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Area of the Pupil The relationship between the area of the pupil of the eye and the intensity of light was analyzed by B. H. Crawford. Crawford concluded that the area of the pupil is $$ f(x)=\frac{160 x^{-.4}+94.8}{4 x^{-.4}+15.8} \quad(0 \leq x \leq 37) $$ square millimeters when \(x\) units of light are entering the eye per unit time. (Source: Proceedings of the Royal Society.) (a) Graph \(f(x)\) and \(f^{\prime}(x)\) in the window \([0,6]\) by \([-5,20]\) (b) How large is the pupil when 3 units of light are entering the eye per unit time? (c) For what light intensity is the pupil size 11 square millimeters? (d) When 3 units of light are entering the eye per unit time, what is the rate of change of pupil size with respect to a unit change in light intensity?

Step by step solution

01

Understand the Function

The given function is \[ f(x) = \frac{160 x^{-0.4} + 94.8}{4 x^{-0.4} + 15.8}\] which describes the area of the pupil in square millimeters as a function of the light intensity, \( x \), entering the eye.

02

Graphing f(x) and f'(x)

Use a graphing tool or graphing calculator to plot the function \( f(x) \) and its derivative \( f'(x) \) within the given window \( [0, 6] \) by \([-5, 20] \). Note the shape and key features of both graphs.

03

Calculate the Pupil Size when x=3

To find the pupil size when 3 units of light are entering the eye per unit time, substitute \( x = 3 \) into the function: \[ f(3) = \frac{160 \cdot 3^{-0.4} + 94.8}{4 \cdot 3^{-0.4} + 15.8} \].Evaluate the expression using a calculator to obtain the pupil size.

04

Solve f(x) = 11

To find the light intensity \( x \) for which the pupil size is \( 11 \) square millimeters, solve the equation \[ \frac{160 x^{-0.4} + 94.8}{4 x^{-0.4} + 15.8} = 11 \]. Use algebraic techniques or a numerical solver to find the value of \( x \).

05

Find f'(x)

To determine the rate of change of pupil size when \( x = 3 \), first find the derivative \( f'(x) \). Use the quotient rule: \[ f'(x) = \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \], where \[ u = 160 x^{-0.4} + 94.8 and v = 4 x^{-0.4} + 15.8 \]. Evaluate \( f'(3) \) using the derived expression.

06

Calculate f'(3)

Substitute \( x = 3 \) into the derivative expression obtained in the previous step to find \[ f'(3) = \frac{u'(3) v(3) - u(3) v'(3)}{v(3)^2} \]. Calculate this value to determine the rate of change of the pupil size.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Light Intensity and Pupil Size

Understanding the relationship between light intensity and pupil size is crucial for various fields such as medicine and optometry. In essence, as the intensity of light entering the eye varies, so does the pupil's size to regulate how much light gets in.
The given function describes this behavior mathematically. Here's how it looks: \ \f(x) = \frac{160 x^{-0.4} + 94.8}{4 x^{-0.4} + 15.8}\ This function represents the area of the pupil in square millimeters, where \( x \) is the light intensity entering the eye per unit time. Understanding this relationship helps us examine how our eyes naturally adjust to different lighting conditions.

Derivative Application

The derivative of a function helps us understand how the function changes with respect to one of its variables. In this exercise, we want to find the rate of change of the pupil size based on changes in light intensity.
To do this, we use the Quotient Rule for differentiation because our function is a ratio of two expressions. This rule states: \ \f'(x) = \frac{u'v - uv'}{v^2} \ Here, \( u = 160 x^{-0.4} + 94.8 \) and \( v = 4 x^{-0.4} + 15.8 \). By calculating \( u' \) and \( v' \), and substituting them into the Quotient Rule formula, we can find the derivative \( f'(x) \).\
For instance, evaluating this at \( x = 3 \) gives us the rate at which the pupil size changes when light intensity is 3 units.

Graphing Functions

Graphing functions visually demonstrates the relationship between variables. In this case, we graph \( f(x) \) and its derivative \( f'(x) \) to observe how the pupil size and its rate of change vary with different light intensities.
When graphing these functions, we use a specific window for our axes: \[0,6] \] by \[-5,20\], meaning we consider light intensities between 0 and 6 units and pupil sizes ranging from -5 to 20 square millimeters. Using a graphing tool, we can plot both functions:

• \( f(x) \) gives us the pupil size directly.
• \( f'(x) \) shows how rapidly the pupil size changes as light intensity changes.

Noting the shape and key features, such as any peaks, troughs, or asymptotes, helps us better understand the behavior of the pupil under different lighting conditions.

Solving Equations Numerically

Sometimes, finding exact solutions to equations analytically can be difficult or impossible. In these cases, we use numerical methods to find approximate solutions. For example, to determine the light intensity \( x \) for which the pupil size is 11 square millimeters, we solve: \ \frac{160 x^{-0.4} + 94.8}{4 x^{-0.4} + 15.8} = 11\
Using tools like algebraic techniques or numerical solvers can help us approximate the value of \( x \). These methods iteratively adjust \( x \) to get closer to a solution, making them highly useful for complex equations.
Approximations might be slightly off from exact solutions, but they are usually accurate enough for practical purposes.