Problem 5 - Calculus

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 closest points are at \((\pm\sqrt{\frac{9}{2}},\frac{9}{2})\).

Step 1: Express the distance formula

The square of the distance from a point \(x, y\) on the parabola \(y=x^2\) to the point \(0,5\) can be written as \(D^2 = x^2 + (y - 5)^2\). Since \(y=x^2\), substitute \(y\) in the equation to get \(D^2 = x^2 + (x^2 - 5)^2\).

Step 2: Simplify the expression

Expand \(D^2 = x^2 + (x^2 - 5)^2\) to obtain \(D^2 = x^2 + (x^4 - 10x^2 + 25)\). Simplify this to \(D^2 = x^4 - 9x^2 + 25\).

Step 3: Find the derivative

Take the derivative of \(D^2 = x^4 - 9x^2 + 25\) with respect to \(x\). This gives \(abla(D^2)=4x^3 - 18x\).

Step 4: Solve for critical points

Set the derivative \(4x^3 - 18x\) equal to zero to find the critical points: \(abla(D^2) = 4x^3 - 18x = 0\). Factor this equation to get \(2x(2x^2 - 9) = 0\). Solving for \(x\), we get \(x = 0\) or \(2x^2 = 9\), which solves to \(x = \pm\sqrt{\frac{9}{2}}\).

Step 5: Identify the corresponding y-coordinates

For \(x = 0\), the corresponding \(y\) on the parabola \(y = x^2\) is \(y = 0^2 = 0\). For \(x = \pm\sqrt{\frac{9}{2}}\), calculate \(y = (\sqrt{\frac{9}{2}})^2 = \frac{9}{2}\).

Step 6: Determine the closest point

Evaluate \(D^2 = x^4 - 9x^2 + 25\) for each \(x\): for \(x=0\), \(D^2 = 0^4 - 9(0)^2 + 25 = 25\); for \(x = \pm\sqrt{\frac{9}{2}}\), calculate \(D^2\) which turns out to be \(D^2 = (\frac{9}{2})^2 - 9(\frac{9}{2}) + 25 = \frac{9}{4}\), slightly further simplified due to equivalence in opposite values. Since \(\frac{9}{4} < 25\), the closest points are \(x = \pm\sqrt{\frac{9}{2}}, y = \frac{9}{2}\).

Key Concepts

Distance Formula

In mathematics, the distance formula is an expression used to find the distance between two points in a coordinate plane. This concept helps solve problems related to determining how far apart two locations are, which is often necessary for optimization problems in calculus. The distance formula is derived from the Pythagorean theorem and is written as:\[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.

• The formula calculates the straight-line distance between the two points.
• In optimization problems, we often minimize the square of the distance to simplify the calculus.

In the exercise, we start with expressing the distance in terms of a single variable to find a point on the parabola that is closest to another point. By substituting the equation of the parabola into the distance formula, we streamline our calculations. This is because it's easier to work with a single dimension rather than multiple dimensions.

Parabolas

A parabola is a U-shaped curve that can open upwards or downwards on a graph. It's represented by the quadratic equation \(y = ax^2 + bx + c\). The simplest form is \(y = x^2\), where the parabola opens upwards and has its vertex at the origin (0,0).

• The parabola is symmetric about its vertical axis.
• It can be manipulated by adjusting the coefficients \(a\), \(b\), and \(c\) to move its vertex or widen/narrow its opening.

In this exercise, the parabola \(y = x^2\) is used to determine which points on it are closest to another point. By understanding the geometry of parabolas, you can apply the derivative to find where the minimum distance occurs. This transformation from understanding the shape to applying calculus is key in solving such problems.Another important aspect is recognizing that for such optimization, you often analyze critical points, which, in this case, involve the parabola's inherent properties.

Critical Points

Critical points are the x-values where the first derivative of a function is either zero or undefined. Finding these points is essential because they help locate maxima, minima, or points of inflection. They are particularly useful in optimization problems.To find critical points:

• Take the derivative of the function with respect to the variable,\( f'(x) \).
• Set the derivative equal to zero, \( f'(x) = 0 \), to find the points where the slope is zero.
• Solve for \(x\) to get the critical points.

In our exercise, we found critical points by setting the derivative of the squared distance formula equal to zero and solving for \(x\). These were then used to determine the corresponding points on the parabola. Critical points help find where a function’s rate of change is zero, meaning it's at a turning point. Thus, understanding how and why they occur leads you to the best solution for the problem.