Problem 12 - Calculus for Biology and Medicine

Question

(a) The curve with equation \(y^{2}=x^{2}-x^{4}\) is shaped like the numeral eight. Find \(\frac{d y}{d x}\) at \(\left(\frac{1}{2}, \frac{1}{4} \sqrt{3}\right)\). (b) Use a graphing calculator to graph the curve in (a). If the calculator cannot graph implicit functions, graph the upper and the lower halves of the curve separately; that is, graph $$ \begin{array}{l} y_{1}=\sqrt{x^{2}-x^{4}} \\ y_{2}=-\sqrt{x^{2}-x^{4}} \end{array} $$ Choose the viewing rectangle \(-2 \leq x \leq 2,-1 \leq y \leq 1\).

Step-by-Step Solution

Answer

\( \frac{dy}{dx} \) at \( \left( \frac{1}{2}, \frac{1}{4}\sqrt{3} \right) \) is \( \frac{1}{\sqrt{3}} \).

Step 1: Differentiate Implicitly

To find \( \frac{d y}{d x} \), we will apply implicit differentiation to the given curve equation \( y^2 = x^2 - x^4 \). Differentiating both sides with respect to \( x \) gives us \( 2y \frac{dy}{dx} = 2x - 4x^3 \).

Step 2: Solve for dy/dx

From the differentiated equation \( 2y \frac{dy}{dx} = 2x - 4x^3 \), solve for \( \frac{dy}{dx} \) by dividing through by \( 2y \). This yields \( \frac{dy}{dx} = \frac{x - 2x^3}{y} \).

Step 3: Substitute Point into dy/dx

Substitute the point \( \left(\frac{1}{2}, \frac{1}{4} \sqrt{3}\right) \) into \( \frac{dy}{dx} = \frac{x - 2x^3}{y} \). Plug in \( x = \frac{1}{2} \) and \( y = \frac{1}{4} \sqrt{3} \) to get \( \frac{dy}{dx} = \frac{\frac{1}{2} - 2 \left( \frac{1}{2} \right)^3}{\frac{1}{4} \sqrt{3}} = \frac{\frac{1}{2} - \frac{1}{4}}{\frac{1}{4} \sqrt{3}} = \frac{\frac{1}{4}}{\frac{1}{4} \sqrt{3}} = \frac{1}{\sqrt{3}} \).

Step 4: Prepare Graphing Calculator

Since graphing calculators may not directly graph implicit functions, prepare to graph the curve using functions \( y_1 = \sqrt{x^2 - x^4} \) and \( y_2 = -\sqrt{x^2 - x^4} \).

Step 5: Graph Upper Half

Graph the upper half \( y_1 = \sqrt{x^2 - x^4} \) using the calculator within the window \(-2 \leq x \leq 2\), \(-1 \leq y \leq 1\). This should plot the top part of the 8-shaped curve.

Step 6: Graph Lower Half

Now graph the lower half \( y_2 = -\sqrt{x^2 - x^4} \) within the same viewing rectangle. This completes the graph by showing the bottom part of the 8-shaped curve.

Key Concepts

Graphing Implicit Functions

Implicit functions are equations where the variable you solve for is not isolated. Unlike explicit functions, you can't write them in the form \( y = f(x) \). In this case, the implicit function is \( y^2 = x^2 - x^4 \). This represents a curve which, in our example, forms a shape like the number eight.

Graphing implicit functions can be tricky. Many calculators can't directly graph them. However, you can break them into two explicit parts to visualize them better. For this exercise, the curve is split into two:

• The upper half: \( y_1 = \sqrt{x^2 - x^4} \)
• The lower half: \( y_2 = -\sqrt{x^2 - x^4} \)

By graphing \( y_1 \) and \( y_2 \) separately within the given rectangle, you can piece together the full picture. These explicit forms help visualize where the curve lies in relation to the axes.

Calculus Problem Solving

Calculus is a powerful tool for solving complex problems like finding the rate of change in implicit functions. With implicit functions, you differentiate both sides of the equation with respect to one of the variables, often \( x \), to find the rate of change of \( y \) with respect to \( x \), known as \( \frac{dy}{dx} \).

For our eight-shaped curve, we start by differentiating implicitly. The given equation is \( y^2 = x^2 - x^4 \). Differentiating both sides with respect to \( x \) yields \( 2y \frac{dy}{dx} = 2x - 4x^3 \).

Next, rearrange the equation to solve for \( \frac{dy}{dx} \):

• Isolate \( \frac{dy}{dx} \) by dividing both sides by \( 2y \).
• The result is \( \frac{dy}{dx} = \frac{x - 2x^3}{y} \).

This derivative gives insight into the slope of the curve at any given point as you substitute the values of \( x \) and \( y \).

Derivatives in Calculus

Derivatives measure how a function changes as its input changes. For implicit functions, derivatives are essential to find slopes and tell how one variable depends on another through differentiation.

In our exercise, after differentiating the equation \( y^2 = x^2 - x^4 \) with respect to \( x \), and solving for \( \frac{dy}{dx} \), you find:

• At the specific point \( \left(\frac{1}{2}, \frac{1}{4} \sqrt{3}\right) \), by substituting these values into the derivative, \( \frac{dy}{dx} = \frac{1}{\sqrt{3}} \).
This calculation tells us the rate of change, or slope, of the curve at that point. It shows how sensitive \( y \) is to changes in \( x \).

Derivatives in calculus are powerful; they not only tell us about slopes and rates of change but also help analyze implicit relationships in functions that are not straightforwardly solved for one variable.