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Chapter 1: Problem 46

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 \(x=1\).

Short Answer

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

Step by step solution

01

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.
02

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.
03

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 \( C_1 + C_2 = 3 \).
04

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 \( C_1 = -5 \).
05

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 \( C_2 = 8 \).
06

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 \( x = 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider the differential equation \(d y / d x=y(a-b y)\), where \(a\) and \(b\) are positive constants. (a) Either by inspection, or by the method suggested in Problems 33-36, find two constant solutions of the DE. (b) Using only the differential equation, find intervals on the \(y\)-axis on which a nonconstant solution \(y=\phi(x)\) is increasing. On which \(y=\phi(x)\) is decreasing. (c) Using only the differential equation, explain why \(y=a / 2 b\) is the \(y\)-coordinate of a point of inflection of the graph of a nonconstant solution \(y=\phi(x)\). (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the \(x y\)-plane into three regions. In each region, sketch the graph of a nonconstant solution \(y=\phi(x)\) whose shape is suggested by the results in parts (b) and (c).

Learning Theory In the theory of learning, the rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized. Suppose \(M\) denotes the total amount of a subject to be memorized and \(A(t)\) is the amount memorized in time \(t\). Determine a differential equation for the amount \(A(t)\).

Verify that the indicated pair of functions is a solution of the given system of differential equations on the interval \((-\infty, \infty)\). $$ \begin{aligned} &\frac{d x}{d t}=x+3 y \\ &\frac{d y}{d t}=5 x+3 y \\ &x=e^{-2 t}+3 e^{6 t} \\ &y=-e^{-2 t}+5 e^{6 t} \end{aligned} $$

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Qualitative information about a solution \(y=\phi(x)\) of a differential equation can often be obtained from the equation itself. Before working Problems \(55-58\), recall the geometric significance of the derivatives \(d y / d x\) and \(d^{2} y / d x^{2}\). Consider the differential equation \(d y / d x=5-y\). (a) Either by inspection or by the method suggested in Problems \(33-36\), find a constant solution of the \(\mathrm{DE}\). (b) Using only the differential equation, find intervals on the \(y\) -axis on which a nonconstant solution \(y=\phi(x)\) is increasing. Find intervals on the \(y\) -axis on which \(y=\phi(x)\) is decreasing. (c) Explain why \(y=0\) is the \(y\) -coordinate of a point of inflection of a solution curve. (d) Sketch the graph of a solution \(y=\phi(x)\) of the differential equation whose shape is suggested by parts (a)-(c).
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