Ch. 2.4 Direct Variation AND Ch. 2.5 Inverse Variation

Transcription

1 Ch. 2.4 Direct Variation AND Ch. 2.5 Inverse Variation Learning Intentions: Learn the properties of direct & inverse variation equations. Graph direct & inverse variation equations. Read direct & inverse variation graphs to find missing values in a table. Use a direct & inverse variation equation to etrapolate values from given data. Develop an intuitive understanding of slope & linear equations. Direct Variation: y = k ~ the quantities represented by & y are directly proportional, & k is the constant of variation. ~ as increases -> y increases ~ as decreases -> y decreases Inverse Variation: y = k ~ the quantities represented by & y are inversely proportional, & k is the constant of variation. ~ as increases -> y decreases ~ as decreases -> y increases

2 Direct Variation Inverse Variation Neither Given: y y y If you re-organize the ordered pairs, it might make the comparison easier As increases, y increases y = k Let (,y)= (2,24) 24 = k(2) 12 = k y = 12 y As increases, y decreases y = k Let (,y)= (2,12) 12 = k 2 24 = k y = 24 y Direct: (,y) must be (0,0) y y = 9(0.5) Inverse: neither variable can equal 0

3 Ch. 2.4 p.117 #(3, 4 & 5) #3.) Find the missing values of the table given: y = 1.6 where represents distance in miles; y in terms of km. Round to the nearest tenth. Distance (mi) Distance (km) #4.) Describe how to solve each equation for. Then solve. a.) 14 = 3.5 b.) 8 = 45(0.62) c.) What does k = 1.6 really mean?? = d.) 12 = 0.8

4 SOLUTIONS: Ch. 2.4 p.117 #(3, 4 & 5) #3.) Find the missing values of the table given: y = 1.6 where represents distance in miles; y in terms of km. Round to the nearest tenth. Distance (mi) Distance (km) #4.) Describe how to solve each equation for. Then solve. Multi. each by Divide each side by 3.5 Divide each by 8 Multi. each by 7 Divide each by 0.8 a.) 14 = 3.5 b.) 8 = 45(0.62) c.) = = 8 8 = = Let y = 4.5 ; solve for y = = = = Let y = 1500 ; solve for Let = 7.8 ; solve for y y = 1.6 y = 1.6(7.8) y = 12.5 Let = 650 ; solve for y = d.) 12 = 0.8 (7) 7 = 0.375(7) = k = 1.6 means 1.6 km = 1 mile or 1.6km is the rate of change 1 mile () 12 = 0.8() 12 = =

5 Ch. 2.4 p.117 #(3, 4 & 5) #5.) The equation c = 1.25f shows the direct variation relationship between the length of fabric and its cost. The variable f represents the length of the fabric in yards; c represents the cost in dollars. a.) How much does 2½ yards of fabric cost? b.) How much fabric can you buy for $5? c.) What is the cost of each additional yard of fabric? d.) Graph each possible ordered pair (f, c). What does this model?

6 SOLUTIONS: Ch. 2.4 p.117 #5.) The equation c = 1.25f shows the direct variation relationship between the length of fabric and its cost. The variable f represents the length of the fabric in yards; c represents the cost in dollars. a.) How much does 2½ yards of fabric cost? Let f = 2.5 ; solve for c c = 1.25f Thus, 2½ yds. costs $3.13 c = 1.25(2.5) c = b.) How much fabric can you buy for $5? Let c = 5 ; solve for f c = 1.25f Thus, 4 yds. costs $5 5 = 1.25f = f c.) What is the cost of each additional yard of fabric? slope of a line = m = y c If c = 1.25f is written in terms of y = m + b or y =1.25 then m = 1.25 or which means $1.25 per 1 yard. d.) c = 1.25f models a linear equation with a constant increase (m = 1.25) f

7 Ch. 2.5 p.127 #(2, 5, 8 & 9) 2.) Two quantities, & y, are inversely proportional. When = 3 and y = 4. Find the missing coordinates for each point. Hint: Solve for k given y = k OR y = k. a.) (4, y) b.) (, 2) c.) (1, y) d.) (, 24) 5.) The amount of time it takes to travel a given distance is inversely proportional to how fast you travel. a.) How long would it take you to travel 90 miles at 30 mi/h? b.) How long would it take you to travel 90 miles at 45 mi/h? c.) How fast would you have to go to travel 90 miles in 1.5 h?

8 SOLUTIONS: Ch. 2.5 p.127 #(2, 5, 8 & 9) 2.) Two quantities, & y, are inversely proportional. When = 3 and y = 4. Find the missing coordinates for each point. Hint: If y = k Given = 3 & y = 4; y = k or y = k (3)(4) = k 4 = k 3 12 = k 12 = k then y = k. Thus, the inverse variation equation is: y = 12 or y = 12 a.) (4, y) b.) (, 2) c.) (1, y) d.) (, 24) Let = 4 Let y = 2 Let = 1 Let y = 24 y = 12 y = 12 4 y = 12 (2) = 12 y = 12 y = 12 (1) y = 12 (24) = 12 y = 3 = 6 y = 12 = ½

9 SOLUTIONS: Ch. 2.5 p.127 #(2, 5, 8 & 9) 5.) The time it takes to travel a given distance is inversely proportional to your speed. Let = time ; y = speed y = k or y = k k = distance (or 90 miles) a.) How long would it take you to travel 90 miles at 30 mi/h? Let = time ; k = 90 miles ; y = 30mph y = 90 (30) = 90 = 3 hrs. b.) How long would it take you to travel 90 miles at 45 mi/h? Let = time ; k = 90 miles ; y = 45mph y = 90 (45) = 90 = 2 hrs. c.) How fast would you have to go to travel 90 miles in 1.5 h? Let = 1.5 hrs. ; k = 90 miles ; y = mph y = (y) = 90 y = 60 mph

10 Ch. 2.5 p.127 #(2, 5, 8 & 9) 8.) Emily & her little brother Sid are playing on the seesaw at the park. Sid weighs 65 lbs. The seesaw balances when Sid sits on the seat 4 ft. from the center & Emily sits on the board 2½ ft. from the center. a.) About how much does Emily weigh? b.) Sid s friend Joe sits with Sid at the same end of the seesaw. They weigh about the same. Can Emily balance the seesaw with both boys on it? If so, where should she sit? If not, eplain why not. HINT: distance*weight = k because your weight is inversely related to the distance you must sit from center in order to balance. (i.e. the more you weigh, the closer you have to sit to center)

11 SOLUTIONS: Ch. 2.5 p.127 #(2, 5, 8 & 9) 8.) Emily & her little brother Sid are playing on the seesaw at the park. Sid weighs 65 lbs. The seesaw balances when Sid sits on the seat 4 ft. from the center & Emily sits on the board 2½ ft. from the center. a.) About how much does Emily weigh? Emily weights 104 lbs. Let = Emily s weight 4(65) = = b.) Sid s friend Joe sits with Sid at the same end of the seesaw. They weigh about the same. Can Emily balance the seesaw with both boys on it? If so, where should she sit? If not, eplain why not. Let d = the distance Emily would need to sit from center to balance 4( ) = d(104) 4(130) = d(104) Emily would need to sit 5ft. from center, but since 5 = d the seats are only 4ft. from center, the boys would need to move up. If Emily sits on the seat, Sid 4ft. 2.5ft Emily y(130) = 4(104) they would need to sit y = 3.2 ft. from the center.

12 Ch. 2.5 p.127 #(2, 5, 8 & 9) 9.) To use a double-pan balance, you put the object to be weighed on one side then put known weights on the other side until the pans balance. a.) Eplain why it is useful to have the balance point halfway between the two pans. b.) Suppose the balance point is off-center, 15 cm. from one pan & 20 cm. from the other. There is an object in the pan closest to the center. The pans balance when 7 kg. is placed in the other pan. Write an equation & solve for the weight of the unknown object.

13 SOLUTIONS: Ch. 2.5 p.127 #(2, 5, 8 & 9) 9.) To use a double-pan balance, you put the object to be weighed on one side then put known weights on the other side until the pans balance. a.) Eplain why it is useful to have the balance point halfway between the two pans. If the balance point is at the center, then the weight of an unknown will be eactly the same as the weight that balances it on the other side. If off-center, you must know the two distances & do calculations. b.) Suppose the balance point is off-center, 15 cm. from one pan & 20 cm. from the other. There is an object in the pan closest to the center. The pans balance when 7 kg. is placed in the other pan. Write an equation & solve for the weight of the unknown object. Let = weight of unknown object 15 = 20(7) = kg. 15cm. 20cm. 7kg.