91Ó°ÊÓ

In calculus, differentiable functions are a type of mathematical function that has a derivative at each point in its domain. This means that you can calculate the rate of change or the slope of the function at any given point. Differentiability implies continuity; however, a continuous function isn't necessarily differentiable. In the context of our problem, both \(x\) and \(y\) are differentiable functions of \(t\), which means that their rates of change over time, or their derivatives, can be calculated. This is crucial when we want to understand how the distance \(s\) between two points changes, as we're differentiating \(s = \sqrt{x^2 + y^2}\) with respect to \(t\) to determine \(\frac{ds}{dt}\). If a function is differentiable, it provides us with a smooth curve that has a tangent line at each point, allowing us to accurately study changes along this curve.

The important points to remember about differentiable functions are:

• Differentiable functions have derivatives wherever they're defined, offering a precise way to measure how fast variables change.
• Such functions are continuous; hence there are no sharp breaks or holes in their graphs, making them predictable and easy to analyze.

The derivative is a key tool in calculus that measures how a function changes as its input changes. For any differentiable function, its derivative at a certain point tells you the slope of the tangent line to the function's curve at that point. Mathematically, if \(x\) and \(y\) are functions of \(t\), then \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) refer to their respective derivatives with respect to \(t\). These derivatives represent the speed or velocity of \(x\) and \(y\) over time.

In questions about related rates, such as those in this exercise, the derivative \(\frac{ds}{dt}\) tells us how fast the distance \(s\) is changing with respect to time. This is crucial in understanding the interplay between different rates of change. In our different scenarios, these derivatives help us comprehend:

• The direct relationship between \(\frac{ds}{dt}\) and \(\frac{dx}{dt}\) when \(y\) is constant.
• The combined effect of both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) on \(\frac{ds}{dt}\) when neither \(x\) nor \(y\) is constant.
• And the interrelated changes in \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) when \(s\) remains constant.

Overall, derivatives allow us to quantify motion, growth, and change in practically all dynamic processes.

The distance formula in the coordinate plane allows us to calculate the space between two specific points. In two dimensions, this is typically given by the formula \(s = \sqrt{x^2 + y^2}\), which is derived from the Pythagorean theorem. This formula computes the straight-line distance between the point \((x, 0)\) and \((0, y)\).

Understanding this helps solve our exercise, where \(s\) represents the distance between these points. When \(x\) or \(y\) change with respect to time, this affects \(s\) accordingly. Calculating \(\frac{ds}{dt}\) involves taking the derivatives of \(x\) and \(y\), which represent their rates of change:

• When \(y\) is constant, \(\frac{ds}{dt}\) is solely affected by \(\frac{dx}{dt}\).
• When both \(x\) and \(y\) change, we need to consider both rates of change to understand how \(s\) changes.
• When \(s\) is constant, a change in \(x\) necessitates an opposite change in \(y\) to maintain the distance.

By applying these insights, we gain a deeper understanding of how different variables interact spatially within the coordinate plane.