Table of Contents Related Rate Word Problems

Transcription

1 00 D.W.MacLean: Related Rate Word Problems -1 Table of Contents Related Rate Word Problems General Strategy 003 Doug MacLean Expanding Balloon A Sand Pile Docking AirCraft Carrier Sliding Ladder Moving Ships Flag Raising A Lighthouse A Point Moving On a Curve Inscribed Rectangle 1 Inscribed Rectangle Filling a Trough Pulling Carts People Walking Hands On a Clock: A Short Quiz Planetary Orbits Piston Movement

2 00 D.W.MacLean: Related Rate Word Problems - Related Rate Word Problems How to Solve Related Rate Word Problems 003 Doug MacLean Step 1: Draw a sketch, if appropriate. Step : Make a list of variables. Don t expect to get it right immediately, you may have to come back and add more. Step 3: Make a list of symbols. Step 4: Make a list of relations. Again, you may have to add more as the problem develops. Now is a good time to review the collection of formulas on the web site. Step 5: given. Step 6: Step 7: Identify the variable whose rate of change is to be found, and the variable whose rate of change is Differentiate the relations. Solve the mathematical problem by inserting the given value of the variables. This must NOT be done BEFORE differentiating! Step 8: Translate the problem back into English, and be sure that it is a reasonable solution.

3 00 D.W.MacLean: Related Rate Word Problems -3 Expanding Balloon A weather balloon is filled with helium gas at the rate of litres /second. How fast is the radius increasing when it is 50 cm? How fast is the surface area increasing at that time? 003 Doug MacLean Solution: Step 1: Sketch: r Steps &3: The radius r(t), the volume V(t), and the Surface area S(t) are the only variables. Step 4: Relations: V(t) = 4 3 πr (t)3 and S(t) = 4πr (t) Step 5: Identification: The rates of change r of r(t) and S (t) of S(t) are to be found when r(t) = 50 cm, given that V(t) is changing at the rate of V = 000 centimetres3. second Step 6: Differentiate: V (t) = 4πr (t) r (t) and S (t) = 8πr (t)r (t). Step 7: Solve: Solve for r (t) when r(t) = 50 centimetres:

4 00 D.W.MacLean: Related Rate Word Problems -4 r (t) = V (t) 4πr (t) = 000 centimetres 3 second 4π(50centimetres) = 10 π centimetres second and then find S (t) = 8πr (t)r (t) = 8π(50centimetres) 10 π centimetres second = 4000 centimetres second = 0.4 metres second Step 8: Translate: The radius is increasing at the rate of 10 π centimetres per second and the surface area is increasing at the rate of 4000 square centimetres per second.

5 00 D.W.MacLean: Related Rate Word Problems -5 A Sand Pile Sand is poured onto a pile at the rate of 100 litres/minute. The radius of the base is always 1.5 times the height of the pile. How fast is the radius increasing when it is 15 metres? 003 Doug MacLean Solution: Step 1: Sketch: h r Steps &3: The radius r(t) and the volume V(t) are the only relevantvariables. Step 4: Relations: V(t) = 1 πr (t)3 3 Step 5: Identification: The rates of change r of r(t) is to be found when r(t) = 15 metres, given that V(t) is changing at the rate of V = 0.1 metres3 minute. Step 6: Differentiate: V (t) = πr (t) r (t). Step 7: Solve: Solve for r (t) when r(t) = 15 metres:

6 00 D.W.MacLean: Related Rate Word Problems -6 r (t) = V (t) πr (t) = 0.1 metres 3 minute π(15metres) = 0.1 5π metres minute millimetres minute Step 8: Translate: The radius is increasing at the rate of about millimetres per minute.

7 00 D.W.MacLean: Related Rate Word Problems -7 Docking AirCraft Carrier An aircraft carrier is being moored to a wharf. When it is 30 metres(horizontally) from the wharf a cable is attached to the wharf at point O and is pulled onto the bow of the ship s flight deck at point A, which is 30 metres above the point of attachment to the wharf, at the rate of 10 cm per minute. How fast is the ship approaching the dock when it is 0 metres(horizontally) from the point of attachment? y Solution: Step 1: Sketch: Doug MacLean 30 A O x Steps &3: The horizontal distance x(t) of A from O, and the distance s(t between O and A are the only variables. Step 4: Relations: x(t) + 30 = s(t) Step 5: Identification: The rate of change x of x is to be found, given that s is changing at the rate of s = 10 centimetres. minute Step 6: Differentiate: xx + 0 = ss. Step 7: Solve: Solve for s(t) when x(t) = 0 metres: s(3) = = (metres).

8 00 D.W.MacLean: Related Rate Word Problems -8 Insert the values s(t) = 10 13, x(t) = 0, s = 10 centimetres minute x(t)x = s(t)s becomes (0)x = (10 ( 13) ( ) x = (10 13) 10 centimetres minute (0) = 5 13 centimetres minute 10 centimetres minute ( into the differentiated equation: ), which has solution: 18 centimetres minute Step 8: Translate: The ship is approaching the wharf at about the rate of 18 centimetres per minute. )

9 00 D.W.MacLean: Related Rate Word Problems -9 Moving Ships A ship leaves a port at noon and steams North at the speed of 5 lilometres per hour. An hour later, another ship leaves port and steams East at the speed of 0 kilometres per hour. How fast is the distance between the ships changing at 3 P.M.? Solution: Step 1: Sketch: y x Doug MacLean Steps &3: The distance y(t) of the first ship North of the port, the distance x(t) of the second ship East of the port, and the distance s(t between the two ships are the only variables. Step 4: Relations: x(t) + y(t) = s(t) Step 5: Identification: The rate of change s of y is to be found, given that y is changing at the rate of y = 5 kilometres hour and x is changing at the rate of x = 0 kilometres hour Step 6: Differentiate: xx + yy = ss. Step 7: Solve: Solve for x(t), y(t), and s(t) when t = 3 hours: x(3) = 3hours 5 kilometres hour kilometres, y(3) = hours 0 kilometres = 40 kilometres, hour = 75

10 00 D.W.MacLean: Related Rate Word Problems -10 x(3) + y(3) = s(3) s(3) = kilometres = 85kilometres. Insert the values x(3) = 75, y(3) = 40, s(3) = 85, x = 5 kilometres, and x = 0 kilometres hour hour differentiated equation: into the x(3)x + y(3)y = s(3)s becomes (75)5 kilometres hour (75)5 kilometres s = hour + (40)0 kilometres hour (85) kilometres hour + (40)0 kilometres hour = (85)s, which has solution: Step 8: Translate: The distance between the two ships is increasing at about the rate of 31.5 kmh at 3 P.M.

11 00 D.W.MacLean: Related Rate Word Problems -11 SlidingLadder A ladder 5 metres long is leaning against a vertical wall. The bottom of the ladder is pushed away from the wall at the speed of 0.1 metres per second. How fast is the top of the ladder descending when it is 4 metres above ground level? y Solution: Step 1: Sketch: Doug MacLean Steps &3: The distance x of the bottom of the ladder from the wall and the height y of the top of the ladder from the ground are the only variables. Step 4: Relations: x + y = 5 Step 5: Identification: The rate of change y of y is to be found, given that x is changing at the rate of x = 0.1 metres. sec Step 6: Differentiate: xx + yy = 0. Step 7: Solve: Solve for x when y = 4 metres: x + 4 = 5 x = 3 metres. Insert the values x = 3, y = 4, x = 0.1 metres into the differentiated equation: sec xx + yy = 0 becomes (3)0.1 metres sec + (4)y = 0, which has solution: x

12 00 D.W.MacLean: Related Rate Word Problems -1 y (3)0.1 metres = sec (4) = metres sec Step 8: Translate: The top of the ladder is descending at the rate of metres per second.

13 00 D.W.MacLean: Related Rate Word Problems -13 Flag Raising: A flag is being raised on a 0 metre high flagpole at the rate of 0.6 metres per minute. An observer is standing 10 metres away, with his eyes metres above ground level. How fast is the angle α of elevation of the flag increasing when the flag is 1 metres above ground level? y Solution: 0 Step 1: Sketch: x Steps &3: The y-coordinate y(t) and the angle α(t) of elevation of the flag are the only variables. α 003 Doug MacLean Step 4: Relations: tan α(t) = y(t). 10 Step 5: Identification: The rates of change α is to be found, given that y(t) is changing at the rate of y = 0.6 metres minute. Step 6: Differentiate: sec α(t)α (t) = 1 10 y or α (t) = cos α(t) 1 10 y.

14 00 D.W.MacLean: Related Rate Word Problems -14 Step 7: Solve: Insert the values cos α(t) =, y = 0.6 metres minute into the equation α (t) = cos α(t) 1 10 y : α (t) = ( ) 1 metres minute = 0.03 radians minute Step 8: Translate: The angle α of elevation is increasing at the rate of 0.03 radians per minute.

15 00 D.W.MacLean: Related Rate Word Problems -15 A Lighthouse: A lighthouse is located on an island 5 kilometres offshore from a perfectly straight beach. The beam of light it projects rotates 5 times per minute. How fast is the beam B moving along the beach when it is one kilometer from the closest point P on the beach? 003 Doug MacLean Solution: Step 1: Sketch: L θ Steps &3: The x-coordinate x(t) of B and the angle θ(t) PLB are the only variables. Step 4: Relations: tan θ(t) = x(t) 5. P B Step 5: Identification: The rates of change x is to be found, given that θ(t) = 5(π) = 10πradians. Step 6: Differentiate: sec θ(t)θ (t) = 1 5 x or (t) = 5 sec θ(t)10π km/min. Step 7: Solve: Insert the value x(t) = 1: (t) = 5 sec θ(t)10π = 50π 6 5 = 5π km/min. Step 8: Translate: The beam B moving along the beach at the speed of 5π km/min.

16 00 D.W.MacLean: Related Rate Word Problems -16 Step 7: Solve: Insert the value t = 0: s (0) = [ 4 cos π(0) 400 [ ][ 4 cos 0 cos π 4 [ ][ [ π 50 ][ ] [ ][ ] π(0+1.5) cos 50 4 sin π(0) π π(0+1.5) π sin sin π(0) π(0+1.5) 400 sin 50 4 cos π(0) π π(0+1.5) π cos [ ] 4 cos π(0) π(0+1.5) [ ] = 400 cos sin π(0) π(0+1.5) 400 sin 50 ] [ ][ ] 4 sin 0 π sin π π sin 0 sin π 4 4 cos 0 π 400 cos π π 4 50 [ ] 4 cos 0 cos π [ ] = sin 0 sin π 4 ] [ ][ π π ] 400 π 50 [ ] [ ] = 4 + ] [ ] 4 1 km/day Doug MacLean Step 8: Translate: The distance is changing at the rate of 10 6 π , 751, 490 km/day.

17 00 D.W.MacLean: Related Rate Word Problems -17 A Point Moving On a Curve: A point is moving toward the origin on the parabola x + y = 0 with its x-coordinate increasing at the rate of units/ms. What is the rate of change of the y-coordinate as it passes through the point (-1,1)? How fast is it approaching the origin? Solution: Step 1: Sketch: y 003 Doug MacLean x -1 Steps &3: The x-coordinate x and the y-coordinate y of the point and its distance s from the origin are the only variables. - Step 4: Relations: x + y = 0 and s = x + y. Step 5: Identification: The rates of change y of y and s of s are to be found, given that x is changing at the rate of x = units ms. Step 6: Differentiate: x + yy = 0, and ss = xx + yy.

18 00 D.W.MacLean: Related Rate Word Problems -18 Step 7: Solve: Insert the values x = 1, y = 1, x = units ms ( units ) ms to get y = 1 units ms ( + (1)y = 0, ss = ( 1) ( and ( 1) + 1 s = ( 1) units ms into the differentiated equations: units ) ms + (1)y. ) ( + (1) 1 units ms ). or s = 6 units ms,sos = 6 units ms = 3 units ms. Step 8: Translate: The y coordinate is decreasing at the rate of one unit per millisecond, while the distance from the origin is decreasing at the rate of 3 units per millisecond.

19 00 D.W.MacLean: Related Rate Word Problems -19 Inscribed Rectangle 1: A rectangle of has its base on the x-axis and is inscribed inside the parabola y = 8 x. If its height is decreasing at the rate of 4 units/sec, how fast are its width and area changing when its height is 4 units? y Solution: Step 1: Sketch: Doug MacLean Steps &3: The width w and the height h of the rectangle and its area A are the only variables. x Step 4: Relations: h = 8 (w/) = 8 w /4 and A = wh. Step 5: Identification: The rates of change w of w and A of A are to be found, given that h is changing at the rate of h = 4 units sec.

20 00 D.W.MacLean: Related Rate Word Problems -0 Step 6: Differentiate: h = (w/)w, and A = wh + w h. Step 7: Solve: Insert the values w = 4, h = 4, h = 4 units sec 4 units sec = (4/)w, gives us w = units sec into the differentiated equations: and thus ( A = (4units) 4 units ) sec + (units) units sec 4 = units 8 sec. Step 8: Translate: The width coordinate is increasing at the rate of two units per second, while the area is decreasing at the rate of eight square units per second.

21 00 D.W.MacLean: Related Rate Word Problems -1 Inscribed Rectangle : A rectangle is inscribed in a right triangle with legs of lengths 6 cm and 8 cm. with two sides of the rectangle lie along the legs. If the area of the rectangle is increasing at the rate of one square cm per second, how fast are the height and width changing when its area is twelve cm? Solution: Step 1: Sketch: 6 y Doug MacLean Steps &3: The width w and the height h of the rectangle and its area A are the only variables. x Step 4: Relations: h = w and A = wh = w ( w ) = 6w 3 4 w. Step 5: Identification: The rates of change w of w and h of h are to be found, given that A is changing at the rate of A = 1 cm sec. Step 6: Differentiate: h = 3 4 w, and A = 6w 3 ww = (6 3 w)w.

22 00 D.W.MacLean: Related Rate Word Problems - Step 7: Solve: First we need to find the value of w when A = 1. To do this we must solve the quadratic equation 1 = 6w 3 4 w for w = 4 Insert the values w = 4cm, h = 3cm, A = 1 cm sec into the second differentiated equation: A = (6 3 w)w becomes 1 cm sec = (6 3 4)w = 0, so w = h = 0! Step 8: Translate: Something weird is happening here! Neither height nor width is changing, but the area is! Can you find the division by zero?

23 00 D.W.MacLean: Related Rate Word Problems -3 Filling a Trough: A trough is ten metres long and its ends have the shape of isosceles trapezoids that are 80 cm across at the top and 30 cm across at the bottom, and has a height of 50 cm. If the trough is filled with water at a rate of 0. min m3, how fast is the water level rising when the water is 30 cm deep? Solution: Step 1: Sketch: Doug MacLean w 30 w h Step : Variables: The width of the horizontal sides of the two triangles and their height will be needed to express the area of the wet end of the trough and the volume of water in the trough. Step 3: Symbols: w and h will be as depicted. w T and w B will denote the widths of the top and bottom of the blue trapezoid, and A will be its area. V will denote the volume of water in the trough. Step 4: Relations: V = 10A, A = w B + w T w B = 3 10, w T = w = h, so V = 10A = h, w h = 5 50 = 1,soh = w. ( h + 3 ) 10 h = 5(h )h = 5h + 3h, orv = 5h + 3h

24 00 D.W.MacLean: Related Rate Word Problems -4 Step 5: Identification: The rate of change h of h is to be found, given that V is changing at the rate of 1 5 Step 6: Differentiate: V = 10hh + 3h = (10h + 3)h. Step 7: Solve: Insert the values h = 3 10, V = m 3 min = ( )h = 6h to get h = 1 30 m min m 3 min into the differentiated equation V = (10h + 3)h : Step 8: Translate: The water level is increasing at the rate of 1/30 of a metre per minute. m 3 min

25 00 D.W.MacLean: Related Rate Word Problems -5 Pulling Carts: Two carts, A and B, are connected by a rope 39 feet long that passes over a pulley P. The point Q is on the floor 1 feet directly beneath P and between the carts A and B. Cart A is being pulled away from Q at a speed of sec ft. How fast is cart B moving toward Q at the instant when cart A is 5 feet from Q? Solution: Step 1: Sketch: P 003 Doug MacLean x z 1 A w y B Q Steps &3: Let w and x be the distance of A from Q and P respectively, and let y and z be the distance of B from Q and P respectively. Step 4: Relations: There are three obvious relations: w + 1 = x, x + y = 39, and z + 1 = y. Step 5: Identification: We must find dz dw, given that dt dt = ft, at the point in time when w = 5ft. sec Step 6: Differentiate: ww = xx, x + y = 0, and zz = yy, which we may reduce to:

26 00 D.W.MacLean: Related Rate Word Problems -6 z = y z y, y = x, and x = w x w, or z = y z w x w. Step 7: Solve: We now find the values of x, y, and z when w = 5: x = = 169 = 13, y = 39 x = = 6, z = y 1 = 6 1 = 133. ( )( )( ) 6 5 ft Thus z = = 10 ft sec 133 sec. Step 8: Translate: Cart B is moving towards Q at the rate of feet per second

27 00 D.W.MacLean: Related Rate Word Problems -7 People Walking: Two people start from the same point. One walks east at 3 miles hour northeast at miles. How fast is the distance between the people changing after 15 minutes? hour Solution: Step 1: Sketch: and the other walks 003 Doug MacLean y z x Steps &3: Let x be the distance of the first person from the initial point at time t, let y be the distance of the second person from the initial point at time t, and let z be the distance between the two at time t. Step 4: Relations: By the Law of Cosines, z = x + y xy cos π 4 = x + y xy. Step 5: Identification: We must find z, given that x = 3 miles hour and y = miles hour when t = 0.5hour. Step 6: Differentiate: zz = xx + yy (x y + xy ) or z = xx + yy (x y + xy ) z ( Step 7: Solve: After 15 minutes, x = 3 4 mile, y = 1 ) 3 ( ) 1 mile, so z = + ( )( ) 3 1 mile= 4 4

28 00 D.W.MacLean: Related Rate Word Problems mile= mile. 4 Thus z = ( ) ( ) 1 ( ( ) 1 + ( ) ) 3 4 miles hour = Step 8: Translate: The two people are separating at the rate of 13 6 miles hour miles per hour.

29 00 D.W.MacLean: Related Rate Word Problems -9 Hands On a Clock: The minute hand on a watch is 8mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o clock? 003 Doug MacLean Solution: Step 1: Sketch: Steps &3: We let x be the distance between the tips of the hands, θ h and θ m be the angles made by the hour and minute hands with the vertical line through 1, and we let θ be the angle from the minute hand to the hour hand. Step 4: Relations: x = (4)(8) cos θ = cos θ, θ = θ h θ m. Step 5: Identification: We must find x, and to do this we will need to find θ m, θ h and θ :

30 00 D.W.MacLean: Related Rate Word Problems -30 We have θ m = θ = θ h θ m = ( π 6 π radians 1 hour = π radians hour π) radians hour = 11π 6 and θ π h = radians 1 hour radians hour. = π 6 radians hour,so Step 6: Differentiate: xx = 64( sin θ)θ = 64 sin θθ,sox 3 sin θ = θ, θ = θ h x θ m. Step 7: Solve: At one o clock, θ = π 6, x = cos π 6 80 = 64 3 = = ( ) and thus x = 3 1 ( π ) π = mm 3 hour Step 8: The two tips are getting closer at the rate of π 3 5 mm per hour 3

31 00 D.W.MacLean: Related Rate Word Problems -31 Planetary Orbits: Two planets P 1 and P revolve around a star S in circular orbits lying in the same plane of radii r 1 = 100, 000, 000 and r = 400, 000, 000 km. The closest planet P 1 makes a complete orbit in 100 Earth days, while the most distant planet P takes 800 days. How fast is the distance P 1 P between the two planets changing when the angle P 1 SP made by the three bodies is 45? 003 Doug MacLean Solution: Step 1: Sketch: y P1 S P x

32 00 D.W.MacLean: Related Rate Word Problems -3 Step &3: We set time 0 to a the time when the angle P 1 SP is 45 or π, and we take the x-axis to be the line 4 SP at time 0. Then we let θ 1 (t) be the angle that the line SP 1 makes with the positive x-axis, and the θ (t) be angle that that the line SP makes with the positive x-axis. We also let x 1 (t), y 1 (t), x (t), y (t) be the x- and y-coordinates of P 1 and P respecitively, and we let s(t) be the distance P 1 P at time t. Step 4: Relations: θ 1 (t) = π 100 t + π π(t + 1.5) = radians, and θ (t) = π t = πt 400. x 1 (t) = r 1 cos θ 1 (t), y 1 (t) = r 1 sin θ 1 (t), x (t) = r cos θ (t), y (t) = r sin θ 1 (t), s(t) = [x (t) x 1 (t)] + [ y (t) y 1 (t) ] Step 5: Identification: The rates of change s (0) is to be found. Step 6: Differentiate: s(t)s (t) = [x (t) x 1 (t)][x (t) x 1 (t)] + [ y (t) y 1 (t) ][ y (t) y 1 (t) ] 003 Doug MacLean or s (t) = [x (t) x 1 (t)] [ x (t) x 1 (t)] + [ y (t) y 1 (t) ][ y (t) y 1 (t)] = s(t) [r cos θ (t) r 1 cos θ 1 (t)] [ r ( sin θ (t))θ (t) r 1( sin θ 1 (t))θ 1 (t)] + [r sin θ (t) r 1 sin θ 1 (t)] [ r cos θ (t)θ (t) r 1 cos θ 1 (t)θ 1 (t [r cos θ (t) r 1 cos θ 1 (t)] + [r sin θ (t) r 1 sin θ 1 (t)] 10 8 [4 cos θ (t) cos θ 1 (t)] [ 4 sin θ (t)θ (t) + sin θ 1(t)θ 1 (t)] + [4 sin θ (t) sin θ 1 (t)] [ 4 cos θ (t)θ (t) cos θ 1(t)θ 1 (t)] [4 cos θ (t) cos θ 1 (t)] + [4 sin θ (t) sin θ 1 (t)] = 10 8 [ 4 cos πt 400 [ 10 8 π 4 cos πt ][ π(t+1.5) cos 50 ][ π(t+1.5) cos 50 4 sin πt π [ 4 cos πt sin πt 400 [ 4 cos πt sin π(t+1.5) 50 π 50 cos π(t+1.5) 50 ] [ π(t+1.5) + 8 sin + 50 cos π(t+1.5) 50 ] [ + 4 sin πt 400 ] [ + 4 sin πt sin πt 400 ] + [ 4 sin πt 400 ][ ] π(t+1.5) sin 50 4 cos πt π π(t+1.5) π cos ] = π(t+1.5) sin 50 ][ ] π(t+1.5) sin 50 4 cos πt π(t+1.5) 400 cos 50 sin π(t+1.5) 50 ]

33 00 D.W.MacLean: Related Rate Word Problems -33 Step 7: Solve: Let t = 0: [ ][ ] [ ][ ] s (0) = 10 8 π 4 cos 0 cos π 4 4 sin sin π sin 0 sin π 4 4 cos 0 cos π [ ] 4 cos 0 cos π [ ] = sin 0 sin π 4 [ ][ ] [ ][ ] 10 6 π [ ] [ ] = π [ ] 8 4 = Doug MacLean π (8 ) Step 8: Translate: The distance between the planets is changing at the rate of π (8 ) km/day.

34 00 D.W.MacLean: Related Rate Word Problems Piston Movement: A piston at point A moves in a vertical cylinder. It is connected by a connecting rod AB of length 15 cm to a crankshaft of radius 5 cm and centre O which5 is revolving at 5000 rpm. How fast is the piston moving when the angle θ is 45? B Step 1: Sketch: Solution: 003 Doug MacLean 0 O θ(t) x -5

35 00 D.W.MacLean: Related Rate Word Problems -35 Step &3: We set time 0 to be a the time when the point B lies on the x-axis. Then we let θ(t) be the angle that the line OB makes with the positive x-axis. 003 Doug MacLean We let x(t) and y(t) be the x- and y-coordinates of B, and we let Y(t) be the distance OA at time t. Step 4: Relations: θ(t) = 5000(π)t = 10000πt radians. x(t) = 5 cos θ(t), y(t) = 5 sin θ(t), Y(t) = y(t) + 15 x(t) = 5 sin θ(t) cos θ(t) Step 5: Identification: The rates of change Y (t) is to be found when θ(t) = 45 = π 4 radians. Step 6: Differentiate: Y (t) = y (t) [ 5( cos θ(t))( sin θ(t))θ (t) ] = 5 5 cos θ(t) 5 cos θ(t)θ (t) + 5(cos θ(t))(sin θ(t))θ (t) = [ 5 5 cos θ(t) ] sin θ(t) 50000π cos θ(t) cos θ(t) Step 7: Solve: When θ(t) = π 4, we have Y (t) = 50000π 1 + ( ) = π = 50000π = π Step 8: Solve: The piston is travelling at a speed of 5000π cm/min, or 15π kmh.

36 00 D.W.MacLean: Related Rate Word Problems -36 A Fast Airplane: An airplane is known to be flying at an altitude of 1000 metres. When it is directly overhead (at point A) an observer a point O on the ground, the sound of its engines appears to be coming from y a point B 10 behind the airplane. How fast is it going? Assume the speed of sound to be 33 m/s 100 kilometres per hour. A B Solution: Step 1: Sketch: 003 Doug MacLean θ(t) O x The distance AB =1000metres tan metres, and the distance OB = 1000 cos 10 metres. The time it 1 cos 10 kilometre takes for the sound of the engines to reach O is thus t = 1000 = 3 seconds so the 33m/s 33 cos 10 speed of the airplane is about 176 m/sec or about 10 kmh. 3