Question

(Calculator) Find a point on the parabola \(y=\frac{1}{2} x^{2}\) that is closest to the point (4,1)

Short answer

Answer: The point on the parabola \(y=\frac{1}{2}x^2\) that is closest to the point (4,1) is (4, 8).

Step-by-step solution

Write down the distance formula

The distance formula between two points (x1, y1) and (x2, y2) is \(d=\sqrt{(x2-x1)^2+(y2-y1)^2}\).

Plug in the given point (4,1) and the general point (x, y) on the parabola

Using the given point (4,1) and a point on the parabola \((x, \frac{1}{2}x^2)\), we can write the distance formula as \(d=\sqrt{(x-4)^2+(\frac{1}{2}x^2-1)^2}\).

Square the distance formula to get rid of the square root

Since we are only interested in minimizing the distance, we can remove the square root by squaring both sides, which will make it easier to work with: \(d^2 = (x-4)^2+(\frac{1}{2}x^2-1)^2\)

Take the derivative of the distance squared formula with respect to x

We will minimize the distance by finding the critical points of the derivative. Therefore, we need to find the derivative of the distance squared with respect to x: \(\frac{dd^2}{dx} = 2(x-4)+ (\frac{1}{2}x^2-1)(2x)\)

Set the derivative equal to zero and solve for x

To find the critical points, set the derivative equal to zero: \(\frac{dd^2}{dx}=0 \Rightarrow 2(x-4)+(\frac{1}{2}x^2-1)(2x)=0.\) To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, the equation simplifies to \(x^3 - 8x^2 + 16x = 0\), and we can factor an x out, getting \(x(x^2-8x+16)=0\). The solutions are \(x=0\) and \(x=4\).

Check which solution minimizes the distance

Plug in the critical points, x=0 and x=4, into the distance equation to find which critical point minimizes the distance. If x=0, we get \(d=\sqrt{(4)^2+(1)^2}=\sqrt{17}\), and if x=4, we get \(d=\sqrt{(4-4)^2+(0-1)^2} = 1\). Since the distance is minimized when x=4, the point on the parabola closest to (4,1) is the one with x=4.

Find the y-coordinate of the closest point on the parabola

Plug in x=4 into the equation of the parabola \(y=\frac{1}{2} x^2\) to find the y-coordinate of the closest point: \(y=\frac{1}{2}(4)^2 = 8\).

Write the final solution

The point on the parabola \(y=\frac{1}{2} x^2\) that is closest to the point (4,1) is (4, 8).

Key concepts

Parabola Analysis

A parabola is a simple, yet fascinating curve. It's the graph of a quadratic function, typically taking the form \( y = ax^2 + bx + c \). In our exercise, we analyze the parabola \( y = \frac{1}{2}x^2 \). This specific type of parabola is known as a "vertical parabola," as it opens upwards.
Additionally, the vertex, which is the highest or lowest point, is located at the origin (0,0) for this equation, as there are no \( bx \) or \( c \) terms to shift it.

• The symmetry of a parabola about the vertical line through its vertex helps in solving distance problems.
• By understanding the behavior of the parabola, such as its direction and vertex position, we can apply suitable mathematical techniques to determine points on it that meet certain conditions.

Distance Formula

The distance formula is a fundamental tool in geometry, used to measure the distance between two points in a plane. It stems from the Pythagorean theorem, which establishes the relationship between the sides of a right triangle.
The formula to calculate the distance \(d\) between two points, \((x_1, y_1)\) and \((x_2, y_2)\), is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
In our problem, this formula is applied to determine how far a general point \((x, \frac{1}{2}x^2)\) on the parabola is from a given fixed point, (4,1).

• The direct application of this formula helps us set up an equation to minimize and find the closest point on the parabola.
• By squaring the distance formula, we simplify derivation while preserving the location of minimum distance.

Derivative Application

The calculus concept of derivatives is integral to solving optimization problems, such as finding the closest point on a curve. In this case, we apply the derivative to minimize the distance.
Once we have the equation of the distance squared, \(d^2 = (x-4)^2 + (\frac{1}{2}x^2 - 1)^2\), we take its derivative with respect to \(x\).
By finding \( \frac{d(d^2)}{dx} \), we ascertain how the squared distance changes as \( x \) changes.

• Taking the derivative helps in identifying critical points, which indicate potential minimum or maximum values.
• This is efficient because calculating slope changes helps pinpoint where changes from decreasing distance to increasing distance occur.

Critical Points Calculation

Critical points are essential in determining where functions reach their maximum or minimum values. They occur where the derivative equals zero or is undefined.
In our context, finding the critical points involves setting the derivative of the squared distance formula to zero: \( \frac{d(d^2)}{dx} = 0 \). This results in a factorable equation: \( x(x^2 - 8x + 16) = 0 \).
Solving it provides us with \( x = 0 \) and \( x = 4 \), representing potential solutions.

• To confirm which point minimizes the distance, we substitute these back into the original distance formula.
• This validation ensures accuracy, allowing us to determine that when \( x = 4 \), the distance (1 unit) is smallest.

Thus, the closest point on the parabola to \((4, 1)\) is \((4, 8)\).