Problem 11 - Calculus for Biology and Medicine

Question

How close does the curve \(y=1 / x\) come to the origin? (Hint: Find the point on the curve that minimizes the square of the distance between the origin and the point on the curve. If you use the square of the distance instead of the distance, you avoid dealing with square roots.)

Step-by-Step Solution

Answer

The curve comes closest to the origin at a distance of \\sqrt{2}\\.

Step 1: Square of the Distance Formula

To find how close the curve comes to the origin, we need to calculate the distance from a point \(x, y\) on the curve to the origin \(0, 0\). The square of the distance is given by the formula \(D^2 = x^2 + y^2\). Now substitute \(y = \frac{1}{x}\) into this formula.

Step 2: Substitute the Curve Equation

Substitute \(y = \frac{1}{x}\) into \(D^2 = x^2 + y^2\) to get \(D^2 = x^2 + \frac{1}{x^2}\). The goal is to minimize this expression \(D^2\).

Step 3: Set Up the Function to Minimize

The function we want to minimize is \(f(x) = x^2 + \frac{1}{x^2}\). We will do this by finding the derivative of \(f(x)\) and setting it equal to zero to find its critical points.

Step 4: Derive the Function

Compute the derivative: \(f'(x) = 2x - 2x^{-3}\). We differentiate \(x^2\) and \(rac{1}{x^2}\). The derivative of \(x^2\) is \(2x\), and the derivative of \(rac{1}{x^2}\) is \(-2x^{-3}\).

Step 5: Find Critical Points

Set \(f'(x) = 2x - 2x^{-3} = 0\) and solve for \(x\). Rearranging gives \(2x = 2x^{-3}\), and simplifying gives \(x^4 = 1\). Thus we find \(x = 1\) or \(x = -1\).

Step 6: Evaluate Distance at Critical Points

When \(x = 1\), \(y = \frac{1}{1} = 1\). Calculate \(D^2 = 1^2 + 1^2 = 2\). When \(x = -1\), \(y = \frac{1}{-1} = -1\). Similarly, \(D^2 = (-1)^2 + (-1)^2 = 2\).

Step 7: Conclusion

The smallest value for the square of the distance \(D^2\) from the origin is 2, which corresponds to a distance of \(\sqrt{2}\) when considering either critical point.

Key Concepts

Distance Minimization

Understanding how to find the closest point on a curve to the origin is a crucial concept in calculus called distance minimization. This involves finding a point on a given curve where the distance to the origin, or any particular point, is the smallest possible. To make the calculations simpler, it's common to use the square of the distance instead of the actual distance formula.
Why do we use the square of the distance? Using the square of the distance equation, instead of the distance equation itself, eliminates dealing with cumbersome square roots. For a point \((x, y)\) on the curve and the origin \((0, 0)\), the square of the distance is calculated as \(D^2 = x^2 + y^2\). In this case, since \(y = \frac{1}{x}\) along the curve \(y=1 / x\), we substitute to get \(D^2 = x^2 + \frac{1}{x^2}\).
The objective is to adjust \(x\) such that the expression for \(D^2\) is as small as possible. This sets the stage for using calculus, where our ultimate goal becomes minimizing this expression through derivative calculations and critical point analysis.

Derivative Calculation

To successfully minimize the distance, we need to calculate the derivative of the function we're aiming to minimize. The function specified for this problem is \(f(x) = x^2 + \frac{1}{x^2}\). The derivative of the function gives us the rate of change of the distance, allowing us to find points where the derivative is zero (potential minimum points).
Let's go step-by-step:

• The derivative of \(x^2\) is straightforward; it is \(2x\).
• The derivative of \(\frac{1}{x^2}\) is a bit more complex. It derives to \(-2x^{-3}\) using the power rule.

Combine these to find the derivative of the full function: \(f'(x) = 2x - 2x^{-3}\). This derivative will be used to find critical points that potentially minimize the distance. Setting \(f'(x)\) to zero and solving for \(x\) will provide those key points.

Critical Points Analysis

Critical points analysis is the process by which we identify candidates for maximum, minimum, or saddle points of a function. After calculating the derivative \(f'(x) = 2x - 2x^{-3}\), the next step is setting this equation to zero to find significant \(x\) values where the smallest distance may occur.
Solve \(2x = 2x^{-3}\), simplifying to get \(x^4 = 1\). This gives us potential solutions for \(x = 1\) and \(x = -1\).
For these critical points, substitute back into the original equation for \(D^2 = x^2 + \frac{1}{x^2}\):

• When \(x = 1\), \(y = 1\) and \(D^2 = 1^2 + 1^2 = 2\).
• Similarly for \(x = -1\), \(y = -1\) and again, \(D^2 = 1 + 1 = 2\).

The conclusion here is that the equation demonstrates a minimum \(D^2\) of 2 for both critical points. Thus, the closest distance from the curve \(y = \frac{1}{x}\) to the origin is \(\sqrt{2}\), validating our calculus-driven solution.