Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the points on the parabola \(y=x^{2}\) that are closest to the point \((0,5) .\) Hint: Minimize the square of the distance between \((x, y)\) and \((0,5)\)

Step-by-Step Solution

Answer

The points closest to (0,5) are \(\left(\sqrt{\frac{9}{2}}, \frac{9}{2}\right)\) and \(\left(-\sqrt{\frac{9}{2}}, \frac{9}{2}\right)\).

Step 1: Define the Distance Formula

To determine the points on the parabola that are closest to the point \((0, 5)\), we start with the distance formula. The distance \(d\) between two points \((x, y)\) and \((0, 5)\) is given by \[d = \sqrt{(x - 0)^2 + (y - 5)^2}\].

Step 2: Substitute the Parabola's Equation

Since the point \((x, y)\) lies on the parabola \(y = x^2\), we substitute \(y = x^2\) into the distance formula. Now the distance formula becomes \[d = \sqrt{x^2 + (x^2 - 5)^2}\].

Step 3: Simplify the Expression

We simplify the expression under the square root: \(x^2 + (x^2 - 5)^2 = x^2 + (x^4 - 10x^2 + 25)\). This results in \[d = \sqrt{x^4 - 9x^2 + 25}\].

Step 4: Minimize the Square of the Distance Function

Instead of minimizing \(d\), we can minimize its square \(d^2\) to avoid dealing with the square root. The function becomes \[f(x) = x^4 - 9x^2 + 25\].

Step 5: Differentiate the Function

Find the derivative of the function to find its critical points. Calculate \(f'(x)\): \[f'(x) = 4x^3 - 18x\].

Step 6: Solve for Critical Points

Set the derivative equal to zero to solve for \(x\): \(4x^3 - 18x = 0\). Factor this expression: \[4x(x^2 - \frac{9}{2}) = 0\].

Step 7: Find Possible Values for x

The solutions are \(x = 0\) or \(x^2 = \frac{9}{2}\). Therefore, the values of \(x\) are \(x = 0\), \(x = \sqrt{\frac{9}{2}}\), and \(x = -\sqrt{\frac{9}{2}}\).

Step 8: Calculate Corresponding y-Coordinates

Calculate the \(y\)-coordinates for each value of \(x\) using the parabola equation \(y = x^2\). For \(x = 0\), \(y = 0\). For \(x = \sqrt{\frac{9}{2}}\), \(y = \frac{9}{2}\). For \(x = -\sqrt{\frac{9}{2}}\), \(y = \frac{9}{2}\).

Step 9: Evaluate Distance at Each Point

Calculate the distance \(d\) for each point: \(d(0, 0) = 5\), \(d\left(\sqrt{\frac{9}{2}}, \frac{9}{2}\right) = \sqrt{\frac{1}{4}} = \frac{1}{2}\), and the same for \(d\left(-\sqrt{\frac{9}{2}}, \frac{9}{2}\right)\).

Step 10: Identify the Points Closest to (0,5)

The smallest distance is \(\frac{1}{2}\), so the points \(\left(\sqrt{\frac{9}{2}}, \frac{9}{2}\right)\) and \(\left(-\sqrt{\frac{9}{2}}, \frac{9}{2}\right)\) are the closest to \((0, 5)\).

Key Concepts

Distance Formula

The distance formula is a crucial tool in calculus optimization problems to measure how far apart two points are in a plane. To find the distance between a point \(x, y\) and another point, say \(0, 5\), we use the formula:\[d = \sqrt{(x - 0)^2 + (y - 5)^2}\]This formula calculates the direct "straight-line" distance. In our problem, since the point \(x, y\) is on the parabola \(y = x^2\), we can substitute this equation into the formula. This substitution simplifies our task as it lets us express everything in terms of a single variable, \(x\). This approach reduces complexity, paving the way to find the closest points effectively.

Critical Points

Critical points are fundamental in finding the optimal solutions in calculus. Once you have a function, such as the squared distance in minimization problems, you need to locate these points where the rate of change of the function slows down or stops. Typically, you find them by setting the function's derivative to zero and solving the resulting equation.

• Find the derivative of the function with respect to the variable.
• Set the derivative equal to zero to identify potential critical points.
• Solve the equation to find values of the variable that make the derivative zero.

In this problem, setting the derivative of \(f(x) = x^4 - 9x^2 + 25\) equal to zero helped locate the \(x\)-values corresponding to the parabola's critical points where the distance is minimized.

Derivative

The derivative is like a journal of how a function behaves. It tells you how a function is changing at any point, allowing us to assess where it speeds up, slows down, or even reverses direction. Finding a derivative involves calculating how small changes in \(x\) affect the function. In optimization, this is critical to understanding and finding minimum or maximum values.To calculate the derivative for the distance squared function \(f(x) = x^4 - 9x^2 + 25\), follow these steps:

• Differentiate each term in the function: \(\frac{d}{dx}(x^4) = 4x^3\) and \(\frac{d}{dx}(-9x^2) = -18x\).
• Combine the results to get \(f'(x) = 4x^3 - 18x\).

This derivative then guides us in finding the points at which the function has the potential to be minimized, helping solve the problem.

Parabolas

Parabolas are elegant curves that arise in various mathematical problems. They're defined by their simple quadratic equation form, such as \(y = x^2\) in our example. Here, the point \(x, y\) lies on a parabola, meaning \(y\) is always equal to \(x^2\). These curves have some unique properties, including:

• A vertex, which is the point where the parabola changes direction.
• Symmetry, typically about a vertical line that runs through the vertex.
• The shape, which looks like a U, can open up or down depending on the sign of the quadratic term.

In optimization problems, such as finding the closest point on a parabola, these properties determine how you'll find and interpret critical points and distances from other points. The parabola's symmetry and shape guide you through correctly understanding its behavior when seeking solutions.