91Ó°ÊÓ

When dealing with parametric equations, we have a pair of equations that express coordinates as functions of a common variable or parameter, usually denoted by \( t \). These parametric equations allow us to represent curves in the plane that can be difficult to describe with only a single equation in terms of \( x \) and \( y \). In the given problem, the equations are \( x = \sinh t \) and \( y = \cosh t \). Here, \( t \) is the parameter that connects \( x \) and \( y \).

This means that as \( t \) changes, the values of \( x \) and \( y \) change accordingly, tracing a path in the coordinate plane. This approach is particularly useful for closed curves or loops, or when dealing with curves where \( y \) cannot be easily expressed as a function of \( x \).

These equations highlight a key feature: if you don't eliminate the parameter, it can often simplify the calculus process, especially when dealing with derivatives. In this exercise, we find derivatives, not in terms of \( x \) and \( y \) directly, but in terms of their derivatives with respect to \( t \).

The chain rule is an essential tool in calculus when you want to differentiate a composite function. In simpler terms, it's a formula for computing the derivative of the composition of two or more functions. In the context of parametric equations, the chain rule is handy to find the derivative of \( y \) with respect to \( x \) even though we initially find derivatives with respect to \( t \).

For our problem, to find \( \frac{dy}{dx} \), you don't differentiate \( y \) directly with respect to \( x \). Instead, you find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), which gives you rates of change with respect to the parameter \( t \).

• First, differentiate \( y = \cosh t \) with respect to \( t \), giving \( \frac{dy}{dt} = \sinh t \).
• Second, differentiate \( x = \sinh t \) with respect to \( t \), giving \( \frac{dx}{dt} = \cosh t \).

With these results, you calculate \( \frac{dy}{dx} \) using \( \frac{dy/dt}{dx/dt} \), which is a real manifestation of the chain rule. This result gives \( \frac{dy}{dx} = \tanh t \). At \( t = 0 \), \( \tanh 0 = 0 \), so the slope of the curve is 0 at that point.

Understanding the second derivative is crucial as it provides information about the curvature of a function. It can tell us whether a function is concave up or down at a particular point, which is important for determining local maxima and minima.

In our parametric problem, to find the second derivative \( \frac{d^2y}{dx^2} \) without eliminating the parameter, we apply derivatives a second time. We've already computed \( \frac{dy}{dx} = \tanh t \). The next step is to differentiate \( \tanh t \) with respect to \( t \), which gives us \( \text{sech}^2 t \). This derivative tells us how the slope \( \frac{dy}{dx} \) is changing as \( t \) changes.

To adjust this result for \( \frac{d^2y}{dx^2} \), we again apply the chain rule:

• Differentiate \( \tanh t \) to get \( \text{sech}^2 t \).
• Divide it by \( \frac{dx}{dt} \), which is \( \cosh t \).

This computation yields \( \text{sech}^3 t \).

Finally, by evaluating at \( t = 0 \), we find that \( \text{sech}(0) = 1 \), and so \( \text{sech}^3(0) = 1 \). This value for \( \frac{d^2y}{dx^2} \) at \( t=0 \) tells us the rate at which the slope \( \frac{dy}{dx} \) changes, reflecting the concavity of the curve at that point.