Problem 20 - The Calculus with Analytic Geometry

Question

At any point \((x, y)\) on a curve, \(D_{x}^{3} y=2\), and \((1,3)\) is a point of inflection at which the slope of the inflectional tangent is \(-2 .\) Find an equation of the curve.

Step-by-Step Solution

Answer

The equation of the curve is \y = \frac{x^3}{3} + C_1 x^2 + (-3 - C_1) x + C_3\.

Step 1: Identify Given Information

Given: The third derivative of y with respect to x is 2, and \((1, 3)\) is a point of inflection with a slope of \(-2\).

Step 2: Integrate Third Derivative

Integrate \(D_{x}^{3} y = 2\) with respect to x to get the second derivative of y: \[D_{x}^{2} y = 2x + C_1\]

Step 3: Integrate Second Derivative

Integrate again to get the first derivative of y: \[D_{x} y = x^2 + C_1 x + C_2\]

Step 4: Integrate First Derivative

Integrate one more time to find the expression for y: \[y = \frac{x^3}{3} + \frac{C_1 x^2}{2} + C_2 x + C_3\]

Step 5: Apply Inflection Point Information

At the point of inflection \(x = 1, y = 3\), substitute into the equation for y: \[3 = \frac{1^3}{3} + \frac{C_1 \(1^2\)}{2} + C_2 \(1\) + C_3\]Simplify to obtain an equation: \[3 = \frac{1}{3} + \frac{C_1}{2} + C_2 + C_3\]

Step 6: Use Slope Information

At the inflection point where \(x = 1\), the slope is given as \(-2\):\[D_x y \bigg|_{x=1} = -2\]Substitute into \[D_x y:\]\[-2 = 1^2 + C_1 (1) + C_2\]Simplify to:\[-2 = 1 + C_1 + C_2\]

Step 7: Solve for Constants

From the slope equation: \C_1 + C_2 = -3\Substitute into the earlier equation for y: \[3 = \frac{1}{3} + \frac{C_1}{2} + C_2 + C_3\]Rewrite as:\[3 = \frac{1}{3} + \frac{C_1}{2} + (-3 - C_1) + C_3\]Solve to find \C_3\.

Key Concepts

Third Derivative

The third derivative is denoted as \(D_{x}^{3} y\). It represents the rate of change of the second derivative or acceleration of acceleration. In this problem, you are given that \(D_{x}^{3} y = 2\). This tells us that the acceleration of the slope's rate of change is constant, and specifically equal to 2. Knowing the third derivative allows us to work backwards through integration to find the lower order derivatives and, eventually, the original function y.

Integration of Constants

Integration is the process of finding the antiderivative of a function. When we integrate a constant, we get a linear term plus a new constant of integration. This is essential to the problem as we start integrating from the third derivative: \[ \int 2 \, dx = 2x + C_1 \] Here, we integrated 2 with respect to \ x \ and introduced a constant \ C_1 \, which we will determine later using boundary conditions. Each subsequent integration follows the same principle, accumulating new constants: \[ \int (2x + C_1)dx = x^2 + C_1 x + C_2 \] Lastly, integrating again gives us: \[ \int (x^2 + C_1 x + C_2) dx = \frac{x^3}{3} + \frac{C_1 x^2}{2} + C_2 x + C_3 \] These integrate to form the polynomial that represents \ y \.

Point of Inflection Analysis

A point of inflection is where the curve changes concavity. For a point at \ (1,3) \, it means the second derivative changes sign at \ x = 1 \. Here, it also provides information about the curve's behavior. Given a slope of \ -2 \ at this point allows us to set up equations to solve for constants: \[ y = \frac{x^3}{3} + \frac{C_1 x^2}{2} + C_2 x + C_3 \] At \( x = 1 \) and \ y = 3 \, we substitute to get: \[ 3 = \frac{1}{3} + \frac{C_1}{2} + C_2 + C_3 \] Likewise, using the slope: \[ -2 = 1 + C_1 + C_2 \] These allow us to solve for the unknowns. Solving these reveals \ C_3 \ thereby giving the full equation that represents the curve passing through these points.