Problem 42 - Calculus for Biology and Medicine

Question

We are not always given the function of interest in explicit form.y\( is related to \)x\( by an implicit equation. Determine using implicit differentiation and the first derivative test whether \)y\( is an increasing or a decreasing function of \)x\( \)x^{2}-y^{2}=1, \quad x>1, y>0$

Step-by-Step Solution

Answer

\(y\) is an increasing function of \(x\) in this region.

Step 1: Differentiate the equation implicitly

Start with the equation \(x^2 - y^2 = 1\). Differentiate both sides with respect to \(x\). The derivative of \(x^2\) with respect to \(x\) is \(2x\) and for \(y^2\) it's \(-2y\frac{dy}{dx}\) due to the chain rule. Differentiating the right side gives zero. Thus, the implicit differentiation gives: \[2x - 2y \frac{dy}{dx} = 0\]

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the equation from Step 1 to solve for \(\frac{dy}{dx}\). First add \(2y\frac{dy}{dx}\) to both sides resulting in \(2x = 2y\frac{dy}{dx}\). Then divide both sides by \(2y\) to isolate \(\frac{dy}{dx}\) on one side. The result is:\[\frac{dy}{dx} = \frac{x}{y}\]

Step 3: Use the given conditions

Given that \(x > 1\) and \(y > 0\), we have that \(\frac{dy}{dx} = \frac{x}{y}\) is positive since both \(x\) and \(y\) are positive. This indicates that \(y\) is an increasing function of \(x\) within this region. Therefore, \(y\) increases as \(x\) increases.

Key Concepts

First Derivative Test

One of the core tools for analyzing the behavior of functions is the first derivative test. This test helps determine where a function is increasing or decreasing and identify potential local maxima and minima. To do this, we first need the derivative of the function. In cases where the function is given implicitly, like in our exercise, we use implicit differentiation to obtain this derivative. Once the derivative, \(f'(x)\), is known, the first derivative test involves the following steps:

• Identify where the derivative is zero or undefined - these points are critical points.
• Test the sign of the derivative around these critical points.

If \(f'(x)\) changes from negative to positive, the function has a local minimum at that point. Conversely, if it changes from positive to negative, the function has a local maximum. If there’s no sign change, the point is neither. In our exercise, determining \(f'(x)\) using implicit differentiation gives us insights into the nature of the function’s slope, which sets the stage for analyzing whether it is increasing or decreasing.

Increasing Function

Functions are labeled as increasing when their output values rise along with the input values, over a specific interval. Mathematically, a function \(f(x)\) is increasing on an interval \((a, b)\) if for every \(x_1, x_2\) in \((a, b)\) where \(x_1 < x_2\), the inequality \(f(x_1) < f(x_2)\) holds true.The derivative comes in handy for testing this condition. If the derivative, \(f'(x)\), is positive over an interval, the function is increasing there. This comes from the mathematical property where the slope of the tangent line to the curve is positive. In the exercise, using implicit differentiation of the equation \(x^2 - y^2 = 1\), we derived \(\frac{dy}{dx} = \frac{x}{y}\). Given \(x > 1\) and \(y > 0\), both \(x\) and \(y\) are positive, which means \(\frac{dy}{dx} > 0\). Therefore, \(y\) is an increasing function of \(x\) in this context.

Decreasing Function

In simple terms, a function is called decreasing when it moves downward as you progress from left to right along the graph. Specifically, for a function \(f(x)\) to be decreasing on an interval \((a, b)\), for any \(x_1, x_2\) in \((a, b)\) where \(x_1 < x_2\), the condition \(f(x_1) > f(x_2)\) is true.The derivative, \(f'(x)\), plays a crucial role in detecting this behavior. If \(f'(x) < 0\) throughout an interval, \(f(x)\) decreases there. The slope of the tangent line shows a negative angle with the horizontal axis, indicating a downward trend. In our problem’s solution, we don't encounter this situation as \(\frac{dy}{dx} = \frac{x}{y}\) remains positive under the given conditions. Thus, the function \(y\) of the variable \(x\) is not decreasing when \(x > 1\) and \(y > 0\). The absence of a negative derivative in the specified region tells us that the function doesn’t descend in value as \(x\) increases.