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

Find critical points by differentiating the distance squared, then use second derivative test to identify closest and farthest points.

Step 1: Understand the problem

We are given a quadratic function describing the path of an object thrown from a cliff. We need to find where this object is closest and farthest from an observer standing 3 feet away from the cliff's base.

Step 2: Write the distance formula

We calculate the distance between a point on the object's path \((x, y)\) and the observer located at \((3, 0)\). The distance formula is: \[D = \sqrt{(x - 3)^2 + (y - 0)^2}\]Substitute the expression for \(y\) from the object's path into the distance formula.

Step 3: Substitute the expression for y

Substitute the expression for \(y\) in the distance formula:\[y = -\frac{2x^2}{25} + x + 42\]So the distance \(D\) becomes:\[D = \sqrt{(x - 3)^2 + \left(-\frac{2x^2}{25} + x + 42\right)^2}\]

Step 4: Optimize the distance

To find the minimum and maximum distance, we need to minimize and maximize the square of the distance for simplicity, hence:\[D^2 = (x - 3)^2 + \left(-\frac{2x^2}{25} + x + 42\right)^2\]Simplifying \(D^2\) using the expression for \(y\), and take the derivative with respect to \(x\) and set it equal to zero to find the critical points.

Step 5: Differentiate and find critical points

Differentiate \(D^2\) with respect to \(x\), solve \(\frac{d}{dx}(D^2) = 0\). This yields the critical points, which indicate potential positions of closest and farthest distances.

Step 6: Calculate second derivative test

Calculate the second derivative \(\frac{d^2}{dx^2}(D^2)\) at the critical points to determine if they are minima or maxima. A positive value at a point indicates a local minimum (closest point), and a negative value indicates a local maximum (farthest point).

Step 7: Confirm and interpret results

Identify the values of \(x\) from the derivative tests that correspond to closest and farthest distances. Calculate corresponding values of \(y\) using the path equation. Interpret these coordinates in the context of the problem to provide the final answer.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that can be expressed in the form: \[ f(x) = ax^2 + bx + c \] where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In the context of this exercise, the path of the object is described by the quadratic function:\[ y = -\frac{2x^2}{25} + x + 42 \] This function models the path of an object thrown from a cliff, which is a parabola opening downwards due to the negative sign in front of \(x^2\). When examining quadratic functions:

• The vertex represents the highest or lowest point of the parabola, which is crucial in optimization problems.
• The roots or zeros are the values where the function crosses the x-axis, though not directly relevant here.

Understanding how to manipulate and interpret quadratic equations is vital in determining positions of minimum and maximum distances in the problem.

Distance Formula

The distance formula is a fundamental tool in geometry and calculus used to calculate the distance between two points in the Cartesian plane. It is derived from the Pythagorean theorem and is given by:\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] In this exercise, we calculate the distance between a generic point \((x, y)\) on the object's path and the observer at the point \((3, 0)\). Thus, the distance formula becomes:\[ D = \sqrt{(x - 3)^2 + (y - 0)^2} \] Substituting for \(y\) using the equation of the object's path, the formula expands to:\[ D = \sqrt{(x - 3)^2 + \left(-\frac{2x^2}{25} + x + 42\right)^2} \] This equation allows us to determine how far the object is from the observer at any point along its trajectory. Adjusting this formula is the key step for the optimization process.

Critical Points

In calculus, critical points are the values of \(x\) where the derivative of a function is zero or undefined. They are essential in finding local minima, maxima, or saddle points of a function. For this problem, we apply critical points to the squared distance formula, \(D^2\), to simplify calculations and optimize:\[ D^2 = (x - 3)^2 + \left(-\frac{2x^2}{25} + x + 42\right)^2 \] To find the critical points, we:

• Compute the derivative \(\frac{d}{dx}(D^2)\)
• Set the derivative equal to zero: \[ \frac{d}{dx} (D^2) = 0 \] Solve this equation for \(x\)

These critical points represent the potential positions where the object could be closest or farthest from the observer. Further analysis, such as the derivative test, helps determine the nature of these points.

Derivative Test

The derivative test is a method used to determine whether a critical point of a function is a minimum, maximum, or a point of inflection. In this exercise, after finding the critical points by differentiating \(D^2\), we use the second derivative test to ascertain which is which.To apply the derivative test, you:

• Calculate the second derivative \(\frac{d^2}{dx^2}(D^2)\).
• Evaluate the second derivative at each critical point:

- If \(\frac{d^2}{dx^2}(D^2) > 0\), the function has a local minimum at that critical point (closest distance).- If \(\frac{d^2}{dx^2}(D^2) < 0\), the function has a local maximum at that critical point (farthest distance).This method helps distinguish whether the object is, at any given critical point, at its minimum or maximum distance from the observer. Properly applying the derivative test thus concludes the optimization process in your calculus problem.