Problem 21 - The Calculus with Analytic Geometry

Question

The equation \(x^{2}=4 a y\) represents a one-parameter family of parabolas. Find an equation of another one-parameter family of curves such that at any point \((x, y)\) there is a curve of each family through it and the tangent lines to the two curves at this point are perpendicular. (HINT: First show that the slope of the tangent line at any point \((x, y)\), not on the \(y\) axis, of the parabola of the given family through that point is \(2 y / x\).)

Step-by-Step Solution

Answer

The equation of the new one-parameter family of curves is \( x^2 + 2y^2 = C \).

Step 1: Differentiate the Given Equation Implicitly

Start with the given equation of the parabola: \[ x^2 = 4ay \]Differentiate both sides with respect to \(x\):\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4ay) \]This results in: \[ 2x = 4a \frac{dy}{dx} \]

Step 2: Solve for the Slope of the Tangent Line

Rearrange the differentiated equation to solve for the slope \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = \frac{2x}{4a} = \frac{x}{2a} \]Given \[ y = \frac{x^2}{4a} \]Replace \(a\) by \( \frac{x^2}{4y} \):\[ \frac{x}{2a} = \frac{x}{2 \cdot \frac{x^2}{4y}} = \frac{2y}{x} \]Thus, the slope of the tangent line at any point \((x, y)\) not on the \(y\)-axis is \(\frac{2y}{x}\).

Step 3: Identify the Slope of the Perpendicular Tangent

To find a family of curves perpendicular to the given family, the product of their slopes must be \(-1\). If the slope of the tangent to the given parabolas is \(\frac{2y}{x}\), the slope of the perpendicular tangent should be:\[ -\frac{x}{2y} \]

Step 4: Integrate to Find the New Family of Curves

Integrate the perpendicular slope to find the equation of the new family of curves. Starting from: \[ \frac{dy}{dx} = -\frac{x}{2y} \]Separate variables and integrate:\[ 2y \ dy = -x \ dx \]Integrate both sides:\[ y^2 = -\frac{x^2}{2} + C \]Rearrange to get:\[ x^2 + 2y^2 = C \]

Key Concepts

Implicit Differentiation

When dealing with equations that define relationships between variables, explicit forms aren't always available. Implicit differentiation helps us find derivatives when the equation involves both variables intermingled.
In the given problem, start with the equation of the parabola: \[ x^2 = 4ay \]Differentiate each term with respect to x:\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4ay)\]This approach allows you to handle terms on both sides of the equation, differentiating them implicitly.

Family of Curves

A family of curves is a set of curves described by an equation involving a parameter. In our case, the given family is parabolas defined by \[ x^2 = 4ay \] Here, 'a' is the parameter that changes the shape of each specific parabola. Each curve in the family differs by the value of 'a'.
Another family of curves can be found by considering the condition that their tangents are perpendicular to those of the given parabolas at any point.
The family of curves we seek is one where for any point (x, y), there will be effects of curves from both families at that point.

Slope of Tangent Line

The slope of a tangent line indicates how a function changes at any given point. It's the derivative of the function concerning x.
Given our parabola equation: \[ x^2 = 4ay \]Differentiate both sides to find the slope: \[ 2x = 4a \frac{dy}{dx} \]Solving for \[ \frac{dy}{dx} \], rearranging terms,\[ \frac{dy}{dx} = \frac{2y}{x} \]
shows the slope at any point (x, y). For perpendicular lines, the product of their slopes must be -1.

Integration

Integration is the process of finding the original function from its derivative. For finding the perpendicular family of curves, we need to integrate the perpendicular slope Integrate \[ \frac{dy}{dx} = -\frac{x}{2y} \] to obtain this family.
Separate variables and integrate both sides: \[ 2y \text{ dy} = -x \text{ dx} \]
This results in:\[ y^2 = -\frac{x^2}{2} + C \].Finally, rewriting gives: \[ x^2 + 2y^2 = C \].
So, these are ellipses that cross the parabolas perpendicularly at any point.