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4.6 Flip Chart

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37 views18 pages

4.6 Flip Chart

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AP ‘Name ns Worksheet Section 4.6 Dae ff get = % 1. The radius r, height h, and volume V of tight circular cylinder are related by the equation” = 77h. (@) Howis —Y reatedto its is constant? dt dt adh (op since = EF we have fat. a Sine at wetne nana, we, Bad etherder (6) Howis reated to “itis constant? ° a, at at Sort) wv qean’ Berane (©) Howis © rautedto < and % stoetner ror his constant? at dt at 2. Ifa and b are the length of two sides ofa age and @ themeasure of the included angle, the area 4 of the twiangle is_4=(1/2)absin @. Howis Fretted oS oe a? ae at ae dA _\fda . oo d. B-1( -b-snora-@-sno+ab- Sind) ail an 1{vino% + ain +abcoso) 3. The length / of a rectangle is decreasing at the rate of cm/sec while the width wis increasing at arate of 2emisec, When != 12om and w =5 cm, find therates of change of (a) the area, (Othe perimeter, and (c) the the diagonal ofthe rectangle. (4) Which of these quantities are decreasing, and which are increasing? ae eae A « 41242)4 (51 7 KD +(K-2) 14 cm? /soe ‘The rate of change of the area is 14 cm? /sec. (b) P=2/+20 ae _ dl ye dt dt dt ¢. 2(-2)+ 2(2) = 0 envsec The rate of change of the perimeter is O cm/sec. 2 a_i tt Pew AD _ (12K~2)+(5)2) 14 aoe ‘The rate of change ofthe length ofthe diameter is 4 ~ yy omen (©) D= cemisee (4) The areas increasing, because its derivative is positive. The perimeter is not changing, because its derivative is zero. The diagonal length is decreasing, because its derivative i negative. 4. Suppose that a droplet of mist is a perfect sphere and that, through condensation, the droplet picks up moisture ata rte proportional tits surface area. Show that under these circumstances the droplets radius increases at a constant rate. Step I: r= radius of spherical droplet S= surface area of spherical droplet V= volume of spherical droplet Step 2: No numerical information is given. Step 3: We want to show tha is constant. I Step 4: dy 4 S=4nr Vern, a Steps 5 and 6: Differentiating V =kS for some constant k 3 mr’, we have #. 4nr? ¢. Substituting kS for and S for 4nr?, we have ks = 5, or 5 5. The mechanics at Lincoln Automotive ae reboring a 6in. deep cylinder to fit a new piston. The machine that are using increases the cylinder’s radius one-thousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.? = CD Faas re il point. 500 "3500 iL ce T “The volume isinceasng atthe rat of approximately eee rt ‘The ylnde shown represents the shape ofthe hole. ‘ruradius of cylinder = volume of eyinder ‘Step: dr 0.0010, ‘Aceinsantn question, = 2001 5 indnin and (since the diameter is 3800 in), r= 1.900 in Step 3: w ewan iotind Step Vane @)= 607 Sep: © et Barer 66. Water is flowing atthe rate of 6m’ ‘min from a reservoir shaped like a hemispherical bowl of radius 13m, shown here in profile. Answer the following questions given that the volume of water in a hemispherical bow! ofradius R is V = (1/3)y"(GR-y) when the water is y units deep, Center of phere Be Wuertvel “ (@) Atwhatrateis the water level changing when the wateris &m deep? (©) Whatis the radius r ofthe water's surface when the water is &m deep? (©) At what rate is theradius r changing when the wateris 8m deep? (a) Step 1: 17 deph of water in bow Step 5: Ve volume of water in bow! 8 <6ny- ay % Suep2: : 6: [At the instant in question, ©Y = -§ m/min and “ “y a ~6=[26m(8)—n06)]} y=8m ot Sep3: . rad We want to find tne value of 2, $a - 001326 minin Step: ox Bent 326 cn vaEyFos-y 0 Vetses Ey 6, Wateris flowing atthe rate of 6m’ min from a reservoir shaped like a hemispherical bowl of radius 13m, shown here in profile. Answer the following questions given that the volume of water in a hemispherical bow! ofradius Ris V = (x7/3)y*(3R —y) when the water is y units deep. eer hee Seep ab (y= depth of water (a) Atwhatrateis the water level changing when the wateris & j_"adlusof wer sce (b) Whatis the radius r of the water's surface when the water is (©) At what ate is theradius » changing when the water is $m St¢P 2 ‘tthe instant in question, © = 6 ni, )Sincor? +(13-y? «13%, Meo=(5= yy = aj. tm hereon rom par (a) sip 3: Wwf eae ‘Step 4: From pan (), r= y26y-9". Step: 1 ty iy ty (26-29)2. 2 arsy=9* dt roy» at ‘Step 6: tet (af ot a Poocey a? \=24e ) 12 Zan, 3 => yggg 770200553 minin ia 6, Wate i Bowing arte ate efi oe asesavels ape nk! bowl of 1S ‘ors bein pote Answer ne Sollomins questions sven tha he vlame of water in erohed Sel foftads Ris P(r 39 GR—))whon the water a deep (a) Atwharrteis the waterlerel casas when the waters Su ep? (@) Whats thei reithe water sface wher theaters Bi Seg? (@Ahseeeis thera Sagi when teeter ee? (ose: yoda water oii ate re ‘Vaveane of erin sep acannon, a6 a rot, 7.A dinghyis pulled toward a dock by a rope from the bow through a ring onthe dock 6ft above the bow as. shown in the figure. The rope is hauled in at arate of 2ftse. (@) How fastis theboat approaching the dock when 10ft of rope are out? (©) Atwhatrateis angle@ changing at that moment? Step 1: 1s length of rope x= horizontal distance from boat to dock ‘@= angle between the rope and a vertical line Step 2: at ‘At the instant in question, Step 3: de ag 8. We want to find ihe vatues of - and 2. 2 fee and = 108. ‘Steps 4. 5. and 6: 8. Let y= f(x) =2° —4xIf de/ dt =-2am/sec, find dy/t at the point where @x=3 (b)x=1 (x=4 2 cm/sec, we have Ba 8-6x7 cm/sec. (a) & =8-6(-3)' =-46 emi —6(1)? = 2 cm/sec 2 &) = (©) 2 =8-6(4)? =-88 cm/sec dt 9. A spherical iron ball is coated with a layer of ice of uniform thickness. Ifthe ice melts at a rate of mL/min, how fastis the outer surface of the ice decreasing when the outer diameter (ball plus ice) is 20cm? Step 1: ‘r= radius of balls plus ice ‘S= surface area of ball plus ice V-= volume of ball plus ice Step 2: ‘At the instant in question, 8mL/min = —8em?/minandr = 3020) 10cm. Step 3: as We it to find -—. fe want a Step 4: We have vader and S=4nr®, These equations can be v3 3 combined by noting that (2) +30 s-a*(2) Step 5: Step 6: Note that V = 4 n10)? = 2000", 3 3 34000 -16 2 = (-8) = = -1.6cm?/min 3 ) 9" Foo 16" Since s <0, the rate of decrease is positive. The surface ‘rea is decreasing at the rate of 1.6 cm?/min. 10. A light shines from the top of a pole SOft high. A bal is dropped from the same height from a point 30 ft away from thellght as shown below. How fast is the ball’s shadow moving along the ground ¥ sec later? {sete 6&1 >". Step 1: ‘5 distance ball has fallen listance from bottom of pole to shadow Step 2: 2 Atte instant in question, $= wo} tant} -t6Hne a 2 Step 3: We want to find at Step 4: x-30_ x By similar triangles, . This is equivalent to 50-5 50 50x ~ 1500 = 50x ~ sx, or sx = 1500. We will use = 1500s", Step 5: The shadow is moving at a velocity of -1500 fUsec. 11. On a moming of a day when the sun will pass directly overhead, the shadow of an 80-ft building on level ‘ground is 60 ft long as shown in the figure. At the moment in question, the angle @the sun makes with the ‘ground is increasing at the rate of 0.27°/min. At what rate is the shadow length decreasing? Express your min, to the nearest tenth. (Remember tose radians.) ‘At the instant in question, 1-60 tod 8-027 /nin= 00015 ins sp We want io fin ~24, a Step 5: S2-s0c*6 a dr Step 6: [Note that, atthe moment in question, since tan 0 = 2 mao

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