Math_137_Winter_2010_Solution_3

# Math_137_Winter_2010_Solution_3 - Math 137 Winter 2010...

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Math 137 Winter 2010 Assignment 3 Due Friday, January 29 All solutions must be clearly stated and fully justified. Use the format given on UW-Ace under Content, in the folder Assignments; it is the file Math 137 Assignment Templates . Text problems: Appendix D: 83. Prove the Law of Cosines: If a triangle has sides with length a, b and c and θ is the angle opposite side c, then c 2 = a 2 + b 2 – 2ab cos θ . Consider the triangle with sides a, b, and c as depicted in the drawing. The altitude between P and A creates right triangle OAP. Clearly, for point P = (x,y), x = b cos θ and y = b sin θ . Then using the formula for distance between two points, plus the fact that the distance between (x,y) and (a,0) is c, c 2 = (a – x) 2 + (0 – y) 2 = (a – b cos θ ) 2 + (b sin θ ) 2 = a 2 – 2ab cos θ + b 2 cos 2 θ + b 2 sin 2 θ = a 2 + b 2 – 2ab cos θ . Section 1.6: 58. A recharge function is Q(t) = Q 0 (1 – e –t/a ) where Q 0 is maximum charge and t is in seconds. a) Find the inverse of this function and explain its meaning. Q(t) = Q 0 (1 – e –t/a ) ) ln( ) ln( / ) ln( 1 0 0 0 0 0 0 / 0 0 0 Q Q Q a Q Q Q a t a t Q Q Q e Q Q Q Q Q a t = = = = = This depicts the length of time it takes to recharge the camera to charge level Q. b) How long does it take to recharge to 90% of maximum charge if a = 2? 605 . 4 ) 10 ln( 2 ) 1 . 0 ln( 2 ) 9 . 0 ln( 2 0 0 0 0 0 = = = Q Q Q Q Q t 64. Find the exact value of the following: a) tan(arcsec 4) The domain of arcsec θ , {x | |x| 1}, includes 4; 0 < arcsec(4) < π /2 where the tangent is positive. Using the identity tan 2 θ + 1 = sec 2 θ , we have tan θ = 15 1 4 1 ) 4 sec( sec 1 sec 2 2 2 = = = arc θ b) sin(2 arcsin (3/5)) The domain of arcsin θ , [–1, 1], contains 3/5, and 0 < arcsec(4) < π /2, where the cosine is positive. Thus sin(2 arcsin (3/5)) = 2 sin(arcsin (3/5)) cos(arcsin (3/5))

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25 24 5 4 5 6 ) 5 3 ( 1 5 6 )) 5 3 (arcsin( sin 1 5 3 2 2 2 = = = = Section 2.3: 10. a) What is wrong with saying 3 2 6 2 + = + x x x x ? The left-hand expression is not defined for x = 2, while the right-hand one is.
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## This note was uploaded on 02/21/2010 for the course MATH 137 taught by Professor Speziale during the Spring '08 term at Waterloo.

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Math_137_Winter_2010_Solution_3 - Math 137 Winter 2010...

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