Lesson 6: Manipulating Equations

Transcription

1 Lesson 6: Manipulating Equations Manipulating equations is probably one of the most important skills to master in a high school physics course. Although it is based on familiar (and fairly simple) math concepts, it is still a stumbling block for most new physics students. Manipulating an equation means that you rearrange the equation so that the unknown you are trying to calculate is on its own on one side of the equation. Later in the course it will also give you the power to combine formulas (which is necessary) to solve more complicated problems. Learning how to manipulate formulas now (while the formulas are still easy ones!) will pay off later. At all times remember two basic rules from math.... To move something to the other side, just do the opposite math operation to it.. If you do it to one side, do it to the other. Example : The basic formula for calculating the velocity of an object is v d / t, where "v" is the velocity, "d" is the displacement, and "t" is the time. This formula is great "as-is" if we are going to calculate velocity, but what if I need to calculate the displacement and I've been given the velocity and time? Solve the formula to solve for "d". In the formula v d / t, "d" is being divided by "t". To get "t" to the other side, we need to do the opposite... multiply by "t"! v d t v d t t Warning! Do not use any methods like "the triangle" that you learned in Junior High for the formula v d/t. This works great for an easy formula, but try using it for v f v i + ad... it does not work! but what we do to one side, we do to the other... t v d t t the "t" on the right side cancel each other out leaving... (t) v d the last step (and it's basically just a tradition) is to put our unknown on the left side of the equation. So let's flip flop the whole thing to get our final equation! d t v And that's it! This is not a new formula you need to memorize; it is a formula already on your data sheet that you have manipulated to use more easily for a particular question. /8/007 studyphysics.ca Page of

2 Now try to solve the same formula for "t". You should get... t d v Be careful with formulas with addition, subtraction, square roots and squares. You basically need to follow the BEDMAS (Brackets, Exponents, Division, Multiplication, Addition, Subtraction) rule from math, but backwards. Usually take care of any addition and subtraction first, then multiplication and division, and finally exponents (remember, square root is just an exponent). Example : Solve the formula v f v i +ad for v i Before doing anything else, take care of anything being added or subtracted to v i by doing the opposite... v f - (ad) v i +ad (ad) which leaves us with... v f - ad v i flip the whole formula (so v i is on the left) and take the square root of both sides... v i v f ad and you're done! /8/007 studyphysics.ca Page of

3 Equations Warm-up: Rules for manipulating equations Learning objectives: 3.A.3. to be able to rearrange equations using the following rules: add or subtract the same thing to both sides if a b then a + c b + c multiply or divide both sides by the same thing if a b then a x c b x c replace any term or expression by another equal expression if a + b c and b d x e then a + (d x e) c square or square root both sides if a + b c then (a + b) c also if b a c then a b c expand out an equation y(a + x) becomes ya + yx simplify (factorise) ab + ac a(b + c) reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page of 7

4 Now use these rules to answer the following questions. You may want to think about some of these tips. When rearranging an equation, don t be afraid to use a lot of small steps and write down every step. Sometimes it isn t at all clear how best to proceed just start, remembering what it is that you need to make the subject of the equation and eventually you will get there. There can be a lot of different ways of doing it. Brackets are useful because you can move the whole term (ie what is inside the brackets) around as if it is a single item. Questions: Q. onsider the equation v u + at. Make a the subject of the equation. Q. Rearrange s ut + ½ at to make a the subject of the equation. m Q3. Rearrange v to make p the subject p L Q4. Rearrange F to make d the subject. 4πd Q5. Rearrange y to make x the subject. + x 0.69 Q6. If V and k, write an equation for V in terms of and t. k t Q7. Drugs in the blood can be bound to plasma proteins and/or free in solution, in practice there is an equilibrium whereby free + protein <--> bound. bound The percentage of drug bound is given by b 00 where bound + free The fraction of drug in plasma that is free is given by Express f as a function of b. free f. Q8. Rewrite the following so that the brackets are removed. (a )(b 3) 0 Q9. Rewrite the following so that the brackets are removed. (x + 3y + )(x 3) 0 Q0. Factorise 3x x reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page of 7

5 Q. Factorise the following expression: 5 y Q. In pharmacology, the proportion of receptors bound with drug D is given by eqn. DK eqn. proportion bound (K is the affinity constant.) DK + when a competing drug B is added, a higher concentration of drug D is given to get the same number of receptors bound with drug D. D K eqn. proportion bound (K B is the affinity constant for drug B) D K + BK B + since the proportion bound is the same in eqn and eqn we can make the right hand side of eqn equal to the right hand side of eqn. DK D K DK D K + BK + + B Simplify this as much as possible, getting D as a function of B. Answers: A. start Step. Step. Step 3. You want to get a on its own on the left. So start by reversing it. You want a to be on its own so start by subtracting u from both sides To get a on its own, you have to divide both sides by t. v u + at u + at v u u + at v u at v u v u v u a t t t ( ) A. Step. Step. Step 3. You want to get a on its own on the left. So start by reversing it. You want a to be on its own so start by subtracting ut from both sides To get a on its own, you have to multiply both sides by then divide both sides by t ut +½at s ut ut + ½at s ut ½at s ut at (s - ut) ( s ut) ) at t t s ut) a t ( ) reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page 3 of 7

6 A3. Step. Step. Start by squaring both sides. You want p to be on the left, so multiply both sides by p. m v p m pv p pv m Step 3. Now divide both sides by v m p v p A4. Step. Step. Step 3. You need to get d off the bottom. To do this multiply both sides by 4πd Now to leave d on its own, divide both sides by 4πF Now square-root both sides. 4πd 4πd d L F 4πd F L 4πd F 4πF d L 4πF L 4πF L 4πF 4πd A5. Step. Step. Step 3. Step 4. You need to get (+x) off the bottom. To do this multiply both sides by (+x) y( + x) Here you are treating what s inside the brackets (+x) as a single term. y + x You want to get x on its own, so expand out the brackets. y+xy Now subtract y from both sides to leave xy on xy y the left on its own. Now to get x on its own, divide both sides by y. ( + x) ( ) xy y x y or y y y ( + x) y x y these last two expressions are equivalent. reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page 4 of 7

7 A6. t t V A7. bound + free rearranging: free - bound f f f f free b 00 bound bound A8. Step. This is our starting point. (a )(b 3) 0 Step. You have to multiply each term in the first bracket by each term in the second bracket. ab b 3a A9. Step. This is our starting point. (x + 3y + )(x 3) 0 Step. You have to multiply each term in the first x + 3xy + x 3x 9y bracket by each term in the second bracket. Step 3. Then collect similar terms together. x + 3xy + x 3x 9y x + 3xy x 9y A0. The term factorise means to find the terms which were multiplied together to give this. In this case you can take x out of both terms 3x x x(3x ) A. Here you have to remember that the difference of two square numbers is the same as the product of their sum and difference i.e. 5 y (5 y)(5 + y) reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page 5 of 7

8 A. Step. This is our starting point. Our aim is to get all terms with D on the left and all terms with B in them on the right. Step. In order to move things around we need to get them off the bottom (denominator to use the technical). Start by multiplying both sides by (DK+). (OK so it doesn t look a lot better but just wait...) Step 3. Now multiply both sides by (D K + BK B +) I ve used brackets strategically so that I can see it more easily otherwise it can look like a real mess. Setting things out clearly is really important here. DK DK DK D K + BK DK DK ( + ) + B + D K DK + BK ( DK+ ) ( DK + ) ( D K)( DK) + ( DK) DK DK + BK + B B + DK( DK + BKB + ) ( DK)( DK) + ( D K) ( DK)( DK) + ( DK)( BK ) + DK ( DK)( DK) + ( DK) B Step 4. Now have a look and see what terms appear on both sides. See that (DK)(D K) appears on both sides, so you can subtract (DK)(D K) from both sides leaving... (DK)(D K) + (DK)(BK B ) + DK (DK)(D K) + (D K) ( DK)( BK ) DK ( D B + K) Which looks much better. Step 5. Now you have (DK) appearing in both terms on the left hand side. Try simplifying this... Step 6. Now you can see that you can divide both sides by K which will get rid of the K. Step 7. If you want to you can divide both sides by D so you have both D and D on the same side but that is a bit cosmetic. DK(BK B + ) D K DK(BK B + ) D K D(BK B + ) D BK B + D /D reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page 6 of 7

9 looking back, you started with something not too big, then went through something that looked really quite horrible, then ended up with something quite simple. reative ommons Attribution Non-commercial Share Alike Author Dr Jenny A Koenig Page 7 of 7

10 Lesson 7: Graphing Graphing is an essential skill for both Physics 0 and 30. You MUST be able to follow all of the rules of properly drawing a graph, and also be able to do basic interpretation of graphs. When you are presented with a chart of numbers that you are going to graph, you should start by identifying which variable is the manipulated variable and which is the dependent variable. In a lab, you are usually watching to see one thing change on its own, while letting something just plod along in an expected way. The one that you're watching for changes in is called the dependent or responding variable, drawn on the y-axis. The one that is changing in a regular expected pattern is the independent or manipulated variable, drawn on the x-axis. As a rule of thumb, time almost always goes on the x-axis. Graphing Rules Graphs you draw must have the following five basic characteristics. If you miss any part you will lose marks. Title Your title should be short, but still clearly tell what you have graphed. The most common and recommended way to name your graph is to say what your y-axis and x-axis are. One way to say it is Y-Axis versus X-Axis The other common way is to say Y-Axis as a function of X-Axis Labeled Axis Make sure to write out the full name of what you have graphed on each axis, along with the units you used. If you are using any sort of scientific notation for the numbers, make sure you show it here also. A Well hosen Scale The information you plot should always cover at least 75% of the area on your graph. Start by looking at the maximum values you have for both the x and y axis. Then check out how many major ticks you have on each axis of the actual graph. Divide your maximums by the ticks to find out roughly what to label each tick as. Data Plotted orrectly It s too bad when a person does all this work, and then does a sloppy job of plotting their information. Make sure you are as careful as possible when marking your points on your graph, otherwise everything else is a waste of time. You should always put little circles around each dot, since they might be hard to see on the graph paper. It also shows that each data point is a bit iffy. /8/007 studyphysics.ca Page of 4

11 Best Fit Line This step is sometimes optional, since your data might not give you a graph that has a straight line linear relationship. If your graphed data looks like a curved exponential relationship, draw a smooth curved line through your data points instead. When drawing a best fit line do not connect the dots. Instead, you should try to draw a completely straight line that pases through as many of your data points as possible. Try to get as many points above the line as below. This is the line that you must calculate your slope from. Use the formula... slope rise run y y x x You must know this formula and know how to use it! You are not allowed to use the data points you plotted! You must read two points from the best fit line. Example : Determine the slope of the following best fit line d as a function of t Notice that on the graph I have marked off two points on the best fit line... (x, y ) (4., 40) (x, y ) (, 8) I use these to calculate the slope. d (m) 30 0 slope rise run y y x x slope slope m/ s t (s) I am careful to give my final answer as a single value (not a fraction) showing units based on the axis (rise over run is metres over seconds) /8/007 studyphysics.ca Page of 4

12 Line Straightening A special skill involving graphs is a technique called line straightening, sometimes called an averaging technique. Essentially, you will sometimes have data to graph that would give you a curved line. Analyzing this line is very difficult... calculating a single slope given if you are asked to use an averaging technique for a or area under the line is impossible. graphing problem. What we need to do is come up with a way to manipulate the info you have so that, when plotted, it will give you a straight line of best fit (a linear relationship). You'll need to do some thinking to get these done.. Identify the main formula/concept that has something to do with the data you're looking at.. Get rid of the variables in the formula that are not part of the data you're looking at... this will leave you with a relationship, instead of a formula. 3. Look at what is being done mathematically to the variables. Most often one of them will either be squared or square rooted. reate a new column of values where you do this to the values. 4. Plot your new data as a straight line graph. Quite often, the resulting graph will have a slope or an area under the line that has some sort of meaning. Example : A student performs an experiment to measure the how centripetal force changes as the velocity is altered. He collects the data shown on the table below. Using a suitable averaging technique, determine a new set of values that, if graphed, would give a linear graph. Identify the meaning of the slope of this graph, and determine the mass of the object if the radius of the circle is.5 m. F c (N) v (m/s) First we need to identify the main formula (which we haven't studied yet!). Don't worry, questions you'll do will be based on formulas you've seen before. F c m v r Next get rid of the unneeded parts of the formula. The only variables we are concerned about according to the chart of values given are F c and v. F c v Now evaluate what this relationship is telling us. Nothing has been done to F c, but v is being squared. Warning! Do not make the mistake of calculating some sort of average based on the numbers /8/007 studyphysics.ca Page 3 of 4

13 ome up with a new column of values with v squared. No, really, just square v. Be sure to make the units squared as well. F c (N) v (m/s) v (m /s ) The graph would be F c vs v. Try graphing it yourself and then calculate the slope... you should get about 0 N/m /s To figure out what the slope actually means in this case, you would look at your graph this way... slope rise run F c and from the original formula F v c mv r F c v m r slope F c v m r slope m r So if we want to determine the mass of the object... slope m r mslope r m0 x.5 m3 kg /8/007 studyphysics.ca Page 4 of 4