d^2S}{dx^2}\) to verify it's indeed a minimum:\[\frac{d^2S}{dx^2} = 3\]Since \(\frac{d^2S}{dx^2}\) is positive, \(x = 2\) gives a local minimum.

Step 7: Determine Closest Distance

Substitute \(x = 2\) back into the path equation to find \(y\):\[2y^2 - 4 = 8\]\[2y^2 = 12\]\[y^2 = 6\]\[y = \sqrt{6}\] (considering the positive root for geometrical relevance).

Step 8: Calculate Distance

Now, calculate the distance using \(S\) at \(x = 2, y = \sqrt{6}\):\[S = (2 - 3)^2 + 6 = 1 + 6 = 7\]The closest distance is \(\sqrt{7}\).

Key Concepts

Distance Formula

To find how close the airplane comes to the town located at \((3,0)\), we start by using the Distance Formula. This formula helps us calculate the straight-line distance between two points in the coordinate plane. For this problem, if we have a point \((x, y)\) on the hyperbolic path, the distance \(d\) to the point \((3, 0)\) is given by the formula:

• \(d = \sqrt{(x - 3)^2 + y^2} \)

This formula essentially uses the Pythagorean theorem; it considers the horizontal and vertical distances from the point on the hyperbola to the point \((3,0)\) as the two legs of a right triangle. This visualization can be incredibly helpful in understanding why the formula works as it shows the distance as the hypotenuse of the triangle.
However, because finding the minimum distance involves cumbersome square roots, we often look at minimizing the square of the distance instead.

Square of Distance

By working with the square of the distance instead of the distance itself, calculations become more straightforward. This is because minimizing the distance \(d\) is the same as minimizing the square of the distance \(S\), given that the square root function is monotonically increasing. Thus, we define \(S\) as:

• \(S = (x - 3)^2 + y^2 \)

The goal is now to find the minimum value of \(S\), as this will correspond to the minimum distance \(d\) from the point \((3,0)\) to the hyperbolic path.
Moreover, this simplification to \(S\) eliminates the square root, allowing easier calculus manipulations to find the minimum value, especially when derivatives come into play in the next steps.

Minimization Problem

To solve the minimization problem, we have two tasks:

• Express \(S\) in terms of a single variable, \(x\).
• Use calculus to find the minimum value of \(S\).

Given the path equation \(2y^2 - x^2 = 8\), we solve for \(y^2\) as \(y^2 = \frac{x^2 + 8}{2}\). By substituting this into \(S = (x - 3)^2 + y^2\), we rewrite \(S\) solely in terms of \(x\):

• \(S = (x - 3)^2 + \frac{x^2 + 8}{2} \)

Simplifying further, \(S\) becomes

• \(S = \frac{3x^2}{2} - 6x + 13 \)

The minimization task next involves finding the derivative and setting it to zero, then using the value of \(x\) obtained to determine \(S\) and thus the minimum distance.

Second Derivative Test

Once we have the function \(S = \frac{3x^2}{2} - 6x + 13\), finding where \(S\) is minimized involves calculus. By taking the derivative, represented as \(\frac{dS}{dx}\), we identify critical points. First, set the first derivative to zero:

• \(\frac{dS}{dx} = 3x - 6 = 0\)

Solving gives \(x = 2\). To confirm this point gives a minimum, we apply the Second Derivative Test.

• The second derivative of \(S\) is \(\frac{d^2S}{dx^2} = 3\).

Because \(\frac{d^2S}{dx^2} > 0\), \(S\) is concave up at \(x = 2\). This indicates a local minimum. The test affirms not only the efficiency of calculus with algebraic equations but also aids in visualizing adjustments along curves to find extreme values such as this minimum distance to the town.