# A at what interest rate must the gic be earning if

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(a) At what interest rate must the GIC be earning if your investment will double in value after 30 years? State your answer as a percent to 2 decimal places. A ( t ) = A 0 e rt ) A = 12000 e rt Now, A (30) = 24000 ) 12000 e 30 r = 24000 ) e 30 r = 2 ) 30 r = ln 2 ) r = 1 30 ln 2 : = 0 : 0231 = 2 : 31% (b) To 2 decimal places, determine the value of your investment on November 1st, 2016 and the rate at which it is changing. Include units in your answer. A ( t ) = 12000 e 0 : 0231 t ) A 0 ( t ) = 12000(0 : 0231 e 0 : 0231 t ) = 277 : 20 e 0 : 0231 t ) On November 1st, 2016 ( 15 years after initial inversment), its value will be A (15) = 12000 e 0 : 0231(15) : = \$16969 : 31 and it will be changing at a rate of A 0 (15) = 277 : 20 e 0 : 0231(15) : = \$391 : 99 per year. 2. [4 marks ] In baseball, a batter stands 2 feet from home plate as a pitcher throws the ball towards home plate with a velocity of 150 ft/s (or 164 km/h). Suppose: i) a ( t ) represents the batter±s "angle of gaze" (in radians) between the ball and home plate at time t and ii) b ( t ) represents the distance (in ft) between the ball and home plate at time t (see diagram below). Determine the rate at which the batter±s "angle of gaze" must change (i.e. a 0 ( t ) ) in order to "keep his eyes on the ball" as it crosses home plate . tan( a ( t )) = b ( t ) 2 ) sec 2 ( a ( t )) ° a 0 ( t ) = b 0 ( t ) 2 ) a 0 ( t ) = b 0 ( t ) 2 sec 2 ( a ( t )) As the ball crosses home plate, b 0 ( t ) = 150 ft/s and a ( t ) = 0 , so a 0 ( t ) = 150 2 sec 2 (0) = 75 rads/sec. OR a ( t ) = tan ° 1 ° b ( t ) 2 ± ) a 0 ( t ) = 1 1 + ² b ( t ) 2 ³ 2 ° ° 1 2 b 0 ( t ) ± = 2 b 0 ( t ) 4 + ( b ( t )) 2 As the ball crosses home plate, b 0 ( t ) = 150 ft/s and b ( t ) = 0 , so a 0 ( t ) = 2(150) 4 = 75 rads/sec.
3. [4 marks ] During the winter months, humans do not actually feel the temperature of the air directly. What we feel is the temperature of our skin which is determined not only by the temperature of the air (in o C ) but also the speed x (in km=h ) at which the air is moving. This e/ect is called wind chill T . When the air temperature is ± 10 o C and the air speed x ² 5 km=h , Environment Canada has found that the rate of change of wind chill is T 0 ( x ) = 0 : 16 T ( x ) ± 1 : 10480 x : (a) Given that T (15) = ± 16 : 75 o C (i.e. the wind chill is ± 16 : 75 o C when the air speed is 15 km=h and the air temperature is ± 10 o C ) ; °nd the linear approximation L ( x ) of T ( x ) at a = 15 km=h: [Note: Coe¢ cients can be rounded to 3 decimals.] L ( x ) = T (15) + T 0 (15)( x ± 15) = T (15) + 0 : 16 T (15) ± 1 : 10480 15 ( x ± 15) = ± 16 : 75 + 0 : 16( ± 16 : 75) ± 1 : 10480 15 ( x ± 15) = ± 0 : 252 x ± 12 : 965 (b) Use the result in part (a) to approximate wind chill (to 2 decimals) when the air speed is x = 20 km=h: Compare the result to the actual air temperature of ± 10 o C . T (20) ³ L (20) = ± 0 : 252(20) ± 12 : 965 = ± 18 : 00 o C When the air speed is 20 km=h and the air temperature is "only" ± 10 o C; the wind chill is approximately ± 18 o C (i.e. it feels 8 o C colder). 4. [7 marks ] Open the °le named lab5mapletemplate.mw (found in MyLearningSpace) in Maple. Use the template to complete the following exercise. Consider the function f ( x ) = 49 ln ° 28 ± 3 x x ± . (a) De°ne f in the Maple template. (b) Determine each of the following approximations of f about a = 7 . [Note: In Maple, f ( a ) ; f 0 ( a ) , and f 00 ( a ) are evaluated by entering f(a) , D(f)(a) , and D(D(f))(a) , respectively.] i. Linear Approximation.: L ( x ) = ± 28 x + 196 ii. Quadratic Approximation: Q ( x ) = ± 4 x 2 + 28 x (c) Plot each of the approximations in part (b) along with f ( x ) for x 2 [5 ; 9] : plot([f(x),L(x),Q(x)],x=5..9,colour=[black,red,green]) (d) Use each approximation in part (b) to estimate the value of f (6
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