Problem 16 - Calculus

Question

Find numbers \(A\) and \(B\) so that the horizontal line \(y=4\) fits smoothly with the curve \(y=A+B x+x^{2}\) at the point \(x=2\).

Step-by-Step Solution

Answer

The values are \(A = 8\) and \(B = -4\).

Step 1: Understand the Problem

We need to find values of \(A\) and \(B\) so that the curve \(y = A + Bx + x^2\) has the same function value and derivative as the line \(y = 4\) at \(x = 2\). This means the point needs to be on both the line and the curve, and their slopes must match.

Step 2: Matching Function Values

At \(x = 2\), the function of the line is \(y = 4\). So we set the equation of the parabola equal to 4 at \(x = 2\): \[A + B(2) + (2)^2 = 4.\]Simplify this equation: \[A + 2B + 4 = 4,\]\[A + 2B = 0.\] (Equation 1)

Step 3: Matching Derivatives

The derivative of the line \(y = 4\) is 0 because it is a horizontal line. Find the derivative of the curve \(y = A + Bx + x^2\) with respect to \(x\): \[\frac{dy}{dx} = B + 2x.\]At \(x = 2\), set this equal to 0 (the slope of the line): \[B + 2(2) = 0,\]\[B + 4 = 0,\] \[B = -4.\] (Equation 2)

Step 4: Solving the Equations

We have two equations from the previous steps: Equation 1: \(A + 2B = 0\)Equation 2: \(B = -4\)Substitute \(B = -4\) into Equation 1:\[A + 2(-4) = 0,\]\[A - 8 = 0,\]\[A = 8.\]

Step 5: Verify the Solution

Now verify if \(A = 8\) and \(B = -4\) meet both conditions.1. Substitute into the curve equation at \(x = 2\): \[y = 8 - 4(2) + (2)^2 = 8 - 8 + 4 = 4.\]2. The derivative, \(B + 2x = -4 + 4 = 0\), matches the slope of the line.Both conditions are satisfied, confirming the solution is correct.

Key Concepts

Matching Function Values

To solve problems involving matching function values, we aim to make two different functions have the same output at a specific point. This concept was at the heart of our problem, where the challenge was to make the curve (\(y = A + Bx + x^2\)) have the same output as the horizontal line (\(y = 4\)) at \(x = 2\).
In such cases, you substitute the \(x\)-coordinate into both equations and set them equal to each other.

• For the horizontal line, \(y = 4\) simply stays 4.
• For the curve, substitute \(x = 2\) into \(A + Bx + x^2\)
• Set these equal: \(A + 2B + 4 = 4\)

Simplifying gives you an equation that represents the intersection point of the curve and the line, providing the first relationship between \(A\) and \(B\). In our example, Equation 1 turned out to be:\(A + 2B = 0\).Clear understanding of this principle is vital in achieving coherence between disparate mathematical expressions at a shared point.

Matching Derivatives

Matching derivatives means ensuring two functions have the same slope or rate of change at a specific point. The exercise required that the derivative of the curve matches the derivative of the line at \(x=2\).
Since the horizontal line, \(y = 4\), is perfectly flat, its slope, or derivative, is 0 everywhere. For the curve (\(y = A + Bx + x^2\)), the derivative is calculated to show how the curve rises or falls as \(x\) changes.Start by finding the derivative of the curve: \(\frac{dy}{dx} = B + 2x\).

• Substitute \(x = 2\): \(B + 4 = 0\).
• This equation directly gives \(B = -4\), ensuring the curve and the line share the same slope at the point of intersection.

Understanding this concept is essential when you want a smooth transition or connection between curves and lines—common in fields like physics and engineering.

Derivative of a Curve

The derivative of a curve represents its rate of change at each point, essentially showing how steep or flat the curve is. In the parabola given by \(y = A + Bx + x^2\), the derivative gives us crucial information about how the curve behaves as \(x\) varies.
To find the derivative, we differentiate each term of the curve:

• The derivative of \(A\), a constant, is \(0\).
• The derivative of \(Bx\) is simply \(B\)
• The derivative of \(x^2\) is \(2x\).

Combining these results, the derivative is shown as \(\frac{dy}{dx} = B + 2x\). This tells us how the curve's rate of change depends on both the coefficients and the point of interest on the x-axis.

Parabola and Line Intersection

Investigating the intersection of a parabola and a line involves establishing both matching function values and matching derivatives. Both conditions ensure the curve and the line not only touch at a point but also transition smoothly with the same slope, avoiding any harsh turns or corners.
To confirm the point \((x, y)\) of intersection:

• First solve for \(x\) using the equation from matching the function values.
• Determine that the curve's derivative equals the line's derivative at this \(x\) value.

In our example with \(y = 4\) and \(y = A + Bx + x^2\), by calculating and equating the function values, we found \(A + 2B = 0\). Matching derivatives (both \(0\) at \(x=2\)) confirmed that \(B = -4\) and subsequently \(A = 8\).This process proves they intersect smoothly, their slopes precisely matching at the intersection, reflecting a seamless shift at \(x = 2\).