Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object thrown from the edge of a 42 -foot cliff follows the path given by \(y=-\frac{2 x^{2}}{25}+x+42\) (Figure 10 ). An observer stands 3 feet from the bottom of the cliff. (a) Find the position of the object when it is closest to the observer. (b) Find the position of the object when it is farthest from the observer.

Step-by-Step Solution

Answer

Closest position: solve derivative for min, substitute in path equation; Farthest position: solve for max, use path equation.

Step 1: Understand the Problem

We need to find the position of the object on the path described by the equation \( y = -\frac{2x^2}{25} + x + 42 \). The observer is 3 feet from the base of the cliff at the origin (0,0). First, we calculate the distance from the observer \((3,0)\) to any point \((x, y)\) on the trajectory of the object.

Step 2: Distance Formula

To find when the object is closest or farthest from the observer, we use the distance formula between the observer \((3,0)\) and the object \((x, \frac{-2x^2}{25} + x + 42)\). The distance \(D\) is given by:\[ D = \sqrt{(x - 3)^2 + \left(-\frac{2x^2}{25} + x + 42\right)^2} \]

Step 3: Simplify Distance Expression

Since we want to find when \(D\) is minimized or maximized, it's enough to minimize/maximize \(D^2\) instead:\[ D^2 = (x - 3)^2 + \left(-\frac{2x^2}{25} + x + 42\right)^2 \]

Step 4: Set Up Derivative to Find Critical Points

Find the derivative \(\frac{d}{dx}(D^2)\) and set it to zero to locate critical points. This allows us to identify where the distance is minimized or maximized.

Step 5: Calculate the Derivative

Compute the derivative of \(D^2\) with respect to \(x\), apply the quotient and product rules where necessary.

Step 6: Solve the Derivative Equation

Solve \(\frac{d}{dx}(D^2) = 0\) for \(x\) to find critical points. Analyze these points to determine which one corresponds to the minimum distance (closest) and the maximum distance (farthest) by testing them in the second derivative or considering endpoints of the trajectory.

Step 7: Evaluate the y-coordinate

Substitute the \(x\) values found in Step 6 into the equation \(y = -\frac{2x^2}{25} + x + 42\) to find their corresponding \(y\) coordinates.

Step 8: Identify the Positions

Identify the position \((x, y)\) that corresponds to the minimum \(D^2\) for the closest point, and the position with the maximum \(D^2\) for the farthest point.

Key Concepts

Distance Formula

When we want to find out how far two points are from each other, we use the distance formula. It's a mathematical way of computing the distance using coordinates on the grid. To apply it, you need the coordinates of the two points in the form

• For the first point, \((x_1, y_1)\)
• And for the second point, \((x_2, y_2)\)

The formula is:\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
So, if you're finding the distance from the observer ((3,0)) to a point on a path (\((x, y)\)), simply substitute these points into the formula.
The task of finding the closest or farthest point becomes one of minimizing or maximizing this distance, respectively.

Quadratic Equations

Quadratic equations are common in calculus optimization problems and appear in the form of \(ax^2 + bx + c = 0\). They represent parabolic paths.
In our problem, the object's path is a parabola described by the equation: \(y = -\frac{2x^2}{25} + x + 42\). Here,

• \(a = -\frac{2}{25}\)
• \(b = 1\)
• \(c = 42\)

This quadratic nature indicates a parabolic curve. Quadratic equations are key in determining the trajectory of the object. By solving them, we can find specific values of \(x\) that meet certain conditions, such as minimum or maximum points relative to distances.

Derivatives

When we use derivatives, we're looking at how a function changes. They tell us the "rate of change" or slope of the function at any point.
Calculus uses derivatives heavily to find where functions increase or decrease at the fastest rate. For the distance problem, we differentiate the function of \(D^2\), the squared distance, with respect to \(x\). This helps us find crucial values — the points where distance is minimized or maximized.
Understanding how derivatives work can help you identify trends in data or the landscape of a problem area. Learn derivatives, and you'll learn to see functions with a new kind of clarity.

Critical Points

Critical points in calculus are where the derivative of a function is zero or undefined. This is super useful because it tells us where important changes happen in a function's behavior.
In the context of this problem, we find critical points by

• Taking the derivative of the distance squared function (\(D^2\)),
• Setting that derivative equal to zero

By solving for \(x\), we pinpoint where the path is closest or farthest from the observer.
Checking these critical points — using either a second derivative test or plugging them back into the distance formula — helps decide if they correspond to minimum or maximum distances.