Problem 60 - University Calculus: Early Transcendentals

Question

The curves \(y=\) \(x^{2}+a x+b\) and \(y=c x-x^{2}\) have a common tangent line at the point \((1,0) .\) Find \(a, b,\) and \(c\).

Step-by-Step Solution

Answer

\(a = -3, b = 2, c = 1\).

Step 1: Evaluate the Point on Each Curve

First, determine if the point \((1,0)\) lies on both curves. For the curve \(y = x^2 + ax + b\), substitute \(x = 1\) and \(y = 0\):\[0 = 1^2 + a \cdot 1 + b \Rightarrow 0 = 1 + a + b \Rightarrow a + b = -1\]For the curve \(y = cx - x^2\), substitute \(x = 1\) and \(y = 0\):\[0 = c \cdot 1 - 1^2 \Rightarrow 0 = c - 1 \Rightarrow c = 1\]

Step 2: Find Derivatives of the Curves

Find the derivatives of each curve, which represent the slopes of the tangent lines. For \(y = x^2 + ax + b\):\[\frac{dy}{dx} = 2x + a\]For \(y = cx - x^2\):\[\frac{dy}{dx} = c - 2x\]

Step 3: Evaluate Derivatives at the Common Point

The point \((1,0)\) is where the common tangent lies, so the slopes of the tangents (derivatives) should be identical at this point.Substitute \(x = 1\) into the derivative of the first curve:\[\frac{dy}{dx} = 2 \cdot 1 + a = 2 + a\]Substitute \(x = 1\) into the derivative of the second curve (with \(c = 1\) from Step 1):\[\frac{dy}{dx} = 1 - 2 \cdot 1 = 1 - 2 = -1\]Set the two expressions for the slope equal:\[2 + a = -1\]

Step 4: Solve for Variables

Solve the equation from Step 3 for \(a\):\[2 + a = -1\Rightarrow a = -3\]Use the result from Step 1, \(a + b = -1\), to solve for \(b\):\[-3 + b = -1\Rightarrow b = 2\]

Step 5: Present the Results

The values that satisfy both the conditions for the curves to have a common tangent at \((1,0)\) are:\[a = -3, \quad b = 2, \quad c = 1\]

Key Concepts

Derivatives

In calculus, a derivative represents how a function changes as its input changes. Think of it as the function's rate of change at any given point. It's like checking the "speed" of a car at a particular instant.
For instance, if you know how fast you were going when you hit a certain place, you can predict when you'll arrive elsewhere on your route. In our problem, the derivatives are used to find the slope of the tangent line at a specific point. This slope tells us how steep the line is where it touches the curve.

To determine the derivatives of the given curves, we used differentiation:

• The first curve, given by the equation \(y = x^2 + ax + b\), differentiates to \( \frac{dy}{dx} = 2x + a \). This tells us that at any point \(x\) on the curve, the slope is \(2x + a\).
• The second curve, \(y = cx - x^2\), differentiates to \( \frac{dy}{dx} = c - 2x \). Here, the slope at any point \(x\) is \(c - 2x\).

This information helps us understand how steep each curve is at any point and is critical in finding where two curves might share a common tangent, as it gives us both the rate and direction of change.

Tangent Lines

Tangent lines are straight lines that touch a curve at one point and have the same slope as the curve at that point. Imagine a line that just skims the surface of a ball without cutting through it—that's what a tangent line does with a curve.
In our problem, a common tangent line means that both curves share this exact line at the specified point (1,0). They "kiss" the curves right at that spot without crisscrossing.
Finding a common tangent involves setting the slopes of both curves equal at the same point:

• For the first curve, \(y = x^2 + ax + b\), the slope at \(x = 1\) was found to be \(2 + a\).
• For the second curve, \(y = cx - x^2\), with \(c = 1\), the slope at \(x = 1\) was determined to be \(-1\).

By equating these two slopes \(2 + a = -1\), you solve for the variable \(a\) that ensures both curves have the same tangent line at \(x=1\). This confirms where the common tangent line lies on both curves.

Algebraic Equations

Algebraic equations are the maths sentences that tell us about the relationship between numbers and variables. They're the foundation of solving many math problems, capturing relationships using symbols and signs that represent numbers.
In solving this exercise, algebraic equations helped us substitute and solve for unknowns \(a\), \(b\), and \(c\). First, we plugged in the point \((1,0)\) into both curve equations. This helped us write our initial equations:

• For \(y = x^2 + ax + b\), substituting \((1,0)\) led to \(1 + a + b = 0\), simplified to \(a + b = -1\).
• For \(y = cx - x^2\), substituting \((1,0)\) simplified directly to \(c = 1\).

These equations were the starting point to find the values of \(a\), \(b\), and \(c\) needed for the curves to share a common tangent. By equating the derivatives and solving, we found \(a = -3\) and subsequently \(b = 2\).
Therefore, understanding how to manipulate and solve these equations is key to unraveling the conditions for a common tangent line, and it's the interaction of calculus and algebra that brings the solution to life.