1. Read the problem, pull out essential information and identify a formula to be used.

2. Sketch a diagram if possible.

3. Write down any known rate of change & the rate of change you are looking for. 4. Be careful with signs…if the amount is decreasing, the rate of change is negative. 5. Pay attention to whether quantities are constant or varying.

6. Set up an equation involving the appropriate quantities. 7. Differentiate with respect to t using implicit differentiation. 8. Plug in known items (you may need to find some quantities using geometry). 9. Solve for the item you are looking for, most often this will be a rate of change. 10. State your final answer with the appropriate units.

3.10 – Related RatesRelated rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. 

3.10 – Related Rates

1¿ 𝐼𝑓𝑥=𝑦3−𝑦 𝑎𝑛𝑑𝑑𝑦𝑑𝑡

=5 , h𝑡 𝑒𝑛 h𝑤 𝑎𝑡 𝑖𝑠 h𝑡 𝑒𝑣𝑎𝑙𝑢𝑒𝑜𝑓𝑑𝑥𝑑𝑡

h𝑤 𝑒𝑛 𝑦=2?

𝑥=𝑦 3− 𝑦𝑑𝑥𝑑𝑡

=3 𝑦2𝑑𝑦𝑑𝑡−𝑑𝑦𝑑𝑡

𝑑𝑥𝑑𝑡

=3 (2 )2 (5 )−5

𝑑𝑥𝑑𝑡

=55

Example

Related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known. 

3.10 – Related Rates

𝑟+𝑠2+𝑣3=12𝑑𝑟𝑑𝑡

+2𝑠𝑑𝑠𝑑𝑡

+3𝑣2𝑑𝑣𝑑𝑡

=0

𝑑𝑣𝑑𝑡

=0.042

4+2 (1 ) (−3 )+3 (4 )2 𝑑𝑣𝑑𝑡

=0

48𝑑𝑣𝑑𝑡

=2

Example

3.10 – Related Rates

𝑉=13𝑟2h

𝑑𝑉𝑑𝑡

=13𝑟2

h𝑑𝑑𝑡

+132𝑟

𝑑𝑟𝑑𝑡h

3) The radius r and the height h of a right circular cone are related to the cone’s volume V by the equation

How is related to if r is constant? How is related to if h is constant?

How is related to if h and r are not constant?

𝑑𝑉𝑑𝑡

=13𝑟2

h𝑑𝑑𝑡

𝑉=13𝑟2h

𝑑𝑉𝑑𝑡

=132𝑟

𝑑𝑟𝑑𝑡h

𝑑𝑉𝑑𝑡

=23h𝑟

𝑑𝑟𝑑𝑡

𝑉=13𝑟2h

Example

3.10 – Related Rates4) 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 𝑟2

𝑑𝑟𝑑𝑡

=1.5𝑚𝑖/h𝑟

𝐹𝑖𝑛𝑑𝑑𝐴𝑑𝑡

h𝑤 𝑒𝑛𝑟=0.6𝑚𝑖

𝐴=𝜋 𝑟2

𝑑𝐴𝑑𝑡

=𝜋 2𝑟𝑑𝑟𝑑𝑡

𝑑𝐴𝑑𝑡

=𝜋 2 (0.6 ) (1.5 )

𝑑𝐴𝑑𝑡

=5.655𝑚𝑖2/ h𝑟

3.10 – Related Rates5) 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=43𝜋 𝑟3

𝑑𝑉𝑑𝑡

=10 𝑐𝑚3/h𝑟

𝐹𝑖𝑛𝑑𝑑𝑟𝑑𝑡

h𝑤 𝑒𝑛𝑟=5𝑐𝑚

𝑉=43𝜋 𝑟3

𝑑𝑉𝑑𝑡

=43𝜋 3𝑟2

𝑑𝑟𝑑𝑡

10=4𝜋 (5 )2 𝑑𝑟𝑑𝑡

𝑑𝑟𝑑𝑡

=0.032𝑐𝑚 /𝑠𝑒𝑐

𝑑𝑉𝑑𝑡

=4𝜋 𝑟2𝑑𝑟𝑑𝑡

10

4𝜋 (5 )2=𝑑𝑟𝑑𝑡

3.10 – Related Rates6) 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?

80 ftr

40 ft

𝑉=13𝜋𝑟2h

𝑑𝑉𝑑𝑡

=9 𝑓𝑡 3/𝑚𝑖𝑛

𝐹𝑖𝑛𝑑h𝑑

𝑑𝑡h𝑤 𝑒𝑛h=60 𝑓𝑡

𝑛𝑜𝑡𝑔𝑖𝑣𝑒𝑛𝑑𝑟𝑑𝑡

𝑟h¿4080

𝑟=12h

𝑉=13𝜋𝑟2h

𝑉=13𝜋( 12 h)

2

h

𝑉=112

𝜋 h3

𝑑𝑉𝑑𝑡

=112

𝜋 3h2h𝑑

𝑑𝑡

9=𝜋4

(60 )2 h𝑑𝑑𝑡

h𝑑𝑑𝑡

=0.003𝑓 𝑡 /𝑚𝑖𝑛

𝑢𝑠𝑒𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠

3.10 – Related Rates7) 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?

𝑦

𝑥

𝑧

𝑥2+ 𝑦2=𝑧2

𝑧=25 𝑓𝑡𝑑𝑥𝑑𝑡

=3 𝑓𝑝𝑠

𝐹𝑖𝑛𝑑𝑑𝑦𝑑𝑡

h𝑤 𝑒𝑛𝑥=15 𝑓𝑡

𝑥2+ 𝑦2=𝑧2

𝑥2+ 𝑦2=252

2 𝑥𝑑𝑥𝑑𝑡

+2 𝑦𝑑𝑦𝑑𝑡

=0

2 (15 ) (3 )+2 𝑦 𝑑𝑦𝑑𝑡

=0

152+ 𝑦2=252

𝑦=20

2 (15 ) (3 )+2 (20 ) 𝑑𝑦𝑑𝑡

=0

40𝑑𝑦𝑑𝑡

=−90

𝑑𝑦𝑑𝑡

=−2.25𝑓 𝑝𝑠

3.10 – Related Rates8) 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?

𝑦

𝑥

𝑧

𝐴=12h𝑏 =12𝑥𝑦

𝑥=15 𝑓𝑡𝑑𝑥𝑑𝑡

=3 𝑓𝑝𝑠

𝐹𝑖𝑛𝑑𝑑𝐴𝑑𝑡

𝑎𝑡 h𝑡 𝑎𝑡 𝑖𝑛𝑠𝑡𝑎𝑛𝑡 .

𝑑𝐴𝑑𝑡

=12𝑥𝑑𝑦𝑑𝑡

+12𝑑𝑥𝑑𝑡

𝑦

𝑑𝐴𝑑𝑡

=13.125𝑓𝑡2/ 𝑠𝑒𝑐𝑦=20 𝑓𝑡𝑑𝑦𝑑𝑡

=−2.25 𝑓𝑝𝑠

𝐴=12𝑥𝑦

𝑑𝐴𝑑𝑡

=12

(15 ) (−2.25 )+ 12

(3 ) (20 )

𝑦

𝑥

𝑧

3.10 – Related Rates9) 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?

𝑟 𝑎𝑑 /𝑠𝑒𝑐

𝜃

tan𝜃=𝑦𝑥

𝑠𝑒𝑐2𝜃𝑑𝜃𝑑𝑡

=𝑥𝑑𝑦𝑑𝑡−𝑑𝑥𝑑𝑡

𝑦  

𝑥2

1

𝑐𝑜𝑠2𝜃𝑑𝜃𝑑𝑡

=𝑥𝑑𝑦𝑑𝑡−𝑑𝑥𝑑𝑡

𝑦  

𝑥2

tan𝜃=2015

𝜃=𝑡𝑎𝑛−143¿0.927 𝑟𝑎𝑑 .

1

𝑐𝑜𝑠2(0.927)𝑑𝜃𝑑𝑡

=(15 ) (−2.25 )− (3 ) (20 )  

152

tan𝜃=𝑦𝑥

𝑥=15 𝑓𝑡𝑑𝑥𝑑𝑡

=3 𝑓𝑝𝑠

𝐹𝑖𝑛𝑑𝑑𝜃𝑑𝑡

𝑎𝑡 h𝑡 𝑎𝑡 𝑖𝑛𝑠𝑡𝑎𝑛𝑡 .

𝑦=20 𝑓𝑡𝑑𝑦𝑑𝑡

=−2.25 𝑓𝑝𝑠

1

𝑐𝑜𝑠2𝜃𝑑𝜃𝑑𝑡

=𝑥𝑑𝑦𝑑𝑡−𝑑𝑥𝑑𝑡

𝑦  

𝑥2

𝑑𝜃𝑑𝑡

=−0.150

3.10 – Related Rates