3.10 – Related Rates 4) An oil tanker strikes an iceberg and a hole is ripped open on its side. Oil is leaking out in a near circular shape. The radius of the oil spill is changing at a rate of 1.5 miles per hour. How fast is the area of the oil spill changing when the radius is 0.6 mile? 𝐴=𝜋 𝑟 2 𝑑𝐴 𝑑𝑡 =𝜋2𝑟 𝑑𝑟 𝑑𝑡 𝐴=𝜋 𝑟 2 𝑑𝐴 𝑑𝑡 =𝜋2 0.6 1.5 𝑑𝑟 𝑑𝑡 =1.5 𝑚𝑖/ℎ𝑟 𝑑𝐴 𝑑𝑡 =5.655 𝑚𝑖 2 /ℎ𝑟 𝐹𝑖𝑛𝑑 𝑑𝐴 𝑑𝑡 𝑤ℎ𝑒𝑛 𝑟=0.6 𝑚𝑖

3.10 – Related Rates 5) A balloon is being inflated at a rate of 10 cubic centimeters per second. How fast is the radius of a spherical balloon changing at the instant the radius is 5 centimeters? 𝑉= 4 3 𝜋 𝑟 3 𝑑𝑉 𝑑𝑡 = 4 3 𝜋3 𝑟 2 𝑑𝑟 𝑑𝑡 𝑉= 4 3 𝜋 𝑟 3 𝑑𝑉 𝑑𝑡 =4𝜋 𝑟 2 𝑑𝑟 𝑑𝑡 𝑑𝑉 𝑑𝑡 =10 𝑐𝑚 3 /ℎ𝑟 10=4𝜋 5 2 𝑑𝑟 𝑑𝑡 𝐹𝑖𝑛𝑑 𝑑𝑟 𝑑𝑡 𝑤ℎ𝑒𝑛 𝑟=5 𝑐𝑚 10 4𝜋 5 2 = 𝑑𝑟 𝑑𝑡 𝑑𝑟 𝑑𝑡 =0.032 𝑐𝑚/𝑠𝑒𝑐

3.10 – Related Rates 6) A water authority is filling an inverted conical water storage tower at a rate of 9 cubic feet per minute. The height of the tank is 80 feet and the radius at the top is 40 feet. How fast is the water level inside the tank changing when the water level is 60 feet deep? 𝑉= 1 3 𝜋 𝑟 2 ℎ 𝑉= 1 3 𝜋 𝑟 2 ℎ 𝑑𝑉 𝑑𝑡 =9 𝑓𝑡 3 /𝑚𝑖𝑛 𝑉= 1 3 𝜋 1 2 ℎ 2 ℎ 𝐹𝑖𝑛𝑑 𝑑ℎ 𝑑𝑡 𝑤ℎ𝑒𝑛 ℎ=60 𝑓𝑡 𝑉= 1 12 𝜋 ℎ 3 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛 𝑑𝑟 𝑑𝑡 80 ft r 40 ft 𝑑𝑉 𝑑𝑡 = 1 12 𝜋3 ℎ 2 𝑑ℎ 𝑑𝑡 𝑓𝑖𝑛𝑑 𝑎 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝 𝑤𝑖𝑡ℎ 𝑟 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 ℎ. 9= 𝜋 4 60 2 𝑑ℎ 𝑑𝑡 𝑢𝑠𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠 𝑑ℎ 𝑑𝑡 =0.003 𝑟 ℎ = 40 80 𝑟= 1 2 ℎ 𝑓𝑡/𝑚𝑖𝑛

3.10 – Related Rates 7) A 25-foot ladder is leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 3 feet per second. How fast is the top of the ladder moving at the instant the bottom of the ladder is 15 feet away from the wall? 𝑥 2 + 𝑦 2 = 𝑧 2 2 15 3 +2 20 𝑑𝑦 𝑑𝑡 =0 𝑥 2 + 𝑦 2 = 25 2 𝑧 𝑦 40 𝑑𝑦 𝑑𝑡 =−90 2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =0 𝑑𝑦 𝑑𝑡 =−2.25 𝑥 𝑓𝑝𝑠 2 15 3 +2𝑦 𝑑𝑦 𝑑𝑡 =0 𝑥 2 + 𝑦 2 = 𝑧 2 𝑧=25 𝑓𝑡 𝑓𝑖𝑛𝑑 𝑦 𝑤ℎ𝑒𝑛 𝑥=15 𝑎𝑛𝑑 𝑧=25. 𝑑𝑥 𝑑𝑡 =3 𝑓𝑝𝑠 15 2 + 𝑦 2 = 25 2 𝐹𝑖𝑛𝑑 𝑑𝑦 𝑑𝑡 𝑤ℎ𝑒𝑛 𝑥=15 𝑓𝑡 𝑦=20

3.10 – Related Rates 8) A 25-foot ladder is leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 3 feet per second. How fast is the top of the ladder moving at the instant the bottom of the ladder is 15 feet away from the wall? At the same instant, what is the rate of change of the area between the ground, the wall, and the ladder? 𝐴= 1 2 𝑥𝑦 𝑦 𝑥 𝑧 𝑑𝐴 𝑑𝑡 = 1 2 𝑥 𝑑𝑦 𝑑𝑡 + 1 2 𝑑𝑥 𝑑𝑡 𝑦 𝑑𝐴 𝑑𝑡 = 1 2 15 −2.25 + 1 2 3 20 𝐴= 1 2 𝑏ℎ= 1 2 𝑥𝑦 𝑑𝐴 𝑑𝑡 =13.125 𝑓𝑡 2 /𝑠𝑒𝑐 𝑥=15 𝑓𝑡 𝑦=20 𝑓𝑡 𝑑𝑥 𝑑𝑡 =3 𝑓𝑝𝑠 𝑑𝑦 𝑑𝑡 =−2.25 𝑓𝑝𝑠 𝐹𝑖𝑛𝑑 𝑑𝐴 𝑑𝑡 𝑎𝑡 𝑡ℎ𝑎𝑡 𝑖𝑛𝑠𝑡𝑎𝑛𝑡.

3.10 – Related Rates 9) A 25-foot ladder is leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 3 feet per second. How fast is the top of the ladder moving at the instant the bottom of the ladder is 15 feet away from the wall? At the same instant, what is the rate of change of the angle () between the ground and the ladder? tan 𝜃 = 𝑦 𝑥 1 𝑐𝑜𝑠 2 𝜃 𝑑𝜃 𝑑𝑡 = 𝑥 𝑑𝑦 𝑑𝑡 − 𝑑𝑥 𝑑𝑡 𝑦 𝑥 2 𝑦 𝑥 𝑧 𝑥=15 𝑓𝑡 𝑦=20 𝑓𝑡 𝑓𝑖𝑛𝑑 𝜃 𝑤ℎ𝑒𝑛 𝑥=15 𝑎𝑛𝑑 𝑦=20 𝑑𝑥 𝑑𝑡 =3 𝑓𝑝𝑠 𝑑𝑦 𝑑𝑡 =−2.25 𝑓𝑝𝑠 tan 𝜃 = 20 15 𝜃 𝐹𝑖𝑛𝑑 𝑑𝜃 𝑑𝑡 𝑎𝑡 𝑡ℎ𝑎𝑡 𝑖𝑛𝑠𝑡𝑎𝑛𝑡. 𝜃= 𝑡𝑎𝑛 −1 4 3 =0.927 𝑟𝑎𝑑. tan 𝜃 = 𝑦 𝑥 1 𝑐𝑜𝑠 2 (0.927) 𝑑𝜃 𝑑𝑡 = 15 −2.25 − 3 20 15 2 𝑠𝑒𝑐 2 𝜃 𝑑𝜃 𝑑𝑡 = 𝑥 𝑑𝑦 𝑑𝑡 − 𝑑𝑥 𝑑𝑡 𝑦 𝑥 2 𝑑𝜃 𝑑𝑡 =−0.150 1 𝑐𝑜𝑠 2 𝜃 𝑑𝜃 𝑑𝑡 = 𝑥 𝑑𝑦 𝑑𝑡 − 𝑑𝑥 𝑑𝑡 𝑦 𝑥 2 𝑟𝑎𝑑/𝑠𝑒𝑐

3.10 – Related Rates