Problem 88 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The curve is \(y = x^3 + 1\), and there is only one such curve.

Step 1: Find the First Derivative

To find the curve, we start by integrating the provided second derivative. The given condition is \(\frac{d^{2} y}{d x^{2}} = 6x\). Integrate this with respect to \(x\) to find the first derivative, \(\frac{dy}{dx}\). \[\frac{dy}{dx} = \int 6x \, dx = 3x^2 + C\]Here, \(C\) is a constant of integration.

Step 2: Determine the First Constant of Integration

Use the condition that the curve has a horizontal tangent at \(x = 0\). A horizontal tangent means the slope at that point is zero, i.e., \(\frac{dy}{dx} = 0\) when \(x = 0\). Substitute \(x = 0\) and \(\frac{dy}{dx} = 0\) into the equation from Step 1:\[0 = 3(0)^2 + C \rightarrow C = 0\]This simplifies the first derivative to \(\frac{dy}{dx} = 3x^2\).

Step 3: Find the Original Function

Now, integrate the first derivative to find the function \(y = f(x)\). We have \(\frac{dy}{dx} = 3x^2\), so integrate with respect to \(x\):\[y = \int 3x^2 \, dx = x^3 + D\]Here, \(D\) is another constant of integration.

Step 4: Determine the Second Constant of Integration

Use the condition that the curve passes through the point \((0, 1)\). Substitute \(x = 0\) and \(y = 1\) into the function obtained in Step 3:\[1 = (0)^3 + D \rightarrow D = 1\]Thus, the original function is \(y = x^3 + 1\).

Step 5: Conclusion about the Number of Curves

Since the only constants of integration were uniquely determined by the conditions given (the horizontal tangent and the specific point), the function \(y = x^3 + 1\) is unique. Thus, there is exactly one curve that satisfies both given conditions.

Key Concepts

Integration Techniques

Integration is a powerful calculus tool used to find a function when the rate of change is known. In the exercise, we started with a second derivative, given as \(\frac{d^{2} y}{d x^{2}} = 6x\), requiring integration to discover the original function. Integration, in this case, reverses the process of differentiation.

• **Direct Integration**: We integrate \(6x\) to obtain the first derivative \(\frac{dy}{dx}\). This process strictly follows the power rule of integration, where you increase the power of \(x\) by one and divide by the new power.
• **Constant of Integration**: Every time we integrate, we add an unknown constant \(C\). This accounts for any constant that might have been lost during differentiation.
  

These constants are often figured out using initial conditions or additional information about the function.

Differential Equations

A differential equation involves an unknown function and its derivatives. In our problem, \(\frac{d^{2} y}{d x^{2}} = 6x\) is a differential equation that provides information about the acceleration of the function \(y\) with respect to \(x\).

• **Form of Differential Equations**: They can appear in various forms, such as ordinary (dealing with functions of one variable) and partial (involving multiple variables).
• **Solving Method**: The solution involves finding a function \(y = f(x)\) that satisfies the equation. The first step of solving a second order differential equation typically involves integrating twice, as we did here.
• **Practical Usage**: Such equations often represent physical phenomena, like motion under gravity, where acceleration is constant.
  

Understanding differential equations is crucial for modeling real-world systems and predicting their behavior.

Initial Conditions

Initial conditions are specific values given for a function and its derivatives at particular points. They are vital in solving differential equations uniquely. In our problem, initial conditions are given: the curve has a horizontal tangent at the point \((0, 1)\), and the graph passes through this point.

• **Horizontal Tangent**: Indicates where the first derivative = 0, which means the slope of the tangent line is 0. At \(x = 0\), this allows us to determine one of our constants \(C = 0\).
• **Point on the Curve**: By knowing the curve passes through \((0, 1)\), we find the other constant \(D = 1\).
• **Uniqueness of Solution**: Given enough initial conditions, a differential equation provides a unique solution. For instance, the two conditions we had allowed us to precisely define the function \(y = x^3 + 1\).
  

Using initial conditions transforms a general solution into a specific one, making the results fit a given scenario.

Curves in Geometry

Curves like \(y = x^3 + 1\) hold significant geometric properties. In this problem, these properties help to understand how zero-slopes and intercepts translate to the overall shape of the curve in the Cartesian plane.

• **Graph Interpretations**: The function \(y = x^3 + 1\) depicts a cubic curve translated 1 unit upwards.
• **Horizontal Tangents**: Occur where the rate of change of the curve (the derivative) is zero, a critical point in determining the curve's behavior.
• **Singular Curve**: The exercise proved that only one curve existed under these conditions. Its uniqueness roots in satisfying both the horizontal tangent and passing through the specific point \((0, 1)\).

All these insights demonstrate how the expressions in calculus relate to the geometry of curves we see on graphs.