University:
North Central College
Course:
MATH 140 | Precalculus
Academic year:

2019
Views:

23

Pages:

3
Author:

StellarSymphony

Algebra 2 Honors Related Rates Day 2 Warm Up a. Find two numbers whose sum is 25 and the sum of whose reciprocals is 1/6. x and 25-x 1/x + 1/(25-x) = 1/6 (x-15)(x-10) = 6 x = 10, 15 b. The reciprocal of one-third of a number decreased by one third of the reciprocal of the number is 1/3. Find the number. 1/(x/3) - 1/3(1/x) = 1/3 3/x - 1/3x = 1/3 9 - 1 = x x = 8 Example 1 - More Related Rates a. The intake pipe can fill a certain tank in 6 hours when the outlet pipe is closed, but with the outlet pipe open it takes 9 hours. How long would it take the outlet pipe to empty a full tank? t = time outlet takes to empty 1/6 - 1/t = 1/9 9t - 6t = 54 3t = 54 t = 18 hours b. Tim paddled his kayak 12 km upstream against a 3 km/h current and back again in 5 hours and 20 minutes. In that time, how far could he have paddled in still water (no current)? Rate Distance Time r - 3 12 12/(r-3) r + 3 12 12/(r+3) 12/(r-3) + 12/(r+3) = 16/3 36r + 108 + 36r - 108 = 16r^2 16r^2 - 72r - 144 = 0 2r^2 - 9r - 18 = 0 (2r+3)(r-6) = 0 r = 6 km/h Dist = 32 km This worksheet focuses on solving related rate problems, which involve finding the rate of change of one variable with respect to another variable. It includes problems involving work rates, distance-rate-time relationships, and geometric problems. Problem 1: Members of a ski club contributed equally to obtain $1800 for a holiday trip. When 6 members found that they could not go, their contributions were refunded and each remaining member had to pay $10 more to raise the $1800. How many went on the trip? Solution: Let x be the original number of members. Original fee per member: 1800/x New fee per member: 1800/(x-6) 1800/x + 10 = 1800/(x-6) 1800(x-6) + 10x(x-6) = 1800x 10x^2 - 60x + 10800 = 0 (x-30)(x-36) = 0 x = 30 or x = 36 Since the number of members decreased, the original number was 36. So, 30 members actually went on the trip. Problem 2: Ace Airlines changed to planes that flew 100 km/h faster than its old ones. The time of its 2800 km Dallas-Seattle flight was reduced by 30 minutes. Find the speed of the new planes. Solution: Let x be the speed of the old planes. Old speed: x km/h New speed: x + 100 km/h Old time: 2800/x hours New time: 2800/(x+100) hours 2800/x - 2800/(x+100) = 1/2 5600(x+100) - 5600x = x(x+100) -x^2 - 100x + 560000 = 0 (x+800)(x-700) = 0 x = -800 or x = 700 Since speed cannot be negative, the old speed was 700 km/h. So, the new speed is 800 km/h. Problem 3: Maria could average only 40 km/h for the first 20% of her trip, but she averaged 75 km/h for the whole trip. What was her average speed for the last 80% of her trip? Solution: Let d be the total distance of the trip. Part 1: Distance: 0.2d Rate: 40 km/h Time: 0.2d/40 = d/200 Part 2: Distance: 0.8d Rate: r km/h Time: 0.8d/r Average speed: (Total distance) / (Total time) 75 = d / (d/200 + 0.8d/r) 75 = 200r / (r + 160) 375r + 60000 = 1000r 60000 = 625r r = 96 km/h