2019 Related Rates AB Calculus

Known Limits:

Intro: This is a adaptation of IMPLICIT functions graph ILLUSTRATION:  GOAL: to find the rates of change of two (or more) variables with respect to a third variable (the parameter) This is a adaptation of IMPLICIT functions   x and y are implicit functions of t .   ILLUSTRATION: A point is moving along the parabola, Find the rate of change of y when x = 1 if x is changing at 2 units per second. moving time moving graph dy 𝑑𝑦 𝑑𝑡 =? 𝑦= 𝑥 2 +3 𝑑𝑥 𝑑𝑡 =+2 𝑤ℎ𝑒𝑛 𝑥=1 𝑑𝑦 𝑑𝑡 =2𝑥 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 =2 1 2 =4

1). DRAW A PICTURE! – Determine what rates are being compared. PROCEDURE: 1). DRAW A PICTURE! – Determine what rates are being compared.  2). Assign variables to all given and unknown quantities and rates.  3). Write an equation involving the variables whose rates are given or are to be found   ·       Equation of a graph? ·       Formula from Geometry?   The equation must involve only the variables from step 2. – ((You may have to solve a secondary equation to eliminate a variable.))  4). Use Implicit Differentiation (with respect to the parameter t).  5). AFTER DIFFERENTIATION, substitute in all known values   (( You may have to solve a secondary equation to find the value of a variable.)) May plug in a constant as long as it is unchanging

Geometry formulas:   Sphere: Cylinder: Cone: Pythagorean Theorem:

METHOD: Inflating a Balloon - 1 A spherical balloon is inflated so that the radius is changing at a rate of 3 cm/sec. How fast is the volume changing when the radius is 5 cm.? Draw and label a picture. Step 1: List the rates and variables. 𝑑𝑟 𝑑𝑡 =+3 𝑑𝑉 𝑑𝑡 =? Find an equation that relates the variables and rates. (Extra Variables?) When r =5 𝑉= 4 3 𝜋 𝑟 3 Plugin 5 gives vol not rate of change Differentiate (with respect to t.) 𝑑𝑉 𝑑𝑡 =4𝜋( 𝑟 2 ) 𝑑𝑟 𝑑𝑡 𝑑𝑉 𝑑𝑡 =4𝜋( 5 2 )(3) Plug in and solve. 𝑑𝑉 𝑑𝑡 =300𝜋 𝑐𝑚 3 𝑠𝑒𝑐

Ex 2: Ladder w/ secondary equation A 25 ft. ladder is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at a rate of 3 ft./sec., how fast is the top of the ladder sliding down the wall when the bottom is 15 ft. from the wall? # constant does not change ever you can plug in the equation 𝑧=25 𝑑𝑥 𝑑𝑡 =+3 𝑑𝑦 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑥=15 𝑥 2 + 𝑦 2 = 𝑧 2 𝑥 2 + 𝑦 2 = 𝑧 2 𝑥 2 + 𝑦 2 = 25 2 15 2 + 𝑦 2 = 25 2 2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =2𝑧 𝑑𝑧 𝑑𝑡 2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =0 2 15 3+2 20 𝑑𝑦 𝑑𝑡 =0 2 15 3 +2 20 𝑑𝑦 𝑑𝑡 =0 90+40 𝑑𝑦 𝑑𝑡 =0 90+40 𝑑𝑦 𝑑𝑡 =0 𝑑𝑦 𝑑𝑡 = −45 20 = −9 4 𝑑𝑦 𝑑𝑡 = −45 20 = −9 4 The ladder is coming down -2.25 ft/sec The ladder is coming down

II: Similar Triangles

Similar Triangles B D E A F C Similar Triangles may be the whole set up. Similar Triangles may be required to to eliminate an extra variable – or- to find a missing value

Ex 4: A person is pushing a box up a 20 ft. ramp with a 5 ft. incline at a rate of 3 ft.per sec.. How fast is the box rising? derivative 20 ft 5 z y x 𝑑𝑧 𝑑𝑡 =+3 𝑦 𝑧 = 5 20 as 𝑑𝑦 𝑑𝑡 = 15 20 = 3 4 𝑓𝑡 𝑠𝑒𝑐 5𝑧=20𝑦 5 𝑑𝑧 𝑑𝑡 =20 𝑑𝑦 𝑑𝑡 20 𝑑𝑦 𝑑𝑡 =5(3)

Ex 5: Getting smaller Pat is walking at a rate of 5 ft. per sec. toward a street light whose lamp is 20 ft. above the base of the light. If Pat is 6 ft. tall, determine the rate of change of the length of Pat’s shadow at the moment Pat is 24 ft. from the base of the lamppost. The distance of top of shadow from post 𝑑𝑥 𝑑𝑡 −−5 How fast is the tip of Pat’s shadow changing 𝑑𝑦 𝑑𝑡 =?𝑤ℎ𝑒𝑛 𝑥=24 6 x y 20 20 𝑥+𝑦 = 6 𝑦 20𝑦=6𝑥+6𝑦 20 𝑑𝑦 𝑑𝑡 =6 𝑑𝑥 𝑑𝑡 +6 𝑑𝑦 𝑑𝑡 6 y 6 y-x 14 𝑑𝑦 𝑑𝑡 =−30 𝑑𝑦 𝑑𝑡 = −30 14 = −15 7 20 𝑦 = 6 𝑦−𝑥

Ex 6: Cone w/ extra equation 𝑉= 1 3 𝜋 𝑟 2 ℎ Ex 6: Cone w/ extra equation Three variables Water is being poured into a conical paper cup at a rate of cubic inches per second. If the cup is 6 in. tall and the top of the cup has a radius of 2 in., how fast is the water level rising when the water is 4 in. deep? 𝑉= 1 3 𝜋 𝑟 2 ℎ Too many variables need to find r 𝑑𝑉 𝑑𝑡 =+ 2 3 𝑉= 1 3 𝜋 ℎ 3 2 ℎ r changes h changes 𝑑ℎ 𝑑𝑡 =?𝑤ℎ𝑒𝑛 ℎ=4 𝑉= 𝜋 27 ℎ 3 Only two variables 𝑑𝑉 𝑑𝑡 = 𝜋 9 ℎ 2 𝑑ℎ 𝑑𝑡 2 6 = 𝑟 ℎ 2 3 = 𝜋 9 4 2 𝑑ℎ 𝑑𝑡 2ℎ=6𝑟 ℎ 3 = 2ℎ 6 =𝑟 +9 16𝜋 ∗ 2 3 = 𝑑ℎ 𝑑𝑡 3 8𝜋 = 𝑑ℎ 𝑑𝑡

III: Angle of Elevation

hyp sin 𝜃= 𝑜𝑝𝑝 ℎ𝑦𝑝 opp adj sin 𝜃= 𝑜𝑝𝑝 ℎ𝑦𝑝 csc 𝜃= ℎ𝑦𝑝 𝑜𝑝𝑝 cos 𝜃= 𝑎𝑑𝑗 ℎ𝑦𝑝 s𝑒𝑐 𝜃= ℎ𝑦𝑝 𝑎𝑑𝑗 tan 𝜃= 𝑜𝑝𝑝 𝑎𝑑𝑗 cot 𝜃= 𝑎𝑑𝑗 𝑜𝑝𝑝

c a b 5 3 4 Angles of Elevation SOH – CAH - TOA θ Hint: The problem may not require solving for an angle measure … only a specific trig ratio. ie. need sec θ instead of θ 𝜃=? 5 3 sec 𝜃= 5 4 θ 4

Ex 7: A balloon rises at a rate of 10 ft/sec from a point on the ground 100 ft from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 100 ft. above the ground. 100 100 2 When y =100 or 𝑑𝑦 𝑑𝑡 =40 𝑦 100 𝑑𝜃 𝑑𝑡 =?𝑤ℎ𝑒𝑛 𝑦=100 𝑠𝑒𝑐 2 𝜃 𝑑𝑧 𝑑𝑡 = 1 100 𝑑𝑦 𝑑𝑡 tan 𝜃= 𝑦 𝑥 sec 𝜃= 100 2 100 = 2 2 2 𝑑𝑧 𝑑𝑡 = 1 100 10 2 𝑑𝜃 𝑑𝑡 = 1 10 𝑑𝜃 𝑑𝑡 = 1 20 𝑟𝑎𝑑 𝑠𝑒𝑐

Ex 8: A fishing line is being reeled in at a rate of 1 ft/sec from a bridge 15 ft above the water. At what rate is the angle between the line and the water changing when 25 ft of line is out. 𝜃 z 15 𝑑𝑧 𝑑𝑡 =−1 𝑓𝑡 𝑠𝑒𝑐 𝑑𝜃 𝑑𝑡 = 3 5 1 20 𝑑𝜃 𝑑𝑡 =? When z = 25 ft 𝑑𝜃 𝑑𝑡 = 3 100 𝑟𝑎𝑑 𝑠𝑒𝑐 sin 𝜃= 15 𝑧 ↔ csc 𝜃= 𝑧 15 𝑧 sin 𝜃 =15 𝑧 cos 𝜃 𝑑𝜃 𝑑𝑡 + sin 𝜃 𝑑𝑧 𝑑𝑡 =0 cos 𝜃= 20 25 = 4 5 𝜃𝜃 𝜃 15 20 25 25 4 5 𝑑𝜃 𝑑𝑡 + 3 5 −1 =0 sin 𝜃= 15 25 = 3 5 20 𝑑𝑧 𝑑𝑡 − 3 5 =0

Ex 9: A television camera at ground level is filming the lift off of a space shuttle that is rising vertically according to the position function , where y is measured in feet and t in seconds. The camera is is 2000 ft. from the launch pad. Find the rate of change of the angle of elevation of the camera 10 sec. after lift off. 2000 2 + 5000 2 =4000000+25000000= 29000000= 𝑑𝜃 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑡=10 𝑠𝑒𝑐 1000 29 tan 𝜃= 𝑦 2000 𝑦=50 𝑡 2 𝜃 y 2000ft tan 𝜃= 50 𝑡 2 2000 = 𝑡 2 400 = 1 400 𝑡 2 sec 2 𝜃 𝑑𝜃 𝑑𝑡 = 1 200 𝑡 29 2 2 𝑑𝜃 𝑑𝑡 = 1 200 10 𝑦=50 10 2 =5000 y 1000 29 5000 2000 sec 𝜃= 1000 29 2000 29 4 𝑑𝜃 𝑑𝑡 = 1 20 4 29 1 20 = 1 145 𝑟𝑎𝑑𝑠 𝑠𝑒𝑐

IV: Using multiple rates

Ex 11: If one leg, AB, of a right triangle increases at a rate of 2 in/sec while the other leg, AC, decreases at 3 in/sec, find how fast the hypotenuse is changing when AB is 72 in. and AC is 96 in. A C B z y x 𝑑𝑦 𝑑𝑡 =2 𝑑𝑥 𝑑𝑡 =−3 𝑑𝑧 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑦=72 𝑎𝑛𝑑 𝑥=96 2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =2𝑧 𝑑𝑧 𝑑𝑡 𝑥 2 + 𝑦 2 = 𝑧 2 𝑧 2 = 96 2 + 72 2 2 96 −3 +2 72 2 =2 120 𝑑𝑧 𝑑𝑡 𝑧 2 =9216+5184 −576+288=240 𝑑𝑧 𝑑𝑡 𝑧= 14400 𝑑𝑧 𝑑𝑡 = −288 240 =−1.2 𝑖𝑛 𝑠𝑒𝑐 𝑧=120

Ex 12: A metal rod has the shape of a right circular cylinder. As it is being heated, its length is increasing at a rate of 0.005 cm/min and its diameter is increasing at 0.002 cm/min. At what rate is the volume changing when the rod has a length 40 cm and diameter 3 cm.?

5: AP Questions

Example 12: AP Type At 8 a.m. a ship is sailing due north at 24 knots(nautical miles per hour) is a point P. At 10 a.m. a second ship sailing due east at 32 knots is a P. At what rate is the distance between the two ships changing at (a) 9 a.m. and (b) 11 a.m.? 𝑥=32 𝑥 2 + 𝑦 2 = 𝑧 2 𝑦=24 𝑑𝑦 𝑑𝑡 =+24 𝑑𝑥 𝑑𝑡 =−32 2 32 −32 +2 24 24 =2 40 𝑑𝑧 𝑑𝑡 𝑑𝑧 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑧=40 𝑑𝑧 𝑑𝑡 = −896 80 2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =2𝑧 𝑑𝑧 𝑑𝑡 𝑑𝑧 𝑑𝑡 =−11.2

Ex 13: AP Type A right triangle has height 7 cm and the hypotenuse is increasing at a rate of 2 cm/sec. When the hypotenuse is 25 cm, find: a). the rate of change of the base. b). The rate of change of the acute angle at the base, c). The rate of change of the area of the triangle.

AP Question A coffee pot is made up of a conical filter that is 6 in. tall and has a diameter of 6 in. and a cylindrical pot that is 6 in. in diameter. Coffee is draining from the filter into the coffeepot at a rate of 10 in3/sec. a). How fast is the level in the pot rising when the coffee in the cone is 5 in. deep? b). How fast is the level in the cone falling at that instant?

AP Question: Combined Example At noon a sailboat is 20 km south of a freighter. The sailboat is traveling east at 20 km per hour, and the freighter is traveling south at 40 km per hour. When is the distance between the boats a minimum? If the visibility is 10 kilometers, could the people see each other? At what rate is the distance between the two boats changing at that instant?

Last Update 11/12/11 BC: