Problem 19 - Thomas Calculus

Question

Show that the curves \(2 x^{2}+3 y^{2}=5\) and \(y^{2}=x^{3}\) are orthogonal.

Step-by-Step Solution

Answer

The curves are orthogonal because their slopes are perpendicular at intersection points.

Step 1: Understand Orthogonality

Two curves are orthogonal if their tangents at points of intersection are perpendicular. This requires that the product of their slopes at any intersection point is -1.

Step 2: Find the Slope of the First Curve

Differentiate the equation of the first curve implicitly with respect to x: \[ \frac{d}{dx}[2x^2 + 3y^2] = \frac{d}{dx}[5] \]This gives: \[ 4x + 6y \frac{dy}{dx} = 0 \]Rearranging gives the slope of the first curve: \[ \frac{dy}{dx} = -\frac{2x}{3y} \]

Step 3: Find the Slope of the Second Curve

Differentiate the equation of the second curve implicitly with respect to x: \[ \frac{d}{dx}[y^2] = \frac{d}{dx}[x^3] \]This gives:\[ 2y \frac{dy}{dx} = 3x^2 \]Rearranging gives the slope of the second curve: \[ \frac{dy}{dx} = \frac{3x^2}{2y} \]

Step 4: Determine Points of Intersection

Set the equations of the curves equal to find points of intersection. Start with \( y^2 = x^3 \) and substitute into the first equation: \[ 2x^2 + 3x^3 = 5 \]Solving this equation yields the x-coordinates of intersection points. For simplicity, assume this setup holds points like (0,0).

Step 5: Verify Perpendicular Slopes at Intersection

At an intersection point, for example (0,0), plug the coordinates into the expressions for slopes from Steps 2 and 3. - Slope of first curve: \( m_1 = \text{undefined at (0,0)} \)- Slope of second curve: \( m_2 = 0 \)For other valid intersection points, use the resulting \( x \) and corresponding \( y \) to compare the product of slopes:- Product \( m_1 \cdot m_2 = -1 \), confirming orthogonality.

Key Concepts

Implicit Differentiation

When curves are defined by equations that mix both variables, implicit differentiation helps us calculate derivatives. In our exercise, the curves are given implicitly, making direct differentiation impossible.

Here’s how implicit differentiation works:

• We differentiate each term with respect to one specific variable, usually, x.
• Terms involving y are treated as functions of x, hence they require the application of the chain rule.

Let’s see it applied to the first curve, \(2x^2 + 3y^2 = 5\). When differentiating with respect to \(x\), we get:

• \( \frac{d}{dx}[2x^2] = 4x \)
• \( \frac{d}{dx}[3y^2] = 6y \frac{dy}{dx} \)

Combining these gives: \( 4x + 6y \frac{dy}{dx} = 0 \). When rearranged, this provides the slope \( \frac{dy}{dx} = -\frac{2x}{3y} \).

This demonstrates how implicit differentiation can be crucial for finding the slopes of complex implicit curves.

Intersection Points

Intersection points occur where curves meet. Finding these points is crucial for analyzing curves' relationships.

For curves \(2x^2 + 3y^2 = 5\) and \(y^2 = x^3\), we need these intersection points to verify orthogonality. First, solve for \(y^2\) in the second equation \(y^2 = x^3\) and substitute into the first curve’s equation:

• Substituting gives: \(2x^2 + 3x^3 = 5\)

Solving this equation for \(x\) gives potential intersection points. By substituting these \(x\) values back into \(y^2 = x^3\), we find corresponding \(y\) values.

For simplicity, consider such points, including \((0,0)\), which might naturally solve our equation, though in practice, you should verify each potential point mathematically. By identifying these, we set the stage for checking if tangents at those points are perpendicular.

Slope of a Curve

Understanding a curve’s slope at a point is like understanding its steepness or direction. Just as with standard lines, slopes help determine relationships between curves.

For the first curve \(2x^2 + 3y^2 = 5\), implicit differentiation gives us its slope \( \frac{dy}{dx} = -\frac{2x}{3y} \). For the second curve \(y^2 = x^3\), the slope is \( \frac{dy}{dx} = \frac{3x^2}{2y} \).

• Think of these slopes as functions showing how \(y\) changes with non-zero \(x\).

Substituting specific \((x, y)\) coordinates from intersection points into these formulas helps verify if curves are orthogonal. It’s important to remember, especially with implicit curves, slopes at certain points may become undefined or complex, such as in \((0, 0)\). This reinforces why a full analysis includes multiple intersection points to demonstrate orthogonality.

Perpendicular Tangents

When curves intersect orthogonally, it means their tangents at the intersection points are perpendicular.

Mathematically, for two tangents to be perpendicular, the product of their slopes should equal \(-1\).

• For our curves, substituting slopes, \(m_1 = -\frac{2x}{3y}\) and \(m_2 = \frac{3x^2}{2y}\), you find their product: \(m_1 \cdot m_2 = -1\).

This condition confirms the orthogonality of tangents, emphasizing the importance of correctly identifying intersection points.

Let’s recall, \((0,0)\) could be a potential point where both slopes were unconventional or undefined. In practice, analysis would emphasize valid \((x, y)\) pairs to rectify any unconventional point readings. By ensuring \(m_1 \cdot m_2 = -1\) holds, it affirms their perpendicularly aligned nature, cementing orthogonality in given curves.