5 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 = 1if x is changing at units per second.movingtimemovinggraphdy𝑑𝑦 𝑑𝑡 =?𝑦= 𝑥 2 +3𝑑𝑥 𝑑𝑡 =+2 𝑤ℎ𝑒𝑛 𝑥=1𝑑𝑦 𝑑𝑡 =2𝑥 𝑑𝑥 𝑑𝑡𝑑𝑦 𝑑𝑡 = =4
6 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 givenor 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 equationto 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 equationto find the value of a variable.))May plug in a constant as long as it is unchanging
8 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 =+3𝑑𝑉 𝑑𝑡 =?Find an equation that relates the variables and rates. (Extra Variables?)When r =5𝑉= 4 3 𝜋 𝑟 3Plugin 5 gives vol not rate of changeDifferentiate (with respect to t.)𝑑𝑉 𝑑𝑡 =4𝜋( 𝑟 2 ) 𝑑𝑟 𝑑𝑡𝑑𝑉 𝑑𝑡 =4𝜋( 5 2 )(3)Plug in and solve.𝑑𝑉 𝑑𝑡 =300𝜋 𝑐𝑚 3 𝑠𝑒𝑐
9 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 equation𝑧=25𝑑𝑥 𝑑𝑡 =+3𝑑𝑦 𝑑𝑡 =?𝑤ℎ𝑒𝑛 𝑥=15𝑥 2 + 𝑦 2 = 𝑧 2𝑥 2 + 𝑦 2 = 𝑧 2𝑥 2 + 𝑦 2 = 25 2𝑦 2 = 25 22𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =2𝑧 𝑑𝑧 𝑑𝑡2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =0𝑑𝑦 𝑑𝑡 =0𝑑𝑦 𝑑𝑡 =090+40 𝑑𝑦 𝑑𝑡 =090+40 𝑑𝑦 𝑑𝑡 =0𝑑𝑦 𝑑𝑡 = −45 20 = −9 4𝑑𝑦 𝑑𝑡 = −45 20 = −9 4The ladder is coming down ft/secThe ladder is coming down
11 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
12 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?derivative20 ft5zyx𝑑𝑧 𝑑𝑡 =+3𝑦 𝑧 = 5 20as𝑑𝑦 𝑑𝑡 = = 3 4 𝑓𝑡 𝑠𝑒𝑐5𝑧=20𝑦5 𝑑𝑧 𝑑𝑡 =20 𝑑𝑦 𝑑𝑡20 𝑑𝑦 𝑑𝑡 =5(3)
13 Ex 5:Getting smallerPat 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 post𝑑𝑥 𝑑𝑡 −−5How fast is the tip of Pat’s shadow changing𝑑𝑦 𝑑𝑡 =?𝑤ℎ𝑒𝑛 𝑥=246xy2020 𝑥+𝑦 = 6 𝑦20𝑦=6𝑥+6𝑦20 𝑑𝑦 𝑑𝑡 =6 𝑑𝑥 𝑑𝑡 +6 𝑑𝑦 𝑑𝑡6y6y-x14 𝑑𝑦 𝑑𝑡 =−30𝑑𝑦 𝑑𝑡 = −30 14 = −15 720 𝑦 = 6 𝑦−𝑥
14 Ex 6: Cone w/ extra equation 𝑉= 1 3 𝜋 𝑟 2 ℎEx 6: Cone w/ extra equationThree variablesWater is being poured into a conical paper cup at a rate ofcubic 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 ℎToo many variablesneed to find r𝑑𝑉 𝑑𝑡 =+ 2 3𝑉= 1 3 𝜋 ℎ ℎr changesh changes𝑑ℎ 𝑑𝑡 =?𝑤ℎ𝑒𝑛 ℎ=4𝑉= 𝜋 27 ℎ 3Only two variables𝑑𝑉 𝑑𝑡 = 𝜋 9 ℎ 2 𝑑ℎ 𝑑𝑡2 6 = 𝑟 ℎ2 3 = 𝜋 𝑑ℎ 𝑑𝑡2ℎ=6𝑟ℎ 3 = 2ℎ 6 =𝑟+9 16𝜋 ∗ 2 3 = 𝑑ℎ 𝑑𝑡3 8𝜋 = 𝑑ℎ 𝑑𝑡
17 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 θ𝜃=?53sec 𝜃= 5 4θ4
18 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.100100 2When y =100or𝑑𝑦 𝑑𝑡 =40𝑦 100𝑑𝜃 𝑑𝑡 =?𝑤ℎ𝑒𝑛 𝑦=100𝑠𝑒𝑐 2 𝜃 𝑑𝑧 𝑑𝑡 = 𝑑𝑦 𝑑𝑡tan 𝜃= 𝑦 𝑥sec 𝜃= = 2𝑑𝑧 𝑑𝑡 =2 𝑑𝜃 𝑑𝑡 = 1 10𝑑𝜃 𝑑𝑡 = 𝑟𝑎𝑑 𝑠𝑒𝑐
19 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.𝜃z15𝑑𝑧 𝑑𝑡 =−1 𝑓𝑡 𝑠𝑒𝑐𝑑𝜃 𝑑𝑡 =𝑑𝜃 𝑑𝑡 =?When z = 25 ft𝑑𝜃 𝑑𝑡 = 𝑟𝑎𝑑 𝑠𝑒𝑐sin 𝜃= 15 𝑧 ↔ csc 𝜃= 𝑧 15𝑧 sin 𝜃 =15𝑧 cos 𝜃 𝑑𝜃 𝑑𝑡 + sin 𝜃 𝑑𝑧 𝑑𝑡 =0cos 𝜃= = 4 5𝜃𝜃𝜃152025𝑑𝜃 𝑑𝑡 −1 =0sin 𝜃= = 3 520 𝑑𝑧 𝑑𝑡 − 3 5 =0
20 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.= = =𝑑𝜃 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑡=10 𝑠𝑒𝑐tan 𝜃= 𝑦 2000𝑦=50 𝑡 2𝜃y2000fttan 𝜃= 50 𝑡 = 𝑡 = 𝑡 2sec 2 𝜃 𝑑𝜃 𝑑𝑡 = 𝑡𝑑𝜃 𝑑𝑡 =𝑦= =5000y50002000sec 𝜃=29 4 𝑑𝜃 𝑑𝑡 = 1 20= 𝑟𝑎𝑑𝑠 𝑠𝑒𝑐
22 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.ACBzyx𝑑𝑦 𝑑𝑡 =2𝑑𝑥 𝑑𝑡 =−3𝑑𝑧 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑦=72 𝑎𝑛𝑑 𝑥=962𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =2𝑧 𝑑𝑧 𝑑𝑡𝑥 2 + 𝑦 2 = 𝑧 2𝑧 2 =2 96 − = 𝑑𝑧 𝑑𝑡𝑧 2 =− =240 𝑑𝑧 𝑑𝑡𝑧=𝑑𝑧 𝑑𝑡 = − =−1.2 𝑖𝑛 𝑠𝑒𝑐𝑧=120
23 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 cm/min and its diameter is increasing at cm/min. At what rate is the volume changing when the rod has a length 40 cm and diameter 3 cm.?
25 Example 12: AP TypeAt 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=32𝑥 2 + 𝑦 2 = 𝑧 2𝑦=24𝑑𝑦 𝑑𝑡 =+24𝑑𝑥 𝑑𝑡 =−322 32 − =2 40 𝑑𝑧 𝑑𝑡𝑑𝑧 𝑑𝑡 =? 𝑤ℎ𝑒𝑛 𝑧=40𝑑𝑧 𝑑𝑡 = −896 802𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 =2𝑧 𝑑𝑧 𝑑𝑡𝑑𝑧 𝑑𝑡 =−11.2
26 Ex 13: AP TypeA 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.
27 AP Question A coffee pot is made up of a conical filter that is 6 in. tall and has a diameterof 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 thecone is 5 in. deep?b). How fast is the level in the cone falling at that instant?
28 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?