91Ó°ÊÓ

Implicit differentiation is a technique used in calculus to find the derivative of a function when it is not explicitly solved for one variable in terms of another. This is especially useful when dealing with equations where one of the variables is a function of another, but neither is isolated on one side of the equation.
In implicit differentiation, we differentiate both sides of the equation with respect to a given variable, often using the chain rule. For example, in the exercise provided, we differentiate both sides of the equation \( t = \ln(x-t) \) with respect to \( t \). This results in differentiating \( \ln(x-t) \), which requires us to apply the chain rule.
The chain rule states that if a function \( y \) is dependent on \( x \), and \( x \) is dependent on \( t \), then the derivative of \( y \) with respect to \( t \) involves multiplying the individual derivatives of these relationships. This leads to the equation:

• \( \frac{d}{dt}[t] = \frac{d}{dt}[\ln(x-t)] \)
• This solves to \( 1 = \frac{1}{x-t}(\frac{dx}{dt} - 1) \).

In this context, implicit differentiation allows one to find the derivative \( \frac{dx}{dt} \) needed to determine other rates of change for variables within the same system.

A differentiable function is a function whose derivative exists at each point in its domain. Differentiability implies continuity, but a function can be continuous without being differentiable.
For example, in the exercise, we assume that \( x = f(t) \) and \( y = g(t) \) are differentiable functions of \( t \). This assumption allows us to apply calculus techniques to find derivatives.
When dealing with these functions, the parameter \( t \) is the independent variable, and \( x \) and \( y \) are functions of \( t \). Differentiability is crucial because it ensures that these functions change smoothly around the parameter values, making it possible to apply implicit differentiation.
In practical terms, the differentiability of \( y = t \cdot c^2 \) was straightforward because \( y \) is directly expressed as a function of \( t \). Differentiating \( y \) with respect to \( t \) yielded \( \frac{dy}{dt} = c^2 \), a constant rate of change that makes it easy to analyze the behavior and slope of the equation at given points.

Parametric equations provide a way to represent curves in the plane using parameters. They are particularly useful in situations where a curve is naturally expressed in terms of a third variable.
In this exercise, the curve is defined parametrically by \( x = f(t) \) and \( y = g(t) \). Parametric representation allows exploration of complex curves without needing a direct \( y = f(x) \) form.
One of the main advantages of parametric equations is that both \( x \) and \( y \) can depend on another parameter \( t \), which offers flexibility in describing paths that are not functions in the traditional sense.

• For \( t = 0 \), the coordinates are checked and used in calculations.
• The value of a derivative at a particular point provides the slope of the curve at that point.

In our specific example, the slope at \( t = 0 \) was calculated using the parametric values \( \frac{dy/dt}{dx/dt} \), highlighting an essential use of parametric equations in calculus to understand curves and their properties.