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Related Rates - Classwork

Earlier in the year, we used the basic definition of calculus as ;the mathematics of change.? We defined words that meant change: increasing, decreasing, growing, shrinking, etc. Change occurs over time. So, when we talk about how a quantity changes, we are talking about the derivative of that quantity with respect to time. Example 1) Write the following statements mathematically. a) John is growing at the rate of 3 inches/year. b) My mutual fund is shrinking by 4 cents/day. dh dm = !. /da 04 = 3 in/yr dt dt c) The radius of a circle is increasing by 4 ft/hr. d) The volume of a cone is decreasing by 2 in3/sec. dr dV = !2 in 3 /sec = 4 ft/hr dt dt Example 2) A rectangle is 10 inches by 6 inches whose sides are changing. Write formulas for both the perimeter and area and how fast each is changing in terms of L and W. Perimeter Change of perimeter Area Change of area dP dl dw dA dw dl P = 2l + 2 w A = lw =2 +2 =l +w dt dt dt dt dt dt a. its length and width are increasing at the rate b. its length and width are decreasing at the rate rate of 2 inches/sec. of 2 inches/sec. dl dw dl dw = 2 in/sec = 2 in/sec = !2 in/sec = !2 in/sec dt dt dt dt Change of perimeter Change of area Change of perimeter Change of area 2(!2) + 2(!2) = !8 in/sec 10(!2) + 6(!2) = !32 in/sec 2(2) + 2(2) = 8 in/sec 10(2) + 6(2) = 32 in/sec c. its length is increasing at 3 inches/sec and the d. its length is decreasing at the rate of 2 inches/sec width is decreasing at 3 inches/sec. and its width is increasing at .5 inches/sec. dl dw dl dw = 3 in/sec = 3 in/sec = 2 in/sec = !.5 in/sec dt dt dt dt Change of perimeter Change of area Change of perimeter Change of area 2(3) + 2(!3) = 0 in in 2 10(3) + 6(!3) = 12 sec sec 2(2) + 2(!.5) = 3 in sec 10(!.5) + 6(2) = 7 in 2 sec

Example 3) A right triangle has sides of 30 and 40 inches whose sides are changing. Write formulas for the area of the triangle and the hypotenuse of the triangle and how fast the area and hypotenuse are changing Area Change of area Hypotenuse Change of hypotenuse dy 1 dz x dx + y dt dA 1 " dy dx % 2 2 $x ' A = xy = dt = +y z= x + y 2 dt dt 2 # dt dt & x2 + y2 a) the short side is increasing at 3 in./sec b) the short side is increasing at 3 in./sec and the long and the long side is increasing at 5 in/sec. side is decreasing at 5 in./sec. dx dy dx dy = 3 in/sec = 5 in/sec = 3 in/sec = !5 in/sec dt dt dt dt Change of area Change of hypotenuse Change of area Change of hypotenuse 2 30 3 + 40 5 2 ( ) ( ) = 5.8 in 1 30 !5 + 40 3 = !15 in 30(3) + 40(!5) = !2.2 in in 1 30(5) + 40(3) = 135 ( ) () sec sec 2 sec sec 2 50 50

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)

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)

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Stu Schwartz

Example 4. A right circular cylinder has a height of 10 feet and radius 8 feet whose dimensions are changing. Write formulas for the volume and surface area of the cylinder and the rate at which they change. Volume Change of volume Surface area Change of surface area " dh " dh dV dr % dS dr % V = (r 2 h = ($r 2 + 2 hr ' S = 2(rh = 2($r +h ' dt dt & dt dt & # dt # dt a) the radius is growing at 2 feet/min and b) the radius is decreasing at 4 feet/min and the the height is shrinking at 3 feet/min. the height is increasing at 2 feet/min. dr dh = 2 ft/min = !3 ft/min dt dt Change of volume Change of surface area ft 3 ( 64(!3) + 160(2) = 128( min

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)

dr dh = !4 ft/min = 2 ft/min dt dt Change of volume Change of surface area ft 3 ( 64(2) + 160(!4) = !512( min

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)

ft 2( 8(!3) + 10(2) = !8( min

(

)

2

ft 2( 8(2) + 10(!4) = !48( min

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2

To solve related rates problems, you need a strategy that always works. Related rates problems always can be recognized by the words ;increasing, decreasing, growing, shrinking, changing.? Follow these guidelines in solving a related rates problem. 1. Make a sketch. Label all sides in terms of variables even if you are given the actual values of the sides. 2. You will make a table of variables. The table will contain two types of variables - variables that are constants and variables that are changing. Variables that never change go into the constant column. Variables that are a given value only at a certain point in time go into the changing column. Rates (recognized by ;increasing?, ;decreasing?, etc.) are derivatives with respect to time and can go in either column. 3. Find an equation which ties your variables together. If it an area problem, you need an area equation. If it is a right triangle, the Pythagorean formula may work or gerenal trig formulas may apply. If it is a general triangle, the law of cosines may work. 4. You may now plug in any variable in the constant column. Never plug in any variable in the changing column. 5. Differentiate your equation with respect to time. You are doing implicit differentiation with respect to t. 6. Plug in all variables. Hopefully, you will know all variables except one. If not, you will need an equation which will solve for unknown variables. Many times, it is the same equation as the one you used above. Do this work on the side as to not destroy the momentum of your work so far. 7. Label your answers in terms of the correct units (very important) and be sure you answered the question asked. Example 5) An oil tank spills oil that spreads in a circular pattern whose radius increases at the rate of 50 feet/min. How fast are both the circumference and area of the spill increasing when the radius of the spill is a) 20 feet and b) 50 feet?

C = 2(r dC dr = 2( dt dt ft dC = 2((50) = 100( min dt

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A = (r 2 dA dr = 2(r dt dt ft 2 dA = 2((20)(50) = 2000( min dt

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A = (r 2 dA dr = 2(r dt dt ft 2 dA = 2((50)(50) = 5000( min dt

Stu Schwartz

Example 6) A 13 foot ladder leans against a vertical wall. If the lower end of the ladder is pulled away at the rate 2 feet per second, how fast is the top of the ladder coming down the wall at a) the instant the top is 12 feet above the ground and b) 5 feet above the ground?

x 2 + y 2 = 169 x 2 + 144 = 169 ) x = 5 dx dy 2x + 2y =0 dt dt dy =0 2(5)(2) + 2(12) dt dy 6 = = 1.20 ft/sec dt 5

x 2 + y 2 = 169 x 2 + 25 = 169 ) x = 12 dx dy 2x + 2y =0 dt dt dy =0 2(12)(2) + 2(5) dt dy !24 = = !4.8 ft/sec dt 5

Example 7) A camera is mounted 3,000 feet from the space shuttle launching pad. The camera needs to pivot as the shuttle is launched and needs to keep the shuttle in focus. If the shuttle is rising vertically at 800 feet/sec when it is 4,000 feet high, how fast is the camera-to-shuttle distance changing?

x 2 + y 2 = z2

30002 + y 2 = z2 dy dz 2y = 2z dt dt

30002 + 4000 4 = z2 ) z = 5000

2(4000)(800) = 2(5000)

dz dt

dz ft = 640 dt sec

In this problem, how fast is the angle of elevation of the camera changing at that moment in time? d* What variable are we trying to find? Since this is a function of * , we need a trig function. dt There are three trig functions we could use. Let's try all three and determine which is best. y x sin * = cos* = y z z tan * = x dy dz dz 1 dy d* d* z dt ! y dt d* ! x dt sec2 * = cos* = ! sin * = dt 3000 dt dt dt z2 z2 2 " 5 % d* 1 !4 d* !3000(640) 3 d* 5000(800) ! 4000(640) $ ' = (800) = = # 3 & dt 3000 5 dt 5 dt 50002 50002

d* R = .096 dt sec

d* R = .096 dt sec d* + 5.5o sec dt

R d* = .096 sec dt

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Example 8) Two cars are riding on roads that meet at a 60 degree angle. Car A is 3 miles from the intersection traveling at 40 mph and car # is 2 miles away from the intersection traveling at 50 mph. How fast are the two cars separating if a) they are both traveling away from the intersection and b) car A is traveling away from the intersection and car # is traveling towards it? c2 = a 2 + b2 ! 2 ab cos 60 o

c2 = a 2 + b2 ! ab ) c2 = 9 + 4 ! 6 ) c = 7 dc da db db da dc da db db da 2c = 2 a + 2b ! a ! b 2c = 2 a + 2b ! a ! b dt dt dt dt dt dt dt dt dt dt dc dc 2 7 = 2(3)(40) + 2(2)(50) ! 3(50) ! 2(40) 2 7 = 2(3)(40) + 2(2)(!50) ! 3(!50) ! 2(40) dt dt dc 210 dc 210 = = 39.686 mph = = 20.788 mph dt 2 7 dt 2 7

Example 9) Sand is poured on a beach creating a cone whose radius is always equal to twice its height. If the sand is poured at the rate of 20 in3/sec, How fast is the height changing at the time the height is a) 2 inches, b) 6 inches? 1 V = (r 2 h 3 2 1 V = ((2 h ) h 3 4 V = (h 3 3 dV dh = 4(h 2 dt dt

r = 2h

20 = 4( 22

( ) dh dt

in dh 20 = = .398 sec dt 16( 20 = 4( 62

( ) dh dt

in 20 dh = = .044 sec dt 144(

Example 10) Water is draining from a conical tank at the rate of 2 meter3/min. The tank is 16 meters high and its top radius is 4 meters. How fast is the water level falling when the water level is a) 12 meters high, b) 2 meters high? 1 V = (r 2 h 3 1 "1 % V = ($ h ' h 3 #4 &

2

1 r 4 = )r= h 4 h 16

2=

1 dh ( 122 16 dt m 32 dh = = .071 min dt 144(

( )

V=

1 (h 3 48 dV 1 2 dh = (h dt 16 dt

2=

1 dh ( 22 16 dt m dh 32 = = 2.546 min dt 4(

( )

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Stu Schwartz

Related Rates - Homework

1. A circle has a radius of 8 inches which is changing. Write formulas for its circumference and area. Circumference Change of circumference Area Change of area dC dr dA dr C = 2(r = 2( A = (r 2 = 2(r dt dt dt dt 1 a) its radius is growing at the rate of 3 in./min. b) its radius is shrinking at the rate of inch/sec. 4 1 dr dr = 3 in/min = ! in/sec 4 dt dt Change of cirumference Change of area Change of cirumference Change of area 1 dC dA dC dA = 48( in 2 /min = !4 in 2 /sec = 6( in/min = ! in/sec 2 dt dt dt dt c) its diameter is growing at the rate of 4 yd/min. d) its radius is shrinking at the rate of 1 inch/sec. dr dr = 2 yd/min = !1 in/sec dt dt Change of cirumference Change of area Change of cirumference Change of area dC dA dC dA = 4( yd/min = 32( yd 2 /min = !2( in/sec = !16( in 2 /sec dt dt dt dt 2. A sphere has a radius of 9 feet which is changing. Write formulas for its volume and surface area. Volume Change of volume Surface area Change of surface area 4 V = (r 3 3

dV dr = 4(r 2 dt dt

S = 4(r 2

dS dr = 8(r dt dt

3 inch/sec. 4

a) its diameter is growing at the rate of 1 yd/min.

b) its radius is shrinking at the rate of

dr 1 yd = dt 2 min Change of volume Change of surface area 3 yd yd 2 dV dS = 162( = 36( min min dt dt

dr !3 in = dt 4 sec Change of volume Change of surface area 3 in in 3 dV dS = !243( = !54( sec sec dt dt

3. A right circular cone has a height of 10 feet and radius 6 feet, both of which are changing. Write a formula for the volume of the cone. Volume 1 V = (r 2 h 3 a) the radius grows at 6 ft/sec, the height shrinks at 12 in/sec Change of volume ft 3 dV = 684 ( sec dt Change of volume dV 1 " 2 dh dr % = ($r + 2rh ' dt 3 # dt dt & b) the diameter is shrinking at 6 ft/sec and the height is growing at 2 ft/sec. Change of volume ft 3 dV = !96( sec dt

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4. A rectangular well is 6 feet long, 4 feet wide, and 8 feet deep. If water is running into the well at the rate of 3 ft3/sec, find how fast the water is rising (keep in mind which variables are constant and which are changing). dh 3 = 24 V = lwh dt dh 1 V = 24 h = ft/sec dt 8 5. A spherical hot air balloon is being inflated. If air is blown into the ballon at the rate of 2 ft3/sec, a. find how fast the radius of the balloon is changing b. find how fast the surface area is increasing when the radius is 3 feet. at the same time. 4 V = (r 3 3 dV dr = 4(r 2 dt dt 2 = 36(

dr dt

1 dr ft/sec = dt 18(

S = 4(r 2 dS dr = 8(r dt dt

" 1 % dS ' = 24($ dt #18( &

dr 4 2 = ft /sec dt 3

6. A 12 foot ladder stands against a vertical wall. If the lower end of the ladder is being pulled away from the wall at the rate of 2 ft/sec, a) how fast is the top of the ladder coming down the b. how fast is the angle of the elevation of the wall at the instant it is 6 feet above the ground? ladder changing at the same instant?

x 2 + y 2 = 144 x 2 + 36 = 144 ) x = 108 dx dy 2x + 2y =0 dt dt dy =0 2 108 (2) + 2(6) dt dy ! 108 = = !3.464 ft/sec dt 3

x 12 d* 1 dx ! sin * = dt 12 dt 6 d* 1 ! = (2) 12 dt 12 d* !1 = R/sec dt 3

cos* =

7. Superman is in level flight 6 miles above ground. His flight plan takes him directly over Wissahickon High. How fast is he flying when the distance between him and WHS is exactly 10 miles and this distance is increasing at the rate of 40 mph?

dx = 10(40) dt dx = 50 mph dt 8. Two roads meet at an angle of 60o. A man starts from the intersection at 1 PM and walks along one road at 3 mph. At 2:00 PM, another man starts along the second road and walks at 4 mph. How fast are they separating at 4 PM? db da dc da db 2c = 2 a + 2b ! a ! b dt dt dt dt dt 2 2 2 o c = a + b ! 2 ab cos 60 dc 2 73 = 2(9)(3) + 2(8)(4) ! 9(4) ! 8(3) 2 2 2 2 dt c = a + b ! ab ) c = 81 + 64 ! 72 ) c = 73 dc = 3.394 mph dt x2 + y2 = r2 dx dr x =r dt dt x 2 + 36 = 100 ) x = 8

8

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9. A boy flies a kite which is 120 ft directly above his hand. If the wind carries the kite horizontally at the rate of 30 ft/min, at what rate is the string being pulled out when the length of the string is 150 ft? x 2 + y 2 = z2 dx dz =z x dt dt 150 90( 30) = 150 z dz dt

120

dz = 18 ft/min dt 10. The same boy flies a kite which is now 100 feet above the ground. If the string is pulled out at the rate of 10 ft/sec because the wind carries the kite horizontally directly away from the boy, what is the rate of change of the angle the kite makes with the vertical when the angle is 30o. cos* = 100 30o 11. A baseball diamond is a 90-foot square. A ball is batted along the third-base line at a constant rate of 100 feet per second. How fast is its distance changing from first base at the time when a) the ball is halfway to 3rd base and b) it reaches 3rd base. 2nd y dy dz =z dt dt dz 45(100) = dt 10125 dz = 44.721 ft / sec dt y z d* !100 dz ! sin* = 2 dt z dt 1 d* 100 = 2 (10) 100 2 dt

( )

3

d* = .15 R /sec + 8.6 o /sec dt

z 3rd z 90 y x 90 1st

dz 90(100) = dt 16200 dz = 70.711 ft / sec dt

12. A plane is flying west at 500 ft/sec at an altitude of 4,000 ft. The plane is tracked by a searchlight on the ground. If the light is to be trained on the plane, find the change in the angle of elevation of the searchlight at a horizontal distance of 2,000 ft. x y d* 1 dx sec 2 * = dt 4000 dt tan * = " 20 % d* 1 = ' $ (500) # 4 & dt 4000

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2

5 d* 1 = 4 dt 8 d* 1 = R /sec + 5.730 o /sec dt 10

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Stu Schwartz

13. A revolving light located 5 miles from a straight shore line turns with a constant angular velocity. What velocity does the light revolve if the light moves along the shore at the rate of 15 miles per minute when the beam makes an angle of 60o with the shore line? x d* Given : y = 5 (const ) tan * = 22 =3 15 y dt d* 3 radian 42.8 o dx x 5 = + = 15 tan * = min dt 4 min dt 5 o 60 d* 1 dx * = 60 o sec2 * = dt 5 dt d* d* 1 =? sec2 60 o = (15) dt dt 5 14. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 10 cm3/sec at the time when the radius is 2 cm? 4 cm dV dr 5 Given : = 10 = = .199 V = (r 3 3 sec dt dt 8( dV dr = 4(r 2 r=2 dt dt dr dr =? 10 = 4!(4) dt dt 15. How fast does the water level drop when a cylindrical tank of radius 6 feet is drained at the rate of 3 ft3/min? Given : r = 6 (const )

V = (r 2 h V = 36(h dV dh = 36! dt dt

dV =3 dt dh =? dt

3 = 36(

dh dt ft 1 dh = = .027 min dt 12(

16. A hot air balloon, rising straight up from a level field, is tracked by a range finder 500 feet from the lift-off ( point. At the moment the range finder's elevation angle is , the angle is increasing at the rate of 0.14 4 radians/min. How fast is the balloon rising? Given

x = 500 (const)

y !/4 500 ft

tan * = tan * =

y x

*=

( 4

d* .14 r = dt min dy =? dt

y 500 d* 1 dy = sec2* dt 500 dt 1 dy 2(.14) = 500 dt dy ft = 140 dt min

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17. Water runs into a conical tank at the rate of 9 ft3/min. The tank stands vertex down and has a height of 10 feet and a base radius of 5 feet. How fast is the water level rising when the water is 6 feet deep? 1 5 ft Given V = (r 2 h By similar triangles 3

r

r = radius of water surface at time t

10 ft h

r=5 h=6

1 "h% V = ($ ' h 3 #2&

2

5 = 10

r h

V= dV dt

9=

h = depth of water dV ft 3 =9 surface at time t dt min dh =? dt

dh dt

1 3 h (h r= 12 2 ( dh = h2 4 dt ( 2 dh 6 4 dt 1 ft = = .318 ( min

18. Two truck convoys leave a depot at the same time. Convoy A travels east at 40 mph and convoy # travels north at 30 mph. How fast is the distance between the convoys changing a) in 6 minutes b) in 30 minutes Given da c = distance a2 + b2 = c 2 = 40 mph Side Work dt b= between db da db dc North trucks 2 a + 2b a2 + b2 = c 2 = 2c = 30 mph dt dt dt dt 1 dc 4 ( 40) + 3( 30) = 5 a = ( 40) = 4 miles c =5 10 dt a = East 1 dc a = ( 40) = 20 miles = 50 mph 2 dt 1 dc 20( 40) +15( 30) = 25 b = ( 30) = 3 miles a2 + b2 = c 2 10 dt 1 dc b = ( 30) = 15 miles = 50 mph c = 25 2 dt 19. Two commercial bets at 40,000 ft. are both flying at 520 mph towards an airport. Plane A is flying south and is 50 miles from the airport while Plane # is flying west and is 120 miles from the airport. How fast is the distance between the two planes changing at this time? Given a= 50 miles

a = 50 miles b = 120 miles

Airport

b = 120 miles

da = 520 mph dt db = 520 mph dt

c2 = a 2 + b2 Side Work dc da db = 2a 2c + 2b c2 = a 2 + b2 dt dt dt dc 130 = 50( 520 ) + 120( 520 ) c2 = 2500 + 1440 dt dc 130 = 88400 c = 130 dt dc = 680 mph dt

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20. A spherical Tootsie Roll Pop that you are enboying is giving up volume at a steady rate of 0.25 in3/min. How fast will the radius be decreasing when the Tootsie Roll Pop is .75 inches across? Given :

dV = -.025 dt r = .375 dr =? dt

4 V = (r 3 3 dV dr = 4(r 2 dt dt ! .25 = 4!(.375)

2

in dr = !.141 min dt

dr dt

21. The mechanics at Toyota Automotive are reboring a 6-inch deep cylinder to fit a new piston. The machine that they are using increases the cylinder's radius one-thousandth of an inch every 3 minutes. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.80 inches? Given : .001 dr =3 dt r = 1.9 h = 6 (const) dV =? dt V = (r 2 h V = 6(r 2 dV dr = 12(r dt dt .001 dV = 12((1.9) 3 dt 3 of the 8 in 3 dV = .024 min dt

22. Sand falls at the rate of 30 ft3/min onto the top of a conical pile. The height of the pile is always base diameter. How fast is the height changing when the pile is 12 ft. high? Given : dV = 30 dt h = 12 dh =? dt 1 V = (r 2 h 3 1 "16 2 % V= ( h h 3 #9 & dV "16 2 % dh =( h # 9 & dt dt Side Work h= h= 3 (2r) 8

3 4 r)r= h 4 3

16( dh (144) 9 dt 15 ft dh = = .037 min dt 128( 30 =

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23. A rowboat is pulled toward a dock from the bow through a ring on the dock 6 feet above the bow. If the rope is hauled in at 2 ft/sec, how fast is the boat approaching the dock when 10 feet of rope are out? Given z= 10 feet of rope

x 2 + y 2 = z2 x 2 + 36 = z2 dx dz = 2z 2x dt dt dx = 10(2) 8 dt dx ft = 2.5 dt sec

Side Work

z = 10 y = 6 (const.)

y= 6 feet x

x 2 + y 2 = z2 x 2 + 36 = 100 x=8

dz ft =2 dt sec dx =? dt

24. A particle is moving along the curve whose equation is

8 xy 3 = . Assume the x-coordinate is increasing at 2 5 1+ y the rate of 6 units/sec when the particle is at the point (1, 2). At what rate is the y-coordinate of the point changing at that instant. Is it rising or falling? 5 xy 3 = 8 + 8 y 2 Given : x = 1

y =2 dx =6 dt dy =? dt

" dy dx % dy 5$ x , 3 y 2 + y 3 ' = 16 y dt dt & dt # % " dy dy 5$12 + 8 , 6' = 32 dt & # dt 28

dy 60 units =! - falling dt 7 sec

dy = -240 dt

25. A balloon is rising vertically above a level, straight road at a constant rate of 1 foot/sec. Just when the balloon is 65 feet above the ground, a bicycle passes under it going 17 feet/sec. How fast is the distance between the bicycle and balloon increasing 3 seconds later? Given y = 68 feet x = 51 feet y= 68 ft. z dy ft =1 dt sec dx ft = 17 dt sec dz =? dt x 2 + y 2 = z2 dz dx dy 2x + 2y = 2z dt dt dt dz 51(17) + 68(1) = 85 dt dz 932 = dt 85 dz ft = 11 dt sec Side Work x 2 + y 2 = z2 512 + 65 2 = z 2 z = 85

x

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26. Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in3/min. a) How fast is the level in the pot rising when the coffee in the filter is 5 inches deep? b) How fast is the level in the cone falling then? 6 inches Cylinder Cylinder Cone Side Work 1 V = (r 2 h h = 2r 3 1 "h% V = ($ ' h 3 #2& (h 3 V= 12 dV (h 2 dh = 4 dt dt 25( dh 10 = 4 dt dh 8 = dt 5( dh in = .51 dt min

2

r = 3(const)

How fast is this level falling? 5 in. 6 inches dV in 3 = 10 min dt

V = (r 2 h V = 9(h dV dh = 9( dt dt dh 10 = 9( dt dh 10 = dt 9( dh .35 in = dt min

dh =? dt

Cone How fast is this level rising? 6 inches

dV in 3 = 10 dt min h = 5 in. dh =? dt

27. cn a certain clock, the minute hand is 4 in. long and the hour hand is 3 in. long. How fast is the distance between the tips of the hands changing at 4 P.M?

c2 = a 2 + b2 ! 2 ab cos*

12

2

4

c = 16 + 9 ! 24 cos* dc d* 2c = 24 sin * dt dt dc = 9.84 in/hr = .164 in/min. dt

d* 1Re v 1Re v 2( !11( = ! = ! 2( = dt 12 hr 1 hr 12 6 c = 16 + 9 + 12 = 37

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