Related Rates In this topic we will be dealing with two or more quantities that are either increasing or decreasing with respect to time.

y 2 = a 2 + b 3 ( ) 2 = ( ) 2 + ( ) 3 2( ) dy/dt=2( )da/dt + 3( ) 2 db/dt

All variables below are functions of t. 2x 3 + y = 5z 3 6x2x2 dx/dt +dy/dt= 15 z 2 dz/dt

y = sin  dy/dt = cos  d  /dt y and  are functions of t.

Quantities that changes over time are functions of t. Example: The population of a town increase by 500 people every year. This year there are 3000 P = tdP/dt =500

Things to Remember: 1. A rate is a derivative with respect to time. Thus, units such as cm/min, miles per hour and the like implies that a derivative is involve. 2. If a quantity is increasing, then its derivative is positive. If it is decreasing, its derivative is negative.

1. A 25-foot ladder is leaning against a wall. The base of the ladder is sliding away from the wall at the rate of 2 ft per minute. How fast is the top of the ladder sliding down when the base of the ladder is 7 feet away from the wall ?

25 ft x y dx/dt = 2 ft/min dy/dt = ? x = 7 ft x 2 + y 2 = 25 2 (2x)(dx/dt) + (2y)(dy/dt)= 0 7?2

25 ft x y x = 7 ft x 2 + y 2 = 25 2 (2x)(dx/dt) + (2y)(dy/dt)= 0 7?2 y = ? y 2 = y 2 = 625 y 2 = 576 y = 24

25 ft x y x = 7 ft x 2 + y 2 = 25 2 (2x)(dx/dt) + (2y)(dy/dt)= y = ? y 2 = y 2 = 625 y 2 = 576 y = 24

(2)(7)(2) + (2)(24)(dy/dt) = (dy/dt) = 0 48 (dy/dt) = dy/dt = - 7/12 ft/min

2. Two cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing two hours later ? B A

B A x y z dx/dt = 25 mi/h dy/dt = 60 mi/hi  At what rate is the distance between the cars increasing after 2 hours? dz/dt = ? x 2 + y 2 = z 2

A B x y z At what rate is distance between the two cars increasing after 2 hours? dz/dt = ? (2x)(dx/dt)+(2y)(dt/dt)= (2z)(dz/dt) ???

 B A After 2 hours: ? ?

 B 50 miles A 120 miles After 2 hours: z x y = z = z 2 z =

16900 = 260 dz/dt 2x dx/dt + 2y dy/dt = 2z dz/dt 2 (50) ( 25 ) + 2 (120) (60) = 2 (130) dz/dt = 260 dz/dt 260 dz/dt = 65 mi/h ?

3. The height of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm 2 /min. At what rate is the base of the triangle changing when the height is 10 cm and the area of 100 cm 2 ? dh/dt= 1 cm/min dA/dt= 2 cm/min db/dt = ? h = 10 A = 100

dh/dt = 1 cm/min dA/dt = 2 cm 2 /min h = 10 cm A = 100 cm 2 db/dt = ? A = b h 2 A = ½ bh dA/dt = (½)(h)(db/dt)+(dh/dt)(½ b) 2 10 ? 1

dh/dt = 1 cm/min dA/dt = 2 cm 2 /min h = 10 cm A = 100 cm 2 db/dt = ? A = b h 2 A = ½ bh 100 =( b )(10) = 5b b = 20

dh/dt = 1 cm/min dA/dt = 2 cm 2 /min h = 10 cm A = 100 cm 2 db/dt = ? A = b h 2 A = ½ bh dA/dt = (½)(h)(db/dt)+(dh/dt)(½ b) 2 10 ? 1

dh/dt = 1 cm/min dA/dt = 2 cm 2 /min h = 10 cm A = 100 cm 2 db/dt = ? A = b h 2 A = ½ bh dA/dt = (½)(h)(db/dt)+(dh/dt)(½ b)

dA/dt = (½)(h)(db/dt)+(dh/dt)(½ b) = (½) (db/dt)(10) + (1)(20) 2 = 5 (db/dt) – 20 = 5 (db/dt) - 18 = 5 (db/dt) 5 db/dt = cm/min