91Ó°ÊÓ

Implicit differentiation is a technique used when dealing with equations that define a relationship between variables, but not explicitly in terms of one variable. This often happens in equations involving curves, such as circles or ellipses. For example, taking the equation of a circle, like \( x^2 + y^2 = 4 \), derivatives can be found even though \( y \) is not isolated.
To use implicit differentiation, differentiate both sides of the equation with respect to one variable, usually \( x \). Remember to apply the chain rule to terms involving \( y \). For instance, the derivative of \( y^2 \) with respect to \( x \) is \( 2y \frac{dy}{dx} \).

• Start by differentiating both sides of the equation.
• Use the chain rule for terms involving \( y \).
• Solve for \( \frac{dy}{dx} \) to find the slope.

Practice and careful application of this technique make it useful for solving problems relating to orthogonality and more complex functions.

Intersection points are locations where two curves meet, meaning that they have common points on the graph. Finding these points is essential when verifying if curves meet orthogonally. The process starts by solving the given equations simultaneously to see where both curves share the same \( x \) and \( y \) values.
For instance, if two curves are defined by equations like \( x^2 + y^2 = 4 \) and \( x^2 = 3y^2 \), you would set the expressions equal, substituting or solving algebraically to determine common values.

• Solve the set of equations to find common solutions.
• These solutions give the coordinates where the curves intersect.
• These coordinates are vital for further steps, including finding the product of slopes.

Correctly identifying intersection points can provide the exact values needed to verify the orthogonality of the curves.

Orthogonal curves are curves that intersect at right angles. To confirm this mathematically, you need the product of their slopes at each intersection point to be \( -1 \). This involves calculating the derivative at the intersection points you identified.
Once you have the derivatives for each curve, simply multiply these values at the intersection points. In Case (a) of the original problem, the slopes were \( -\frac{x}{y} \) and \( \frac{x}{3y} \). The product \(-\frac{x^2}{3y^2}\) became \(-1\) after substituting values from the equations for \( x^2 \) and \( y^2 \), proving orthogonality.

• Find derivatives at intersection points.
• Multiply the slopes of the two curves.
• Verify if their product is \( -1 \).

This validation step is crucial for confirming whether the two curves meet orthogonally as expected.