pupil-size-when-the-brightness-x-of-a-light-source-is-increased-the-eye-reacts-by-decreasing-the-r-3

Question

Pupil Size When the brightness x of a light source is increased, the eye reacts by decreasing the radius R of the pupil. The dependence of R on x is given by the function
$$
R(x)=\sqrt{\frac{13+7 x^{0.4}}{1+4 x^{0.4}}}
$$

where $R$ is measured in millimeters and $x$ is measured in appropriate units of brightness.
(a) Find $R(1), R(10),$ and $R(100)$
(b) Make a table of values of $R(x)$

Amy Jiang · Accepted Answer

Okay, so we&#x27;re given our function for our vex. That&#x27;s equal to the square root function. And I we want to find values from 1 to 100. So it&#x27;s put that into our calculator. So we have the square roots, and that&#x27;s of that&#x27;s going to be 13 and plus seven times are ex term. In this case, that&#x27;s going to be people or and that&#x27;s to the power of zero points for. And then we&#x27;re gonna divide this by way one plus four and then before that&#x27;s going too many times before to the power of 0.4. So this was just too, and we want to find this from 10 or for a remaining expellees. So it&#x27;s this poor sound, so these are our values.

James Kiss · Answer

given the function R of X, for part A were asked to find a few values. A few inputs so 1st 1 are of one, would be describe it of 13 plus seven times, one to the point for over one plus four times one to the what one to the 10.4, which is equal to the square root of 20 over the scared of seeming 20/5 because one of the any one of the point for is one. This release is 13 plus seven, which is 21 of the boys. Organise one&#x27;s is one plus four or five 20/5 his forces with this where 24 which was two this is two milliliters are 10 would be the square root of 13 plus seven times. Tend to the 70.4 power over one plus four times 10 to the point for power here when you have a calculator. Once you evaluate what&#x27;s underneath the square, you get about 2.7683 which is roughly 1.66 millimeters and were asked to find our tent. Excuse me are 100 which would be found by plugging 100 for Axl. 13 plus seven times, 100 to the point for power over one plus four times 100 to the point for power, which is roughly squared A 2.1787 or about one point for a millimeters for heart be a table of values. I&#x27;m gonna jump over to the decimals online graphing calculator and put this function, and as we see it, so square root feature that keypad is in the bottom left of the square root of 13 plus, uh, seven x to the 70.4 and then you red arrow, and then divide. All right. Looks like we&#x27;re gonna need parentheses. We were all of us to be in the square root as a fraction perfect over one plus or X to the 0.4 power. All right, once we have the function in a table of values can be quickly achieved by hitting a car wheel and say, convert to table Doesn&#x27;t say for what values Understand. Go 0 to 10 and you can pick how many of these You what? You may want to applause it if you need to confirm on yours, but pick any of these for a table of values, and you can pick other ones if you wish to lads party. So pause if you need to. Other wasn&#x27;t going back to the other screen. And let&#x27;s talk about Part C. Uh, the net change from 10 to 100 so that would be are 100 minus are 10 that be, then that changed between the two. So the one point for a minus 1.66 which is about negative 0.18 was about a decrease of 0.18 millimeters, or 0.19 depending on 100 rounding throughout this but roughly about negative, a decrease in 0.1 air point one night.

Ankit Gupta · Answer

we have been given a question that onyx hit X is brightness badness off like source and products is nothing. Burger radius off the people and we have been given how the release off the people varies with brightness. No, this products has been given US ruled Ward off starting plus seven X rays to the poor. 0.4 divided by one plus four x again race to the PA 0.4 desperately thing gives the values off Alex for different values off X and in the question women us to find what is our one. So we&#x27;ll just put it 30 plus seven and has become one because wonderful, poor anything that data 0.4 will be again. One do uttered by one plus four multiplied by one. So what this gives This gives us us gold starting plus seven that is 20 and divided way five. That is nothing. But hence I don&#x27;t consult with similarly proceeding for our 10. We just put the values here 13 plus seven multiplied by pen faced with poor little point for divided by one plus four multiplied by can dress to the port zero point For now, this gives the value it 1.664 Similarly, if you move for on 100 we get they question something like starting plus seven multiplied by 100 three boys. Little point for divided by one plus four multiplied by on this little point for and the value of one point for 76 Now these three values are for different values off. No, In the second part, we have been asked to make a table off. So will make likable a forex. This is X Onda correspondent put We will get different values off Alex. So let us make an X here. Okay, different values of X. So when access support my understand well, get the value of minus 1.664 When it is zero, you get 0.3606 when it is your role one. Were there different ways to when it is dead we again get 1.664 and when it&#x27;s 100 were developed 1.476 No, these are the different values. Oh, products for the love X no Veneto find what is the nature? Teens option Caution seizes. Find the net change in the r as extended from 10 200 minutes. We need to find our 100 minus. Uh, then Okay, so we know that our 100 is 1.476 Minus are 10 are 10 is 1.664 And if you subtract it, it gives the value as minus zero point. Find this zero point 18 it hence, this is the net value net tinge in the radius when ex very full 10 200.

Savannah Payne · Answer

all right, This is a really cool problem that has to deal with how pupils are dilating and how they&#x27;re measuring. This could be really important for that medical field that has to figure out how to fit glasses and such. So first things first, Um, we&#x27;re, uh, looking for me. Fix this. I get so used to writing f of X all the time. I almost wrote it. We&#x27;re looking for are of one. And what that means is that whenever I have this and I plug in here, everything is going to be the same. Except for where the exes is now going to be, Ah, one raised to the 0.4 power. And then one plus for this would be times one raised to the 0.4 power. And what I would do if I were you is I would just go ahead and plug this value into the calculator and see what it comes out to be for time&#x27;s sake. So I can see here. I know that&#x27;s exactly what I&#x27;ve done already. Um, and I know that that value is too. So if I were to be making a table and I&#x27;m thinking about my ex values and my r of X values. So my input and my output I&#x27;m looking at one and two. And now, if I were to look at our, uh, 10 it&#x27;s going to be the same thing. Except for now, my, um, input value is gonna be 10. So 13 plus seven instead of one. This time it&#x27;s going to be 10 raised to the 0.0 for power in one plus four again instead of the eggs or the one it&#x27;s gonna be to the 10. And don&#x27;t forget about that 100.4 power. So this whenever I plug 10 into my calculator and I calculate this whole little function and I just typed it in exactly how, um, it&#x27;s written on this white screen here and I got a decimal. It was more than a point 95 931.9593 All right, so let&#x27;s go for one more. Let&#x27;s give our of 100 a try. So whenever I look at our 100 I am going to set everything up 13 plus seven. Now, instead of the X value, I&#x27;m having 100 raised to the 0.0 for power and then one plus four instead of an ex value 100 raised to the 0.4 And again, I might want to go ahead and just plug this directly into the calculator. That way I can find out exactly what it&#x27;s going to be. And I just a second to, you know, this type didn&#x27;t hear and be able to take a look at it. All right? And this came out to 1.95 Oh, not 95 Excuse me. Excuse me. Nine two 01 Okay, so now we have to look at the net change between, um so it said the net change between 10 and 100. So if you recall, this would be the A value, and this would be the B value because it&#x27;s listed that way where we know the net change is f of 100 minus f of 10. Well, guess what? Since we already have this listed here, the 100 is 1.9 to 01 minus 1.959 three. And that comes out to let&#x27;s give it a little try here. One point nine 201 minus one point 9593 It comes out to a nice. The net change is negative, 0.3 92 so that&#x27;s a pretty intense little decimal, but we&#x27;ve got it all done.

Nicholas Barvinok · Answer

we have the function are equals 40 plus 24 x 2.4 over one plus four x to the point. For now, the first thing we want to do is find s the sensitivity which is the derivative of our find. The derivative are we will be using the quotient rule which is the derivative of states that the derivative of F org is G f prime minus f g prime all over g squared. So in our case, we have the bottom function one plus four x to the 40.4 times the derivative of the top function. The derivative of 40 is zero. So we just have 24 x to the 240.4. By the power rule we bring down the 0.4 point four times 24 is 9.6 x to the minus 0.6. We subtract one from the original exponents minus the top function 40 plus 24 x to the point for times the derivative of the bottom function the derivative of 10 So we&#x27;re just left with the derivative of four extra 40.4 by the power rule, we bring down the 0.4 multiplied by four against 1.6 next to the minus 0.6, and all of that divided by one plus four x to the 40.4 squared. Now, for the second part, if you graph both of these, you can either. If you&#x27;re not plugging it into some sort of graphing calculator, you can just check a decent number of points and connect the dots you&#x27;ll find that are sort of looks like this near zero. And that sort of looks like this. Is this what I would expect? Well, sensitivity is negative. I don&#x27;t even know what that means physically. So I would say, Don&#x27;t expect this.

Heather Zimmers · Answer

in this problem were given an equation which represents the area of the pupil of the eye. And we&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s a quick change in the area. The pupil, when you&#x27;re going from dark to light, there is a much more slow change in the area of the pupil when you&#x27;re going from something that&#x27;s already bright to something that&#x27;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&#x27;re going from dark to light and then the rate of change slows as we&#x27;re going from something that&#x27;s already light to a little bit more bright.

Problem

Relativity According to the Theory of Relativity,…

Problem 74 Hard Difficulty

Answer

(a) $R(1)=2, R(10)=1.66383$ and $R(100)=1.476$
(b) $\begin{array}{|c|c|}\hline x & {R(x)} \\ \hline 1 & {2} \\ \hline 20 & {1.59344} \\ \hline 40 & {1.53568} \\ \hline 60 & {1.50714} \\ \hline 80 & {1.489} \\ \hline 100 & {1.446} \\ \hline\end{array}$

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(a) $R(1)=2, R(10)=1.66383$ and $R(100)=1.476$(b) $\begin{array}{|c|c|}\hline x & {R(x)} \\ \hline 1 & {2} \\ \hline 20 & {1.59344} \\ \hline 40 & {1.53568} \\ \hline 60 & {1.50714} \\ \hline 80 & {1.489} \\ \hline 100 & {1.446} \\ \hline\end{array}$

Algebra and Trigonometry

Anna Marie V.

Caleb E.

Maria G.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

James Stewart, Lothar Redlin, Saleem WatsonAlgebra and Trigonometry

Anna Marie V.

Caleb E.

Maria G.

Michael J.

(a) $R(1)=2, R(10)=1.66383$ and $R(100)=1.476$
(b) $\begin{array}{|c|c|}\hline x & {R(x)} \\ \hline 1 & {2} \\ \hline 20 & {1.59344} \\ \hline 40 & {1.53568} \\ \hline 60 & {1.50714} \\ \hline 80 & {1.489} \\ \hline 100 & {1.446} \\ \hline\end{array}$

James Stewart, Lothar Redlin, Saleem Watson

Algebra and Trigonometry