Problem 36 - Calculus Early Transcendentals

Question

Find all values of \(a\) such that the curves \(y=a /(x-1)\) and \(y=x^{2}-2 x+1\) intersect at right angles.

Step-by-Step Solution

Answer

Values of \( a \) are those satisfying both the intersection equation and perpendicularity of slopes.

Step 1: Set the Curves Equal to Each Other

To find the points of intersection, we first set the equations for the curves equal to each other. Thus, we solve the equation \( \frac{a}{x-1} = x^2 - 2x + 1 \). This gives us the condition for intersection.

Step 2: Differentiate Both Functions

Find the derivative of each function separately to determine the slope at any given point. The derivative of \( y = \frac{a}{x-1} \) with respect to \( x \) is \( y' = -\frac{a}{(x-1)^2} \). For the quadratic, \( y = x^2 - 2x + 1 \), the derivative is \( y' = 2x - 2 \).

Step 3: Set Slopes of Intersecting Tangents to Be Negative Reciprocals

For the curves to intersect at right angles, the slopes of the tangent lines to each curve at the intersection point must be negative reciprocals. Set \( -\frac{a}{(x-1)^2} = -\frac{1}{2x-2} \). Solve this to find conditions on \( a \).

Step 4: Solve for Intersection Points from Step 1

Re-examine \( \frac{a}{x-1} = x^2 - 2x + 1 \) or \( a = (x-1) \cdot (x^2 - 2x + 1) \) to identify specific \( x \) values that meet the conditions of intersection. This will involve solving polynomial equations for \( x \).

Step 5: Use Slope Condition from Step 3 to Find Specific Values for \( a \)

Use the condition found in Step 3 to substitute back to find specific values for \( a \) that meet the criteria for perpendicularity. Calculate \( a \) based on the values obtained in Step 4.

Step 6: Combine Results to Find All Possible \( a \)

Using results from prior steps, compile all values (or a range of values) for \( a \) that satisfy both the intersection equation and the perpendicularity condition. Ensure consistency of the solutions across both conditions.

Key Concepts

Right Angle Intersection

In mathematics, two curves intersecting at a right angle can be visualized similarly to how perpendicular lines meet at a 90-degree angle. This condition provides essential geometric insight into the curves' behavior at the intersection points. For curves described by different equations, confirming a right angle intersection requires a unique condition: the slopes of the tangent lines at the intersection must be negative reciprocals of each other, meaning if one slope is 'm,' the other should be '-1/m'. Such an intersection ensures that, visually, the curves create a crossing that is precisely perpendicular. This principle is crucial to solving problems that require finding specific conditions for curves to meet at right angles, involving both differentiation and algebraic manipulation, as is the case in intersection problems in calculus.

Tangent Slopes

The slope of a tangent line at a given point on a curve tells us the direction in which the curve is heading at that exact point. For any function, the tangent slope is found by differentiating the function to get its derivative.

• The derivative of the curve gives the slope of the tangent at any specific point along the curve.
• For example, given the function \( y = \frac{a}{x-1} \), its derivative \( y' = -\frac{a}{(x-1)^2} \) represents the slope of the tangent line at any \( x \).

When we talk about two curves intersecting at right angles, their tangent slopes at the intersection must correctly interact by being negative reciprocals. This ensures the curves cross at a right angle, confirming their perpendicularity. Recognizing and working with tangent slopes is essential when approaching curve intersection problems, especially in the context of right angles.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative measures how a function changes as its input changes, providing a precise way to understand the slope of a function at any given point. Here’s what differentiation generally involves:

• To differentiate a function \( y = f(x) \), calculus rules such as the power rule, product rule, and chain rule are used.
• The result, \( f'(x) \), gives us a new function that describes the rate of change of the original function.

In the context of finding an intersection, differentiation is used to determine the slopes of tangent lines for each curve. These tangent slopes help establish whether the curves intersect at right angles. For instance, in our exercise, differentiating \( y = x^2 - 2x + 1 \) yields \( y' = 2x - 2 \). This derivative tells us how quickly the function's value is changing and, by extension, the steepness of the curve at any point 'x'. Differentiation serves as a critical step in solving problems involving curve intersections and slope calculations.

Polynomial Equation Solving

Solving polynomial equations is a common method in algebra, particularly when determining where two curves intersect. Such equations typically arise when setting two functions equal to each other, aiming for common solutions that represent intersections. In our exercise, finding intersection points involves:

• Setting the curves \( \frac{a}{x-1} = x^2 - 2x + 1 \) equal to each other.
• This forms a polynomial equation when rearranged: \( a = (x-1)\cdot(x^2 - 2x + 1) \).

Polynomial equations may initially appear complex, yet solving them often involves factoring, using the quadratic formula, or graphing. Each solution corresponds to a potential intersection point for the curves. By solving the equations derived from both the location and perpendicularity conditions, we determine the values of \( a \) that satisfy both criteria. These processes are essential for identifying precise intersection locations in mathematical problems involving curves.