Problem 21 - Single Variable Calculus

Question

Find the point on the line \(y=2 x+5\) that is closest to the origin.

Step-by-Step Solution

Answer

The point on the line \(y = 2x + 5\) that is closest to the origin is \((-2, 1)\).

Step 1: Define the Distance Formula

Recall the distance formula for finding the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\): \[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] In our case, one of the points is the origin \((0,0)\), and the other point is on the line \(y = 2x + 5\), let's call it (x, y).

Step 2: Substitute the Line Equation into the Distance Formula

Since the point (x, y) lies on the line \(y = 2x + 5\), we can substitute this equation into the distance formula: \[D = \sqrt{(x - 0)^2 +((2x + 5) - 0)^2}\] Simplify the expression: \[D = \sqrt{x^2 + (2x + 5)^2}\]

Step 3: Minimize the Distance

In order to minimize the distance, we can take the derivative of the distance function D with respect to x and set it equal to 0, then solve for x. Note that it is easier to minimize \(D^2\) instead of D since minimizing the square of the distance will result in the same x value that minimizes the distance. So, we have: \[D^2 = x^2 + (2x + 5)^2\] Now, take the derivative of \(D^2\) with respect to x: \[\frac{d(D^2)}{dx} = 2x + 4(2x + 5)\]

Step 4: Set the Derivative Equal to Zero and Solve for x

Now, we want to find the x value that minimizes the distance, so we set the derivative equal to 0 and solve for x: \[2x + 4(2x + 5) = 0\] Simplify, and solve for x: \[10x + 20 = 0\] \[x = -2\]

Step 5: Find the Corresponding y Value

Now that we have the x value, we can find the corresponding y value by substituting x into the line equation: \[y = 2(-2) + 5\] \[y = -4 + 5\] \[y = 1\]

Step 6: State the Solution

The point on the line \(y = 2x + 5\) that is closest to the origin is \((-2, 1)\).

Key Concepts

Distance Formula

The Distance Formula is a crucial part of this problem. It helps us find how far apart two points are in a coordinate plane. To use it, you need coordinates for both points. The formula is: \[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Imagine you want to find the distance between the origin \(0,0\) and another point \(x,y\). This point also lies on a line. By plugging in the coordinates, the formula transforms to help us solve for a specific problem: finding the closest point on a specific line to the origin.

Derivative

Derivatives show how a function changes as its inputs change. Here, we use it to minimize the distance.Instead of working directly with the distance \(D\), we minimize \(D^2\), the squared distance. This is because \(D^2\) simplifies our calculations without altering the point of minimum distance. Taking the derivative of \(D^2\) with respect to \(x\) provides a function we can solve. By setting this derivative to zero, we find the precise x-value where distance from the origin is smallest.

Line Equation

A line equation gives us the relationship between x and y for all points on a line. In this exercise, the line equation is \[y = 2x + 5\]This formula tells us that for any x-value we choose, we can find its corresponding y-value on the line. In our context, solving for x with this equation shows the exact location on the line where distance to the origin is minimized. Once x is known, a simple substitution in the line equation gives us the y-value.

Optimization

Optimization involves finding the best or most efficient solution to a problem. In this exercise, finding the closest point on a line to the origin is an optimization task.Our approach involves several steps:

• Express the distance to the origin in terms of x.
• Use the derivative to find where the distance is minimized.
• Confirm the solution meets all the problem's requirements.

Minimizing the squared distance using calculus led us to a point \((-2, 1)\), which achieves the problem's goal effectively.