Problem 92 - Thomas Calculus

Question

a. Find a curve \(y=f(x)\) with the following properties: \begin{equation} \begin{array}{l}{\text { i) } \frac{d^{2} y}{d x^{2}}=6 x} \\ {\text { ii) Its graph passes through the point }(0,1) \text { and has a hori- }} \\\ {\text { zontal tangent there. }}\end{array} \end{equation} b. How many curves like this are there? How do you know?

Step-by-Step Solution

Answer

There is only one such curve: \( y = x^3 + 1 \). The constants are determined uniquely by the conditions.

Step 1: Integrate Second Derivative

We start by integrating the given second derivative \( \frac{d^2 y}{dx^2} = 6x \) to find the first derivative. Integrating with respect to \( x \), we get \( \frac{dy}{dx} = \int (6x) \, dx = 3x^2 + C_1 \), where \( C_1 \) is a constant of integration.

Step 2: Use Horizontal Tangent Condition

A horizontal tangent at \( (0,1) \) means the derivative is zero there. Thus, \( \frac{dy}{dx} \bigg|_{x=0} = 0 \Rightarrow 3(0)^2 + C_1 = 0 \). Therefore, \( C_1 = 0 \). So, \( \frac{dy}{dx} = 3x^2 \).

Step 3: Integrate First Derivative

Next, we integrate the expression for \( \frac{dy}{dx} \) to find the curve equation. This will give us \( y = \int (3x^2) \, dx = x^3 + C_2 \), where \( C_2 \) is another constant of integration.

Step 4: Apply Initial Condition

Use the given point \( (0,1) \) to find \( C_2 \). Substituting \( x = 0 \) and \( y = 1 \) into \( y = x^3 + C_2 \), we get \( 1 = 0^3 + C_2 \), so \( C_2 = 1 \).

Step 5: Final Equation of the Curve

Substitute \( C_2 = 1 \) back into the equation to get \( y = x^3 + 1 \) as the curve that fits all conditions.

Key Concepts

Integration in Curve Sketching

When sketching curves, integration is an essential mathematical process used to find original functions from their derivatives. In problems like the one from our exercise, we often use integration to move backwards from a given derivative, such as a second or first derivative, to find an equation for the curve itself.

For our specific problem, we started with a second derivative given as \( \frac{d^2 y}{dx^2} = 6x \). To find the first derivative \( \frac{dy}{dx} \), we integrate the second derivative with respect to \( x \). This involves finding a function whose derivative would match \( 6x \), resulting in \( \int 6x \, dx = 3x^2 + C_1 \), where \( C_1 \) is the constant of integration.

Once we have the first derivative, we need to integrate again to find the original function \( y = f(x) \). Thus, we perform another integration: \( \int 3x^2 \, dx = x^3 + C_2 \), where \( C_2 \) represents another constant of integration. This layered approach is how integration helps us derive curves from their derivatives.

Understanding Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a crucial role in describing various real-world phenomena, such as motion, heat, or population dynamics.

In our exercise, the differential equation \( \frac{d^2 y}{dx^2} = 6x \) defines the relationship between \( y \) and its derivatives. Here, it tells us that the rate at which the derivative of \( y \) is changing with respect to \( x \) is directly proportional to \( x \) itself, with a proportionality constant of 6.

Solving this, as depicted in the steps, involves treating the differential equation as a puzzle: you find functions \( y \) that fit the relationships set out by both derivatives given and initial conditions provided. This is where solving differential equations interlinks with integration, as one often needs to integrate to move from a derivative back to the initial function.

By understanding and solving differential equations like this, we uncover the hidden functions and behavior patterns in related scenarios.

Role of Initial Conditions

Initial conditions are critical pieces of information that help define the specific solution to a differential equation from a set of potential solutions. They essentially "anchor" the curve, making sure it fits the conditions required by the problem.

In our task, the initial conditions given are: the curve passes through the point \( (0,1) \) and has a horizontal tangent there. These are used to find precise values for the constants of integration.

For instance, a horizontal tangent implies \( \frac{dy}{dx} |_{x=0} = 0 \). Substituting \( x = 0 \) in \( \frac{dy}{dx} = 3x^2 + C_1 \) confirms that \( C_1 = 0 \). Similarly, the point \( (0,1) \) in the final equation \( y = x^3 + C_2 \) allows us to solve \( 1 = 0^3 + C_2 \), yielding \( C_2 = 1 \).

Without these initial conditions, the solution wouldn't be unique, and multiple curves could fit the equation. They're indispensable for accurately understanding and solving differential equations in curve sketching scenarios.