Problem 21 - Thomas Calculus

Question

Suppose \(a\) and \(b\) are positive numbers. Sketch the parabolas $$ y^{2}=4 a^{2}-4 a x \quad \text { and } \quad y^{2}=4 b^{2}+4 b x $$ in the same diagram. Show that they intersect at \((a-b, \pm 2 \sqrt{a b})\) and that each " \(a\) -parabola"" is orthogonal to every "b-parabola."

Step-by-Step Solution

Answer

The parabolas intersect at \((a-b, \pm 2\sqrt{ab})\) and are orthogonal.

Step 1: Understand the Parabolas

The first parabola is given by the equation \( y^2 = 4a^2 - 4ax \). This is a horizontal parabola pointing left because the term \(-4ax\) suggests opening along the negative x-axis. The vertex is at \((a^2, 0)\), but due to simplifications, we identify the vertex as mostly near the intersection points. The second parabola is \( y^2 = 4b^2 + 4bx \), which is a horizontal parabola opening right, with its vertex at \((-b^2, 0)\), again simplified for the nature of the intersection points.

Step 2: Set the Equations to Find Intersections

To find where the parabolas intersect, set their equations equal to each other: \(4a^2 - 4ax = 4b^2 + 4bx\). Rearrange to form \(4a^2 - 4b^2 = 4ax + 4bx\). Divide each term by 4 to simplify: \(a^2 - b^2 = ax + bx\). Factor and rearrange: \(a^2 - b^2 = x(a + b)\), leading to \(x = \frac{a^2 - b^2}{a + b}\).

Step 3: Simplify and Verify Intersection Points

Simplify \(x = \frac{a^2 - b^2}{a + b}\): this can be rewritten as \(x = a - b\) using the difference of squares formula \((a^2 - b^2) = (a - b)(a + b)\). Substitute \(x = a - b\) into either equation to find \(y^2\). Using the first parabola, \(y^2 = 4a(a - b)\). Similarly, using the second parabola, \(y^2 = 4b(b - a)\), ensuring consistency. Solving for \(y\) gives \(y = 2\sqrt{ab}\) or \(y = -2\sqrt{ab}\). Thus, intersections are \((a - b, \pm 2\sqrt{ab})\).

Step 4: Verify Orthogonal Intersection

The slopes of curves can be determined by implicit differentiation. For the first parabola \(y^2 = 4a^2 - 4ax\), differentiate to get \(2y\frac{dy}{dx} = -4a\), which gives the slope \(\frac{dy}{dx} = \frac{-2a}{y}\). For the second parabola \(y^2 = 4b^2 + 4bx\), differentiation yields \(2y\frac{dy}{dx} = 4b\), i.e., \(\frac{dy}{dx} = \frac{2b}{y}\). To check for orthogonality, the product of slopes \(\frac{-2a}{y} \cdot \frac{2b}{y} = -1\) should hold, showing that \(2ab = y^2 = 4ab\), confirming orthogonality at intersections.

Key Concepts

Parabolas

Parabolas are the curves formed by the set of all points equidistant from a point called the focus and a line called the directrix. When exploring equations like \( y^2 = 4a^2 - 4ax \) and \( y^2 = 4b^2 + 4bx \), it’s important to recognize that these describe horizontal parabolas.

• The first equation has the form, suggesting a parabola opening to the left, centered at some shift along the x-axis.
• The second equation describes a parabola opening to the right, indicating another shift in position.

Both parabolas show symmetry with respect to the x-axis. Interestingly, these parabolas exhibit different behaviors despite having a similar structure. Understanding their behavior as geometric shapes allows us to explore complex scenarios like their intersection.

Orthogonal Curves

When two curves intersect at right angles, they are said to be orthogonal. This means that at the point of intersection, the tangents to each curve are perpendicular to each other. The concept of orthogonality is crucial in many applications like physics and engineering.In the context of the parabolas given by \( y^2 = 4a^2 - 4ax \) and \( y^2 = 4b^2 + 4bx \), orthogonality is verified by examining their slopes at the points of intersection.

• If we calculate the slopes, or gradients, of the tangent lines at \( (a-b, \pm 2\sqrt{ab}) \), and find that the product of these slopes is -1, the curves intersect orthogonally.
• Using implicit differentiation, these slopes are given by \( \frac{-2a}{y} \) and \( \frac{2b}{y} \) respectively.
• If \( \frac{-2a}{y} \) and \( \frac{2b}{y} \) multiply together to give -1, this confirms the orthogonal nature of the intersection.

Understanding how to compute and verify the orthogonality of curves illuminates the precise geometric relationships where they meet.

Implicit Differentiation

Implicit differentiation is a technique in calculus used to find derivatives of equations that cannot be easily solved for one variable in terms of another. When both variables are intermixed in an equation, implicitly differentiating involves taking the derivative of each term and solving for the derivative of the dependent variable. With our given equations \( y^2 = 4a^2 - 4ax \) and \( y^2 = 4b^2 + 4bx \), implicit differentiation is necessary because the equations define \( y \) in terms of \( x \) implicitly.

• For the first parabola, differentiate \( y^2 = 4a^2 - 4ax \), leading to \( 2y \frac{dy}{dx} = -4a \).
• You solve for \( \frac{dy}{dx} \), resulting in \( \frac{dy}{dx} = \frac{-2a}{y} \).
• The approach is similar for the second parabola, leading to \( \frac{dy}{dx} = \frac{2b}{y} \).

Implicit differentiation is powerful as it allows you to find instantaneous rates of change for curves not define as explicit functions. Mastery of this technique opens doors to tackle more complex mathematical challenges.