91Ó°ÊÓ

Implicit differentiation is a technique used when dealing with functions that are not easily solved for one variable in terms of the other. It allows us to find the derivative of a function with respect to a variable, even when the function equation is given in implicit form.

For example, given an equation like \( 2x^2 + 3y^2 = 5 \), where both \( x \) and \( y \) are variables, we differentiate each term with respect to \( x \), treating \( y \) as a function of \( x \).

This involves using the chain rule for the derivative of terms containing \( y \).

• The derivative of \( 2x^2 \) is straightforward: \( 4x \).
• However, for \( 3y^2 \), the chain rule must be applied, resulting in \( 6y \frac{dy}{dx} \), where \( \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \).
• By setting up the equation \( 4x + 6y \frac{dy}{dx} = 0 \), we can solve for \( \frac{dy}{dx} \), which is the derivative of \( y \) in terms of \( x \).

This approach is essential for finding slopes of the tangent lines to curves defined by implicit relations.

Slopes of tangent lines represent the inclination of the tangent to the curve at a given point. They give insight into the behavior of the curve at that particular section. To find these slopes, we use the derivative \( \frac{dy}{dx} \) which, in the context of curves, is often found using implicit differentiation.

For the first curve \( 2x^2 + 3y^2 = 5 \), the slope of the tangent line is given by \( \frac{dy}{dx} = -\frac{2x}{3y} \). For the second curve \( y^2 = x^3 \), the slope is \( \frac{dy}{dx} = \frac{3x^2}{2y} \).

These expressions tell us how quickly \( y \) changes with \( x \) for each curve around any point. By analyzing these slopes, we can understand how steep the curves are at any point on respective curves.

• If \( \frac{dy}{dx} = 0 \), the curve has a horizontal tangent at that point.
• If \( \frac{dy}{dx} \) is undefined, it could indicate a vertical tangent.
• Slopes can further determine the relative angle between two intersecting curves.

In the context of intersecting curves, the product of slopes is a crucial concept for determining orthogonality, which means the curves meet at right angles.

If two curves intersect orthogonally, the product of their slopes at the intersection point is \(-1\). This mathematical condition can be described as

• If curve A has a slope \( m_1 \) and curve B a slope \( m_2 \) at their intersection, then \( m_1 \times m_2 = -1 \).

In our problem, we have slopes of \( -\frac{2x}{3y} \) for the first curve and \( \frac{3x^2}{2y} \) for the second curve.

The product is computed as:
\[ \left(-\frac{2x}{3y}\right) \times \left(\frac{3x^2}{2y}\right) = -\frac{6x^3}{6y^2} = -\frac{x^3}{y^2} \].

Only if this expression equals \(-1\) for any \( x \) and \( y \) at the intersection point, the curves are orthogonal. This verifies their relationship in terms of geometric alignment.

Curve intersection involves identifying points where two curves meet or cross each other. At these intersections, important geometric properties such as tangency and orthogonality can be analyzed.

In our exercise, intersection points are determined by finding common solutions to the equations of both curves:

• \( 2x^2 + 3y^2 = 5 \)
• \( y^2 = x^3 \).

Here, substituting \( y^2 = x^3 \) (from the second curve) into the first equation helps verify such points and orthogonality:

• Substitute into the product of slopes to check whether it equals \(-1\).

This process not only demonstrates the occurrence of intersection but also emphasizes the property of orthogonality. When \( y^2 = x^3 \), we see the calculation simplifies to \(-\frac{x^3}{x^3} = -1\), confirming that the curves intersect at right angles, thus proving their orthogonality!

By understanding how curves interact, these concepts help not only in geometrical representations but also in various applications in physics and engineering.