Problem 9 - Calculus for Biology and Medicine

Question

Denote by \((x, y)\) a point on the straight line \(y=4-3 x .\) (See Figure \(5.55 .\) ) (a) Show that the distance from \((x, y)\) to the origin is given by $$ f(x)=\sqrt{x^{2}+(4-3 x)^{2}} $$ (b) Give the coordinates of the point on the line \(y=4-3 x\) that is closest to the origin. (Hint: Find \(x\) so that the distance you computed in (a) is minimized.) (c) Show that the square of the distance between the point \((x, y)\) on the line and the origin is given by $$ g(x)=[f(x)]^{2}=x^{2}+(4-3 x)^{2} $$ and find the minimum of \(g(x)\). Show that this minimum agrees with your answer in (b).

Step-by-Step Solution

Answer

The point closest to the origin is \( \left( \frac{6}{5}, \frac{2}{5} \right) \).

Step 1: Express the Distance from (x, y) to the Origin

The distance from the point \(x, y\) to the origin can be found using the formula for distance between two points. Here, the point is \( (x, y) \) and the origin is \( (0, 0) \). Using the Pythagorean theorem, the distance \( f(x) \) is given by: \f(x) = \sqrt{x^2 + (y - 0)^2} = \sqrt{x^2 + y^2}.\Substituting the equation of the line \( y = 4 - 3x \), we get: \f(x) = \sqrt{x^2 + (4 - 3x)^2}. \This completes part (a) of the exercise.

Step 2: Differentiate f(x) and Find Critical Points

To find the point on the line closest to the origin, minimize \( f(x) \). First, to simplify the work, observe the problem hint. Consider minimizing \( [f(x)]^2 \) which equals \( x^2 + (4 - 3x)^2 \). Let's consider this as \( g(x) \): \g(x) = x^2 + (4 - 3x)^2 = x^2 + 16 - 24x + 9x^2 = 10x^2 - 24x + 16.Differentiate \( g(x) \) with respect to \( x \), and find where its derivative is zero to locate the minimum point. g'(x) = \frac{d}{dx}(10x^2 - 24x + 16) = 20x - 24.Setting \( g'(x) = 0 \) gives: \20x - 24 = 0 x = \frac{24}{20} = \frac{6}{5}.This \( x \) value will be used to determine the y-coordinate of the closest point.

Step 3: Calculate y-coordinate and Verify Closest Point

Substitute \( x = \frac{6}{5} \) back into the line equation \( y = 4 - 3x \) to find \( y \): \y = 4 - 3\left(\frac{6}{5}\right) = 4 - \frac{18}{5} = \frac{20}{5} - \frac{18}{5} = \frac{2}{5}.So, the point on the line closest to the origin is \( \left( \frac{6}{5}, \frac{2}{5} \right) \). This verifies part (b) of the exercise.

Step 4: Verify and Minimize g(x)

Given the expression \( g(x) = x^2 + (4 - 3x)^2 = 10x^2 - 24x + 16 \), compute \( g(x) \) at the critical point \( x = \frac{6}{5} \). g\left(\frac{6}{5}\right) = 10\left(\frac{6}{5}\right)^2 - 24\left(\frac{6}{5}\right) + 16.Calculate it explicitly: = 10\left(\frac{36}{25}\right) - 24\left(\frac{6}{5}\right) + 16 = \frac{360}{25} - \frac{144}{5} + 16 = 14.4 - 28.8 + 16 = 1.6. The minimum value 1.6 agrees with the calculated point coordinates for minimizing distance. Thus, this confirms part (c) matching answer from (b).

Key Concepts

Distance Formula

The concept of finding the distance between two points is an essential tool in geometry and calculus. It allows us to calculate how far apart points are from one another. In this context, we're finding the distance from a point \((x, y)\) on the line \(y = 4 - 3x\) to the origin \((0, 0)\). To calculate this distance, we use the distance formula derived from the Pythagorean Theorem.
Here’s how it works:

• The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
  \[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
• In our example, substituting the points \((x, y)\) and the origin \((0, 0)\), the formula becomes:
  \[\sqrt{x^2 + y^2}\]
• Given the line equation \(y = 4 - 3x\), substitute this for \(y\) in the formula to get the expression:
  \[f(x) = \sqrt{x^2 + (4 - 3x)^2}\]

This expression represents the distance from any point on the line \(y = 4 - 3x\) to the origin, ready for further analysis like finding the closest point.

Derivatives

Derivatives are a fundamental concept in calculus, used to understand change and motion. In this exercise, derivatives help us find the critical points of a function, which indicate locations where the function changes direction. This is important for finding minimum distances, like the shortest distance from a point on our line to the origin.
To begin with:

• We want to minimize \(f(x) = \sqrt{x^2 + (4 - 3x)^2}\). However, it's easier to minimize \([f(x)]^2 = g(x) = x^2 + (4 - 3x)^2\) because it simplifies the derivative process.
  Why minimize \(g(x)\) instead of \(f(x)\)? - The square is easier to handle since it removes the square root.
• Compute \(g(x) = 10x^2 - 24x + 16\), which is a quadratic polynomial.
  
• Find the derivative \(g'(x) = 20x - 24\).
• Set \(g'(x) = 0\) to get critical points:
  \[20x - 24 = 0 \implies x = \frac{6}{5}\]

This derivative process pinpoints \(x = \frac{6}{5}\) as where the smallest distance to the origin occurs, indicating a turning point in the function.

Critical Points

Critical points are the values of \(x\) where a function’s derivative equals zero or doesn't exist. They are crucial in determining where a function might achieve a minimum or maximum value.
Here's how we use them in this context:

• We established that \(x = \frac{6}{5}\) makes the derivative of \(g(x)\) zero, which is indeed a critical point.
  
• This critical point indicates the possible location for the minimum value of \(g(x)\), since \(g'(x)\) changes sign around this point.
  
• By substituting \(x = \frac{6}{5}\) back into the line equation \(y = 4 - 3x\), we find the corresponding \(y\) value:
  \[y = 4 - 3\left(\frac{6}{5}\right) = \frac{2}{5}\]
• Thus, the point \(\left(\frac{6}{5}, \frac{2}{5}\right)\) is where the line is closest to the origin, confirming it's the critical point where \(g(x)\) reaches its minimum.
These calculations illustrate how critical points help us find the smallest distance from the line to the origin.