MATH section 2.7 Related Rates Page 1 of 7

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1 MATH 0100 section.7 Related Rates Page 1 of 7 Unfotunatel, thee isn t much I can infom befoe ou encounte difficulties in this section. Remembe that this section is all wod poblems. You must be able to convet sentences into diagams, equations, values, and values of ates. Onl thing that is simila in this section is that all ates ae in espect to time. So don t foget to appl implicit diffeentiation. Also, the distance measue ( ) pe unit of time (t ) mentioned in wod poblem is d. Mostl, the vaiable t is not pesent in the equation that we must evaluate. Pa attention to the measuement units; the must be same in ode to ou calculations coect. Take a close look at eamples and stat identifing how the woding in the poblem became the constants and equations. Additional eamples: da ) a) A π? 1 da A π π da π b) da da 1 m/s 0 cm 0. m? π π(0.)(1) 0.6π m /s 4) dl 8 cm/s dw cm/s l 0 cm w 10 cm da? A lw da dl dw 1 1 ( w) + ( l) 1 da dl dw w + l da (10)(8) (0)() cm /s + + dv 6) 4 mm/s diamete 80 mm 40 mm? 4 V π dv 4 1 π dv 4π dv 4 π(40) (4) 600π mm /s l w

2 MATH 0100 section.7 Related Rates Page of 7 8) d d d a)? d d 1 () d () 4()() d d b)? d ( ) 4()() () 9 () d d 4 d ) (4,) d 4 cm, cm cm/s? 8 d d d + 0 d d d (4) () 6 cm/s () d? cm/s

3 MATH 0100 section.7 Related Rates Page of 7 1) 1:00PM 4:00PM km/h B 10 A 10 km B A d km/h In this poblem, we must find Time 1:00PM 4:00PM Distance taveled Boat A Boat B t in h t 0 t 4 at 4:00PM (4)() 140 km (4)() 100 km (10 ) + (10 (140)) + (100) 1 d (10 ) 1 (10) + (100) d (10 ) d ± (10 ) (100)(101) (10 ) d (100) (10 (140)) (100)() (10)() (10)() () () () km/h ) d 1.6 m/s m spotlight (1 ) 1 m

4 MATH 0100 section.7 Related Rates Page 4 of 7 d 1.6m/s 4m? (1 ) 1 4 4(1 ) (1 ) d 1 4 1(1 ) 1 4 d (1 ) 4 4 ( 1.6) ( 1.6) ( 1.6) ( 0.) 0.6 m/s (1 (4)) (8) ) Halfwa fom the bases is 4 ft. d a) 4 ft 4 ft/s? + (90) at 4 ft (4) + (90) (4) + ((4)) (4) + () (4) (4) 1 + () (4) () ± ± (4) () 4 () + 4 () ft d + 0 d d d (4) 4 ( 4) ft/s ( 4 ) 90 ft d 4 ft/s b) 4 ft dv du v 4 ft/s? u v + (90) at v 4 ft u (4) + (90) (4) + ((4)) + + (4) () (4) (4) 1 () u u (4) () u ± ± (4) () 4 () u + 4 () ft 90 ft v dv 4 ft/s

5 MATH 0100 section.7 Related Rates Page of 7 0) du dv u v + 0 du d u v du v dv v dv u u d 8 m? + (1) at 8 m (8) + (1) du (4) 4 (4) ft/s ( 4 ) 1m/s dock 1 m ± 6 6 d + 0 d d boat d 6 ( 1) 6 m/s (8) 8 ) π sin π π π A f peiod 4 π f 1 d (,1) 10 cm/s + 1 at point (, 1) (1) ± 10 ± 9 10 π π π + sin + sin + 4sin

6 MATH 0100 section.7 Related Rates Page 6 of 7 d π π π d 4 sin cos + d π π d + π sin cos d d + πsin( π) d π d + sin( π ) d π d + sin( π ) 1 π 1 ( 10 ) + sin π ( 10 ) 1 π 1 cm/s π 8) 00 ft? tanθ 100cotθ tanθ d csc 1 θ d 100 csc θ d 100csc θ sin θ d Note: fo this tiangle, sinθ. So when 1 1 (8) 0.0 ad/s d 8ft/s θ ft sinθ 00 kite 100 ft

7 MATH 0100 section.7 Related Rates Page 7 of 7 0) a 1 cm b 1 cm π π π dc θ 60 ad /min () ad/min? Fo this eecise we need to use the Law of Cosines (see figue) B c c a + b abcosc c (1) + (1) (1)(1) cosθ π when θ ad a 1 ft π c (1) + (1) (1)(1)cos θ b 1 ft C 1 c c (9)(1) c ± 189 ± (9)(1) ± 1 c 1 c (1) + (1) (1)(1) cosθ dc c 0+ 0 (1)(1) sin 1 θ dc c (1)(1) sinθ dc (1)(1) sinθ c π π (1)(1) sin 180 dc π 90 π π π m/min A