Problem 41 - University Calculus: Early Transcendentals

Question

Find the two points where the curve \(x^{2}+x y+y^{2}=7\) crosses the \(x\) -axis, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents?

Step-by-Step Solution

Answer

The curve intersects the x-axis at \((\sqrt{7}, 0)\) and \((-\sqrt{7}, 0)\); both tangents have a slope of \(-2\), hence they are parallel.

Step 1: Identify points on the x-axis

To find curve intersections on the x-axis, substitute \( y = 0 \) in the equation \( x^2 + xy + y^2 = 7 \). This simplifies to \( x^2 = 7 \), giving \( x = \sqrt{7} \) and \( x = -\sqrt{7} \). Thus, the intersection points are \((\sqrt{7}, 0)\) and \((-\sqrt{7}, 0)\).

Step 2: Determine the slope of the tangent

Differentiate the curve equation implicitly. Starting with \( x^2 + xy + y^2 = 7 \), differentiate both sides with respect to \( x \), yielding \( 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \). Solve for \( \frac{dy}{dx} \) to find the slope: \( \frac{dy}{dx} = \frac{-(2x + y)}{x + 2y} \).

Step 3: Compute the slope at intersection points

Substitute \( y = 0 \) into \( \frac{dy}{dx} = \frac{-(2x + y)}{x + 2y} \), reducing it to \( \frac{dy}{dx} = \frac{-2x}{x} = -2 \) for both \( x = \sqrt{7} \) and \( x = -\sqrt{7} \).

Step 4: Verify parallel tangents

The slope of the tangent line at both points is \( -2 \), confirming that the tangents are parallel since they have the same slope.

Key Concepts

Implicit Differentiation

In calculus, **Implicit Differentiation** is a useful method employed when dealing with equations not easily expressed as a function of one variable. Unlike explicit functions, where you can directly solve for one variable in terms of another, implicit functions are intertwined, often requiring a different approach to differentiation.
When applying this technique, you differentiate both sides of the equation with respect to a particular variable, usually "x." This means you treat all other variables as implicit functions of "x."
For instance, if given a curve equation like \( x^2 + xy + y^2 = 7 \), you’ll differentiate both sides with respect to \( x \), resulting in multiple terms involving \( \frac{dy}{dx} \), which represents the derivative of \( y \) with respect to \( x \).

• In the example, it involves: differentiate \( x^2 \) for \( 2x \)
• differentiate \( xy \) using the product rule to get \( x\frac{dy}{dx} + y \)
• differentiate \( y^2 \) to obtain \( 2y\frac{dy}{dx} \).

By appropriately rearranging and solving for \( \frac{dy}{dx} \), you can find the slope of the tangent line at any point on the curve.

Slope of Tangent Line

The **Slope of the Tangent Line** is essential in understanding the direction at which a curve changes. Essentially, it tells you how steep the line is at any given point.
For our exercise, once you have differentiated the equation implicitly, the next task is to solve for \( \frac{dy}{dx} \). This expression will help in finding the slope of the tangent line to the curve at desired points.
The derivative \( \frac{dy}{dx} \) you end up with is \( \frac{-(2x + y)}{x + 2y} \). To find the slope at specific points, you substitute the coordinates of these points into your derivative expression.

• At the intersection points \((\sqrt{7}, 0)\) and \((-\sqrt{7}, 0)\), substituting \( y = 0 \) simplifies \( \frac{dy}{dx} \) to \( \frac{-2x}{x} \).
• This results in a common slope of \(-2\) at both points.

Such a consistent slope implies that the tangents to the curve at these intercepts are indeed parallel.

Intersection Points

To find **Intersection Points** of a curve with the x-axis, remember these are the points where the curve meets the axis, where the value of \( y \) is zero. In our problem, the goal was to find where the equation \( x^2 + xy + y^2 = 7 \) crosses the x-axis.
By setting \( y = 0 \), the equation simplifies, allowing you to solve for \( x \).

• In this example, by substituting \( y = 0 \), you obtain \( x^2 = 7 \).
• Thus, \( x = \sqrt{7} \) and \( x = -\sqrt{7} \), giving the intersection points \( (\sqrt{7}, 0) \) and \( (-\sqrt{7}, 0) \).

These intersection points help establish not just where the curve meets the x-axis but also where to evaluate the slope of the tangent line. The common characteristics of these tangents offer clues to understanding broader geometrical properties like whether they are parallel.