5.2 Partial Variation

Transcription

1 5.2 Partial Variation Definition: A relationship between two variables in which the dependent variable is the sum of a number and a constant multiple of the independent variable. Notice: If we take the words "...the sum of a number and..." out of the above paragraph then we are left with the definition for direct variation from chapter 5.1 in yesterday's lesson. RECALL: k = dpt/ indpt Rearranging this, and balancing both sides: dpt = k (indpt) Now, we are saying that the dependent variable (dpt) is not only a multiple of the independent variable by a factor of k, it is this PLUS a fixed number. dpt = k (indpt) + number 1

2 In many cases in real life, the value of a variable may be made up of two parts. One part is a constant and the other varies directly as another variable. This type of relationship gives rise to partial variation. 2

3 Example - DIRECT VARIATION: The cost of riding in a taxi depends on how far you travel. There is a charge of $1.50 per km. This could be represented by the equation: Let C represent the total cost of the taxi ride in dollars. Let d represent the distance travelled in km. RECALL FROM YESTERDAY: k = dpt/ indpt = C/ d And from the data given, we know that it costs $1.50 to go the distance of 1 km. So if C=1.50, and d=1, then: k = 1.50 /1 = 1.50 This value for k will be constant for all pairs of C and d, if C and d are in direct variation with each other. If k = C/d, and k = 1.50, then equating these, we get: C = 1.50d What about graphing this information? Distance, d [km] Cost, C [$] (1) 1.50 (2) 1.50 (3) 1.50 (3) Cost of a Taxi Ride Varying with Distance Travelled y cost [$] C = 1.50 d x distance [km] 3

4 Example - INDIRECT VARIATION: The cost of riding in a taxi depends on how far you travel plus a fixed cost. Let us say a fixed cost of $3 is charged, no matter how far you go, plus a charge of $1.50 per km. This could be represented by the equation: Let C represent the total cost of the taxi ride in dollars. Let d represent the distance travelled in km. The total cost is now made up of 2 parts: the first part is still the same as in the direct variation example, C 1 =1.50d, and the second part is a fixed cost, C 2 =3. C = C 1 + C 2 C = 1.50d + 3 Distance, d [km] Cost, C [$] (1) (2) (3) (50) The graph is the same as C = 1.50 d, except that 3.00 is added to each C value. y Cost of a Taxi Ride Varying with Distance Travelled, both with and without a Flat Fee of $3.00 cost [$] C = 1.50 d C = 1.50 d distance [km] x 4

5 Example Question #1 A car rental agency charges $50 per day plus $0.20 per kilometre. a) Write an equation to find the overall cost of renting a car for a day. b) Find the cost of renting a car having driven 200 km, 300km and 400km in one day. c) Graph the relation. d) Using your graph and then using the equation, find the cost of renting a car that needs to be driven 550 km in one day. Step 1: Identify the dependent and the independent variables. Step 2: Write Let statements. Step 3: Write out an expression for k. Recall k = dpt/ indpt. First write this equation by substituting in the dpt and indpt variables from Step 2. Step 4: Find a value for k. You will need actual numerical values of at least one corresponding pair of variables. Substitute in these values into the expression from Step 3. Step 5: Rearrange k = dpt/ indpt > dpt = k (indpt) Step 5: Now account for the 'fixed' portion of the dependent variable. For a Direct Variation: dpt = k (indpt) + 0 For a Partial Variation: dpt = k (indpt) + constant Step 6: Step 5 gives us a general expression for renting a car. If we want to find the answers to part b) of this question, we can work with this general expression and substitute in exact values. Note: If you are substituting in an exact value for the dependent variable, you will then use your algebra skills to rearrange the equation to find the value of the independent variable. Sidenote: It is also possible for some questions, that we could begin by not knowing k, but we might know the value of the constant, and a sample corresponding pair of values for the dpt and indpt variables. In this case, we could solve for k. Step 7: For part d) of this question, note that you will be finding the answer two different ways: graphically (by visually 'reading' the graph that you have made) and algebraically (working with the algebraic expression). 5

6 Example Question #2 A plumber charges $50 for a service call to your home plus $60 per hour. a) How much would you pay if the plumber worked 4 hours? b) How many hours did he work, if you had to pay $215? 6

7 Homework Read pages 246 to 249 Pages 250 to 253 Questions 1, 2, 4, 6, 7, 12 7