💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # It $R$ denotes the reaction of the body to some stimulus of strength $x.$ the sensitivity $S$ is defined to be the rate of change of the reaction with respect to $x.$ A particular example is that when the brightness $x$ of a light source is increased, the eye reacts by decreasing the area $R$ of the pupil. The experimental formula$R = \frac {40 + 24x^{0.4}}{1 + 4x^{0.4}}$has been used to model the dependence of $R$ on $x$ when $R$ is measured in square millimeters and $x$ is measured in appropriate units of brightness, (a) Find the sensitivity.(b) Illustrate part (a) by graphing both $R$ and $S$ as functions of $x.$ Comment on the values of $R$ and $S$ at low levels of brightness. Is this what you would expect?

## a) $S=\frac{-54.4}{\left(1+4 x^{0.4}\right)^{2} x^{0.6}}$b) see solution

Derivatives

Differentiation

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### Video Transcript

in this problem were given an equation which represents the area of the pupil of the eye. And we're looking at the sensitivity to light as the brightness increases. And the way we find the sensitivity is by taking the derivative of the area of the Radius. And so we're going to use the quotient rule to get this derivative since we have a quotient. So here we have the bottom times, the derivative of the top, minus the top times, the derivative of the bottom, over the bottom squared. And now we can simplify that. So what we do is we distribute the 9.6 x to the negative 0.6 power, and we distribute the 1.6 X to the negative 0.6 power and and the next step, we noticed that we can cancel these opposite terms, and then we can add these, like, terms together. And so, for the derivative, we end up with negative 54.4 over X to the 0.6 power times 1.1 plus for two times X to the 0.4 squared. That's a mouthful. Okay, so now what we want to do is use a graphing calculator and look at the graph of R and look at the graph of s together. So I take her calculator and we type both functions in, and then we need to change the window dimensions so that we get a good view. And so I decided to let my ex values go up to 20 and my wife values go from negative 20 to 20. But you could mess around with those numbers and find something that you like. Okay, looking at the graph now, the blue one represents our that was the area of the pupil, and the red one represents s That was the rate of change of the area of the pupil. And that's the sensitivity on one thing we can notice for small values of X that would be values where the amount of light is low as exchanges. There's a more rapid change in the area of the pupil. You can see that the area function is decreasing more rapidly at first, so there's a quick change in the area. The pupil, when you're going from dark to light, there is a much more slow change in the area of the pupil when you're going from something that's already bright to something that's a little bit brighter and similarly in the rate of change of the area, the rate of change happens much more quickly at first as we're going from dark to light and then the rate of change slows as we're going from something that's already light to a little bit more bright. Oregon State University

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Differentiation

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