Working with Formulas

Transcription

1 Working with Formulas Introduction Most mathematical and engineering knowledge is represented using formulas, so it is important to know how to work with them. Example The Area A of a circle with radius r is given by the formula A = πr 2. In this formula π is a number, and A and r are variables or unknowns. A is called the subject of the formula. The subject of a formula is the variable or unknown we are interested in. Working with formulas involves: performing calculations eg. What is the area of a circle with radius r = 5 cm? solving equations eg. If the area of a circle is 10cm 2, what is its radius? changing the subject eg. If A = πr 2, changing the subject to r and gives r = A/π These notes look at changing the subject of formulas. You may need to revise Equations before beginning. Click here to load down a pdf version of these materials. 1

2 2 Self-Test Try the following questions to refresh your memory and check your skills. If you have any difficulty follow the links in the answers. Use the navigation tabs to go to other examples or Topics. Questions 1. The perimeter of a rectangle with length L and width W is given by the formula Change the subject of this formula to L. = 2(L + W ). 2. When a voltage V (volts) is applied to a wire with resistance R (ohms), then the current I (amps) which will flow through the wire can be found from the formula Change the subject of this formula to I. V = RI. 3. The efficiency η of an engine compares the work output W O (Joules) of a machine to the work input W I (Joules), and can be calculated from the formula Change the subject of this formula to W I. η = W O W I. 4. The Area A of a circle with radius r is given by the formula A = πr 2. Show that A r = π. Answers... page 13

3 3 Why change the name of a formula? We change the subject of a formula to create a new formula that has the variable or unknown we are most interested in as its subject. Example Temperatures are measured in degrees Celsius (C) in Australia and in degrees Fahrenheit (F ) in the USA. In Australia we use the formula to interpret US temperature data. C = 5 (F 32) 9 In the USA, the formula F = 9 5 C + 32 would be used to interpret Australian temperature data. Both formulas contain the same information but have different subjects.

4 4 How to change the name of a formula Changing the subject of a formula is very similar to solving an equation. The examples below show different ways of changing (or solving for) the subject of a formula. Linear equations... page 5 Equations with products... page 7 Equations with quotients... page 9 Equations with quotients... page 11

5 5 1. Linear equations Changing the subject of a formula is very similar to solving an equation. Example The perimeter of a rectangle with length L and width W is given by the formula = 2(L + W ). (a) If we had to solve the equation 100 = 2(L + 20), we would find the unknown L by first expanding the brackets 100 = 2L + 40 then rearranging terms so the numbers are on one side and the unknown is on the other = 2L finally swapping sides and dividing both sides by 2 to get: L = [Ans: L = 30] (b) When changing the subject of the formula = 2(L + W ) to L, we think of L as being the unknown that we have to find, and think of and W as representing numbers. first expand the brackets = 2L + 2W then rearrange terms so the numbers are on one side and the unknown is on the other 2W = 2L finally swap sides and divide both sides by 2 get : L = 2W 2 This answer can also be written as L = 1 ( 2W ). 2

6 6 Exercise 1 Change the subject to V in the following formulas. (a) U = V + W (b) U = 2V + W T (c) U = 3(V + W ) + T 2 Answers... page 13

7 7 2. Equations with products Example When a voltage V (volts) is applied to a wire with resistance R (ohms), then the current I (amps) which will flow through the wire can be found from the formula V = RI. (a) If we had to solve the equation 100 = 20I, we would find the unknown I by first swapping sides so the unknown I is on the left and the numbers are on the right 20I = 100 then dividing both sides by 20 to get I: [Ans: I = 5 amps] I = (b) If we wanted to change the subject of the formula V = RI to I, then we can think of this as solving the equation V = RI for I. swap sides so that unknown I is on the left divide both sides by R to get I: RI = V RI \ R\ = V R I = V R

8 8 Exercise 2 Change the subject to Q in the following formulas. (a) = QR (b) = RQ (c) = Q + R (d) Q = R (e) Q = R (f) Q = R 2 Answers... page 14

9 9 3. Equations with quotients Example The efficiency η of an engine compares the work output W O (Joules) of a machine to the work input W I (Joules), and can be calculated from the formula η = W O W I. (a) If we had to solve the equation 0.8 = 20 W I, we would find the unknown W I by multiplying both sides by W I to bring W I out of the denominator of the fraction W I 0.8 = W 20 I W I 0.8 = 20 then dividing both sides by 0.8 to get W I : [Ans: W I = 25 Joules] W I = W I (b) The method for changing the subject of the formula η = W O as solving for W I. W I to W I is the same multiply both sides by W I to bring W I out of the denominator of the fraction W I η = W W I O W I η = W O then divide both sides by η to get W I by itself : W I η η = W O η W I = W O η W I

10 10 Exercise 3 Change the subject to Q in the following formulas. (a) = R Q (b) Q = R (c) = Q R Answers... page 15

11 11 Example 4: Equations with squares When an equation has a squared variable, solve for the squared variable first then take the square root. Example The Area A of a circle with radius r is given by the formula A = πr 2. (a) If we had to solve the equation 100 = πr 2, we would find the unknown r by first solving for r 2 r 2 = 100 π then finding r by taking the square root of both sides: 100 r = π [Ans: r = ] (b) When changing the subject of the formula A = πr 2 to r, we again solve for r 2 first solve for r 2 r 2 = A π then find r by taking the square root of each side: A r = π

12 12 Exercise 4 Change the subject to B in the following formulas. (a) 2A = B 2 (b) A = 2B 2 (c) A = 1 2 B2 (d) A = B 2 C Answers... page 16 (e) A = B2 C (f) A = B 2 + C

13 13 Answers to Self-Test 1. L = 2W see Linear for more information. 2. I = V R see roducts for more information. 3. W I = W O η see Quotients for more information see Squares for more information. Answers to Exercises Exercise 1 (a) (b) (c) U = V + W U W = V V = U W U = 2V + W T U W + T = 2V 2V = U W + T V = 1 2 (U W + T ) U = 3(V + W ) + T U = 3V + 3W + T U 3W T = 3V 3V = U 3W T V = 1 (U 3W T ) 3

14 14 Exercise 2 (a) = QR R = QR R R = Q Q = R (b) = RQ R = RQ R R = Q Q = R (c) = Q + R R = Q Q = R (d) Q = R Q = R Q = R

15 15 (e) Q = R Q = R Q = R (f) Q = R 2 Q = R2 Q = R2 Exercise 3 (a) = R Q Q = R Q = R Q = R (b) Q = R = RQ R = RQ R Q = R

16 16 (c) = Q R R = Q Q = R Exercise 4 (a) 2A = B 2 B 2 = 2A B = 2A (b) A = 2B 2 2B 2 = A B 2 = A 2 B = A 2 (c) A = 1 2 B2 2A = B 2 B 2 = 2A B = 2A

17 17 (d) A = B 2 C A C = B2 B 2 = A C B = A C (e) A = B2 C AC = B 2 B 2 = AC B = AC (e) A = B 2 + C A C = B 2 B 2 = A C B = A C