# Equations, Lenses and Fractions

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1 46 Equations, Lenses and Fractions The study of lenses offers a good real world example of a relation with fractions we just can t avoid! Different uses of a simple lens that you may be familiar with are using a magnifying glass to read small print, focusing a camera lens to take a picture, and using an overhead projector. The focal length of a lens is the distance from the lens at which parallel light rays focus. It is easy to find the focal length for a magnifying glass by holding the lens out in bright sunlight and moving it to and from a surface until you see one small point of light. The focal length is the distance from the lens to the point of light, as seen in the following figure below. All you need to do is measure that length. But be careful. The point of light will be quite hot and can burn you. It can set paper or wood on fire. A lens can focus an object on one side of the lens onto a surface on the other side of the lens. This is what a camera and an overhead projector do. Note in the following figure that the image is upside down. An overhead projector uses a mirror to flip the image right-side up. In the following figure, we let d represent the distance of the object from the lens and let i represent the distance from the lens to a surface on which the image is projected. For the image to be in focus, d, i, and f must satisfy the equation = +. Your professor f d i adjusts an overhead projector by moving the lens closer to or further from the object (the transparency) until the distances satisfy this equation. You adjust a camera by moving the lens closer to and further away from the film until i, the distance of the lens from the film, satisfies this equation.

2 47 Focal lengths for camera lenses are usually 20 mm to 70 mm. Low focal lengths are good for taking pictures of large landscapes. Larger focal lengths are good for close-up pictures. The lens of a disposable camera usually has a focal length of about 30 mm. Suppose you have a camera with a 30 mm lens and you want to take a picture of a flower that is about 600 mm from you (almost 2 feet). To find the distance the lens needs to be from the film, you can solve the equation = + for i. 1. Where did the 30 and the 600 come from? How do we know each one is in the correct position in the equation? 2. Approximate the solution to the equation = + numerically on your TI-83. Write the equations exactly the way you would input them in a calculator: Y 1 = Y 2 = Now use the TABLE feature of your calculator to approximate the solution to the nearest 10 mm. Explain how you do this and why it works. Tell why it is clear from the equation that the solution to the nearest 10 mm is not the exact solution. Use TABLE to find the solution to the nearest mm. Then to the nearest tenth of a mm. 3. Graph Y 1 and Y 2 on your calculator. Observing the y-values in the Table can help you set Ymin and Ymax in WINDOW. How can you use the graphs to solve the equation? Explain why what you described gives the solution to the equation. From your graph, solve = +, accurate to 4 decimal places. 600 i 4. Next we study the algebraic approach. The difficulty with this equation is that it has fractions in it. We wish it didn t! We can make our wish come true. Consider that a fraction is just a division problem. 2 x means x divided by 2. i 1 means 1 divided by i. What operation un-does division? (What is the opposite of division?) Consider as an 1 1 example the equation = 5+. It has two fractions, so it has two divisors (numbers that 3 x are doing the dividing). We could multiply both sides of this equation by 3x.

3 x = 3x 5 +. Why is this a good idea? 3 x Now we have x = 15 x+ 3. Do you see what happened?

5 Solve + = numerically. Explain what you do and why it works. What equations do you put into your calculator? What do you do then? In what column of the table do you find the solution? What is the solution? 11. Solve + = graphically. Explain what you do and why it works. What equations do you put into your calculator? What do you do then? What window did you use? How did you decide what window would be useful? What is your result? 12. Research question. The distance from the lens of the eye to the back of the retina would give us a slight underestimate of the focal length of the eye. See if you can find out what that distance is, on average. Given the focal length of the eye and the problem we saw in question 8, it might seem that we would be unable to see anything that was up close to our eye. Find out why it is that most of us can see things that are closer than the focal length of the average eye suggests.

6 51 Teaching Guide for Equations, Lenses and Fractions Introduction: The mathematical content of this lesson is the solution of equations that involve fractions. The goal is to focus the student s attention on the meaning of solution of an equation more than to teach techniques for algebraic manipulation. Thus, the lesson includes solution of the equations by graphical, numerical, and algebraic methods. Using all three permits the student to focus on what the equation means and what it means to solve it. Take a magnifying glass to class if you can. You can then show students how to find the focal length. Take a camera to class, too, and be sure that an overhead projector will be on hand. Remark for Teachers Only: To derive the relationship among i, d, and f for a lens, observe the figure below, assuming that lines that look parallel are parallel. From the figure, you get, by d i i f f similar triangles, = and =. Solve for a in the first equation and substitute into the a b b a second equation. The b s cancel out. The resulting equation can be rewritten as = +. f d i You shouldn t present this to your class, but it is here for your information and to share with any student who inquires. Answers and teaching suggestions: 1. The 30 and the 60 come from the information in the paragraph before the question. We have a lens with a 30 mm focal length, f. The flower we want to photograph is at a distance of 600 mm, d. 2. Y 1 = 1/30 Y 2 = 1/ /x Now use the TABLE feature of your calculator to approximate the solution to the nearest 10 mm. Explain how you do this and why it works. Set TableStart at 0 (even though that asks for division by 0) or any multiple of 10. Set Table to 10. Then go to the table and look for the row where the two values for Y, Y 1 and Y 2, differ the least. We ll see X Y 1 Y

7 From this we estimate that the solution to the equation is i = 30, to the nearest 10. This is because 30 is the value of x=i that makes the value of 1/30 and the value of 1/ /i most nearly equal. However, the observant student will note that this gives us = +, which is clearly not true. It is useful to discuss this equation. It isn t a true statement because we didn t find the exact value of the solution, but it is approximately so; indeed, we know that the difference between the left side and the right side is exactly 1/600. Use TABLE to find the solution to the nearest mm. Then to the nearest tenth of a mm. Looking at the table again, we see that the value X that would make Y 1 = Y 2 appears to be between 30 and 40 since the Y 2 values are decreasing and the value for Y 1 is greater than the value Y 2 when X=30, but Y 1 is less than Y 2 when X= 40. So we reset the table with TblStart=30 Tbl = 1 This time we see that the solution is between 31 and 32. X Y 1 Y Thus, we revise the Table Setup once again, using TblStart=31 Tbl = 0.1 We see X Y 1 Y So we conclude that the solution to the equation is between 31.5 and These results are the point of the table search, but we can also answer the question that was actually asked. It is now clear that the solution to the nearest whole number was 32. However, to avoid obscuring the points we are making about what it means to solve an equation, either 31 or 32 should be accepted as estimates. We may feel confident that 31.6 is the solution to the nearest tenth, and it is acceptable to just say so. To test that with the table we would use TableStart = 31.5, Tbl =.01. This would show us that x=31.58 gives both Y 1 = and Y 2 =.03333, assuring us that the solution is 31.6 to the nearest tenth. 3. Graph Y 1 and Y 2 on your calculator. Students have observed from the table that we need to be able to see X-values above 30, and that the Y values are very small and positive. They should use this to decide on a useful window. They might, for example, set Xmin = 0 Xmax = 50

8 53 Ymin = 0 Ymax = 0.1 How can you use the graphs to solve the equation? The solution is the x-coordinate of the point of intersection. We can see that by a TRACE, ZOOM, TRACE, ZOOM process; or we can take advantage of the calculator s CALC menu. Hitting 2 nd CALC 5, and then Enter, Enter, Enter (to answer the calculator s questions as Yes, Yes, Okay) will get us Intersection X= , Y= Students should explain that this means that at the point on the graph where X= (to 4 decimal places), both 1/30 and 1/ /x are equal to (about) Thus, the solution to the equation is X= , to 4 decimal places Discussion of the equation = 5+ is given to provide an example and help the students 3 x develop insight so that they can solve our lens equation independently. It is not necessary to 1 1 complete the solution of = 5+ beyond where we take it here. Be sure students discuss and 3 x understand the concepts that (1) fractions are division operations and (2) the inverse of division is multiplication, so that they understand that they can un-do division by multiplying by the divisors. Be sure, too, that they discuss and understand that if two equal expressions are to remain equal, then we will need to multiply both of them by the same anount. The problem tell students we will multiply both sides of this equation by 3x x = 3x 5 +. Why is this a good idea? Students should tell you that multiplying by 3 undoes the division by 3 in the fraction x, and multiplying by x un-does the division by x in the fraction x1. Now we have x = 15 x+ 3. Do you see what happened? With both of those divisions un-done, we have no fractions left in the equation and we can easily solve it. 5. In solving = +, what should you multiply both sides by? Solve = +, giving your answer as a fraction, reduced to lowest terms, and then as a decimal to 5 decimal places. In solving 1/30=1/600+1/i algebraically, some students may multiply both sides of the equation by (30)(600)i. After students have solved this equation, discuss the method of solution with the class. See if anyone realized that it could be solved by multiplying by 600i. Letting them struggle first with a larger number will help them appreciate the value of using the least common denominator. It is important that they realize they don t NEED to use the least common denominator, only that it makes computations easier. To 5 decimal places, i= Look at a camera and estimate how far, in mm, the film is from the lens. (Look at a metric ruler if you need help with this estimate.) Is your solution to the equation in question 5 consistent with your estimated distance? Students really should see the camera and consider the actual distances so that they know this problem models the world they live in. Our solution,

10 55 Checking, besides (1) reinforcing the meaning of a solution to an equation, also (2) reviews addition of fractions and (3) gives the student insight into what was going on in the algebra we did to solve ( 13) 9 (2) 5 = + = 2( 13) 2 ( 13) 13 (2) 2( 13) = Does it satisfy the equation? Yes, we can see that it does. 10. Questions 10 and 11 each offer information in support of the other. It is important that students see and discuss this interaction of the numerical and graphical methods. Solve + = numerically. Explain what you do and why it works. What equations do you put into your calculator? Y 1 = ½ + 9/x and Y 2 = 5/(2x). This is an opportunity for students to discover or for you to tell them that some calculators will read 5/2x as (5/2)x. Since the denominator of the fraction is 2x, they will need parentheses around 2x. What do you do then? Set the table. For example: TableStart = 0 and Table = 10. When we go to the table, we see that the values of Y 1 and Y 2 do not seem to get close together as x increases. Students may want to look at the graph at this point to see what is going on. Or, they may immediately think of going in the other direction to look at negative values of x in the table. Doing so, they see X Y 1 Y ERROR ERROR This suggests that there could be a solution between -20 and -10 since Y 1 > Y 2 when x = -20 but Y 1 < Y 2 when x = -10. So, we revise the Table Setup. Perhaps TableStart = -20 and Table = 1. (TableStart could be either -10 or -20 and Table could be either +1 or -1 since we can move either up or down in the table.) Exploring the table, we find X Y 1 Y In what column of the table do you find the solution? The solution is in the X column. What is the solution? X = Solve + = graphically. Explain what you do and why it works. What equations do you put into your calculator? Y 1 = ½ + 9/x and Y 2 = 5/(2x), as in question 10. What do you do then? Set the window in some useful way. Then find a point where the two graphs intersect.

11 56 What window did you use? How did you decide what window would be useful? Table values help to reduce the guesswork in this process. One window that would work is Xmin = -40 Xmax = 40 Ymin = -1 Ymax = 1 What is your result? The graphs intersect at (-13, ); thus the solution to the equation is x = -13. Observing the graph, students know they have solved an equation where the expressions were not linear. It is worth pointing out to them that the numerical and graphical methods that they have learned to use are powerful tools that permit them to solve many equations beyond the scope of their algebraic knowledge and that they are empowered to use these methods.

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