Problem 46 - Advanced Engineering Mathematics

Question

Find a function \(y=f(x)\) whose second derivative is \(y^{\prime \prime}=\) \(12 x-2\) at each point \((x, y)\) on its graph and \(y=-x+5\) is tangent to the graph at the point corresponding to \(x=1\).

Step-by-Step Solution

Answer

The function is \( y = 2x^3 - x^2 - 5x + 8 \).

Step 1: Integrate the Second Derivative

We start by finding the first derivative of the function by integrating the given second derivative. The second derivative is \( y'' = 12x - 2 \). Integrating this with respect to \( x \) gives us the first derivative: \[ y' = \int (12x - 2) \, dx = 6x^2 - 2x + C_1 \], where \( C_1 \) is an integration constant.

Step 2: Find the Original Function

Next, we find the original function \( y = f(x) \) by integrating the first derivative we found in the previous step. So we integrate \( y' = 6x^2 - 2x + C_1 \): \[ y = \int (6x^2 - 2x + C_1) \, dx = 2x^3 - x^2 + C_1x + C_2 \], where \( C_2 \) is another integration constant.

Step 3: Use the Tangent Information

We know that \( y = -x + 5 \) is tangent to the function at \( x = 1 \). This means at \( x = 1 \), the function value and its first derivative equal those of the line. For the function: \( y(1) = 2(1)^3 - (1)^2 + C_1\cdot1 + C_2 = 1 + C_1 + C_2 \) and from the line, \( y = -1 + 5 = 4 \). Thus, \( 1 + C_1 + C_2 = 4 \), which simplifies to \( C_1 + C_2 = 3 \).

Step 4: Use the Derivative Information

Since \(-1\) is the slope of the tangent line at \( x = 1 \), the first derivative of the function at \( x = 1 \) must also be \(-1\). From \( y' = 6x^2 - 2x + C_1 \), substitute \( x = 1 \): \( y'(1) = 6(1)^2 - 2(1) + C_1 = 4 + C_1 = -1 \). Solve for \( C_1 \) to get \( C_1 = -5 \).

Step 5: Determine the Constants

With \( C_1 = -5 \), substitute into the equation \( C_1 + C_2 = 3 \): \(-5 + C_2 = 3 \). Solve for \( C_2 \) to get \( C_2 = 8 \).

Step 6: Write the Final Function

Substitute the values of \( C_1 \) and \( C_2 \) back into the function: \[ y = 2x^3 - x^2 - 5x + 8 \]. This is the function whose second derivative is \( 12x - 2 \) and is tangent to \( y = -x + 5 \) at \( x = 1 \).

Key Concepts

Integration Techniques

When we talk about integration techniques, we're diving into the world of calculus that helps us find functions from their derivatives. In this exercise, we began with the second derivative, \( y'' = 12x - 2 \).

To discover the original function, we integrate this derivative step by step.

• Step 1: First, integrate the second derivative to get the first derivative. This involved working with \( 12x - 2 \), yielding \( y' = \int (12x - 2) \, dx = 6x^2 - 2x + C_1 \).
  This is our first integration constant, represented by \( C_1 \).
• Step 2: Next, we integrate the first derivative to fetch the original function. By integrating \( y' = 6x^2 - 2x + C_1 \), we get \( y = \int (6x^2 - 2x + C_1) \, dx = 2x^3 - x^2 + C_1x + C_2 \). Now, there's another constant, \( C_2 \).

Integration here forms the backbone of solving differential equations, guiding us to find the specific function that satisfies the given conditions.

Tangent Lines

Understanding tangent lines is crucial since they tell us how a curve behaves at a specific point. In this exercise, we know the line \( y = -x + 5 \) is tangent to our function at \( x = 1 \). This offers two vital pieces of information:

• The function's value at \( x = 1 \) matches that of the line.
  This means \( y(1) = -1 + 5 = 4 \). Therefore, from the function, \( 1 + C_1 + C_2 = 4 \), simplifying to \( C_1 + C_2 = 3 \).
• The slope of the function at \( x = 1 \) is the same as the tangent line's slope. The line's slope \(-1\) implies \( y'(1) = -1 \). By plugging in \( x = 1 \) into the derivative, \( y' = 6x^2 - 2x + C_1 \), we calculate \( 4 + C_1 = -1 \), finding \( C_1 = -5 \).

Tangent lines offer a way to pinpoint these integration constants by connecting the function's behavior to the conditions laid out in the problem.

Initial Conditions

Initial conditions help us find the specific form of a function among its family of solutions. Here, we are using these conditions to solve for the constants \( C_1 \) and \( C_2 \) in our integrated equations.

• Matching function point: The equation \( C_1 + C_2 = 3 \) comes from matching the function's value to the tangent line at \( x = 1 \).
• Slope requirement: From the slope of \(-1\) at \( x = 1 \), we derived \( C_1 = -5 \).
• Solving for constants: With \( C_1 = -5 \), substitute in \( C_1 + C_2 = 3 \) to find \( C_2 = 8 \).

These initial conditions allow us to refine our general solution into the particular one, \( y = 2x^3 - x^2 - 5x + 8 \), which adheres to all specified behaviors of the problem.