91Ó°ÊÓ

Implicit differentiation is a technique used in calculus when you have an equation that defines a function implicitly, rather than explicitly. This means you don't solve for one variable directly in terms of another. Instead, the variables are intertwined in such a way that they cannot be easily separated. For example, in the equation of the curve given by \[ x^3 - 2xy + y^3 = 0 \]both \( x \) and \( y \) are mixed together.
When we need to find the derivative \( \frac{dy}{dx} \), we perform implicit differentiation. This involves differentiating both sides of the equation with respect to \( x \). Each time you differentiate a term involving \( y \), you apply the chain rule and multiply by \( \frac{dy}{dx} \).
For instance, in the differentiated form of our curve's equation:\[3x^2 - 2y - 2x \cdot \frac{dy}{dx} + 3y^2 \cdot \frac{dy}{dx} = 0\]You can spot this usage of the chain rule with terms like \( 3y^2 \cdot \frac{dy}{dx} \). Ultimately, implicit differentiation helps us find slopes or rates of change even when equations don't provide \( y \) as a direct function of \( x \).

The ordinate is a fancy term for the \( y \)-coordinate in the coordinate plane. When considering points on a graph, we often refer to them in the form \((x, y)\), where \( x \) is the abscissa (or x-coordinate), and \( y \) is the ordinate.
In the given exercise, we are tasked with finding the ordinate, \( y_0 \), for a specific \( x \), noted as \( x_0 = 10 \).We achieve this by substituting \( x_0 \) into the equation of the curve and solving for \( y_0 \).
Given:\[ x^3 - 2xy + y^3 = 0 \] Substitute \( x_0 = 10 \):\[ 1000 - 20y + y^3 = 0 \]Solving this equation gives us the specific value for \( y_0 \) that lies on the curve along with \( x_0 = 10 \). Thus, the point \((x_0, y_0)\) like \((10, 10)\) ensures it lies precisely on the curve.

Derivatives are fundamental tools in calculus used to determine the rate of change of a function. In the context of our original problem, the derivative helps us find the slope of the tangent line at a specific point on the curve.
To find the derivative of our implicit function \[ x^3 - 2xy + y^3 = 0 \]we already performed implicit differentiation to express \( \frac{dy}{dx} \). This expression provides the rate of change of \( y \) with respect to \( x \) on the curve. Our derivative turns out to be:\[ \frac{dy}{dx} = \frac{2y - 3x^2}{-2x + 3y^2} \]This expression is evaluated at the point of interest \((x_0, y_0) = (10, 10)\) to determine the slope of the tangent.
Thus, substituting \( x = 10 \) and \( y = 10 \) yields:\[ \frac{dy}{dx} = \frac{20 - 300}{-20 + 300} = -1 \]So the slope of our tangent line at this point is \(-1\).

The equation of a line represents a straight path defined mathematically. For lines in a two-dimensional plane, we can often use the slope-intercept form, denoted as:\[ y = mx + c \]where \( m \) is the slope, and \( c \) is the y-intercept. In our exercise, we need the equation of the tangent line at the point \((10, 10)\) on the curve.
We already found the slope \( m = -1 \) from differentiating the curve. To get the line's equation through this point, we apply the point-slope form of a line:\[ y - y_1 = m(x - x_1) \]Plugging in \( m = -1 \), \( x_1 = 10 \), and \( y_1 = 10 \) gives us:\[ y - 10 = -1(x - 10) \]Simplify this to arrive at:\[ y = -x + 20 \]This equation tells us the behavior of the line that just "touches" or is tangent to the curve at the given point. Such tangents are crucial in setting a path for minimal or maximal values and understanding local behavior of curves.