Problem 25 - 5 Steps to A 5: AP Calculus AB 2021

Question

(Calculator) Find the shortest distance between the point (1,0) and the curve \(y=x^{3}\).

Step-by-Step Solution

Answer

Answer: The shortest distance between the point (1, 0) and the curve y = x^3 is approximately 0.8145.

Step 1: Set up the distance formula

The distance formula is given by: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) Where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, (x1, y1) = (1, 0) and (x2, y2) = (x, x^3), where x belongs to the curve y = x^3.

Step 2: Simplify the distance function and eliminate the square root

Plug in the coordinates to get the distance formula for this specific problem: \(d=\sqrt{(x-1)^2+(x^3-0)^2}\) To minimize this function, we can instead minimize the square of the distance function, which is easier to work with: \(d^2=(x-1)^2+(x^3)^2\)

Step 3: Find the derivative of the distance function

Differentiate the distance squared function with respect to x: \(\frac{d(d^2)}{dx}=2(x-1)+6x^5\)

Step 4: Find the critical points

To find the minimum, set the derivative equal to zero and solve for x: \(2(x-1)+6x^5=0\) Divide both sides by 2: \((x-1)+3x^5=0\) To solve for x, numerical methods, such as Newton's method or a calculator, must be used. The solution is approximately x ≈ 0.9348.

Step 5: Find the coordinate on the curve y = x^3

Plug in the x-value into the equation of the curve to get the y-value: \(y=(0.9348)^3 \approx 0.8140\) So, the coordinate on the curve that is closest to the point (1, 0) is approximately (0.9348, 0.8140).

Step 6: Calculate the shortest distance between the point and the curve

Now that we have the coordinate on the curve, plug the x-value back into the distance formula: \(d=\sqrt{(0.9348-1)^2+(0.8140-0)^2} \approx 0.8145\) The shortest distance between the point (1, 0) and the curve y = x^3 is approximately 0.8145.

Key Concepts

Distance Formula

The distance formula is a tool used to find the distance between two points in a plane. This can be visualized by thinking about the coordinates of each point as an endpoint of a segment. The formula is derived from the Pythagorean theorem and is expressed 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. In our example, the points are \((1, 0)\) and a point on the curve \(y = x^3\), represented as \((x, x^3)\). By applying these values into the formula, we find the expression for distance in terms of \(x\).
This technique simplifies the problem of finding the shortest distance between a point and a curve.

Derivative

In calculus, the derivative measures how a function changes as its input changes. It is a crucial tool for finding rates of change and optimizing functions. Here, we use the derivative to find where the distance from a point to a curve is minimized. To do this, we focus not on the function for distance itself but its square, as it removes troublesome square roots.

• The derivative of the squared distance function \( (x-1)^2 + (x^3)^2 \) is computed as \( \frac{d(d^2)}{dx} = 2(x-1) + 6x^5 \).

Finding the derivative helps identify where the slope of the distance function is zero, indicating potential minimum values.

Critical Points

Critical points are locations on a graph where the derivative is zero or undefined, marking potential maxima, minima, or points of inflection for the function. Once we have the derivative \( 2(x-1) + 6x^5 \), we find critical points by setting the derivative equal to zero and solving for \(x\).

• This yields \((x-1) + 3x^5 = 0\).

Finding the solutions to this equation provides values of \(x\) where the distance might be minimized. In some cases, these zeros cannot be determined analytically, and numerical methods provide approximate solutions.

Numerical Methods

When derivatives lead to complex equations that can't be solved algebraically, numerical methods, like Newton's method, are used. These methods iterate through calculations to approximate a solution efficiently. For the equation \( (x-1) + 3x^5 = 0 \), numerical methods are employed to find \( x \approx 0.9348 \).

• These methods are powerful tools for finding solutions for problems that resist simple algebraic manipulation.

Once you have the numerical solution, it can be plugged back into the original equations to find specific results, like the shortest distance.