Explanation Given: • Resistance of a 100 − w a t t ( W ) bulb, R = 144 o h m s ( Ω ) • Voltage varies as V = V 0 ⋅ sin ( 2 ⋅ π ⋅ f ⋅ t ) • V 0 = 110 V • f = 60 H z • Energy (in joules) expended by the bulb over a period of T seconds is U = ∫ 0 T P ( t ) ⋅ d t • Power, P = V 2 R ( J / s ) • For 5 hours Key Concepts Used: Many trigonometric functions can be integrated by combining substitution and Integration by parts with the appropriate trigonometric identities. The Integration by Parts Formula: ∫ u ( x ) ⋅ v ' ( x ) ⋅ d x = u ( x ) ⋅ v ( x ) − ∫ u ' ( x ) ⋅ v ( x ) ⋅ d x From the above formula we come to know that it applies to a product u ( x ) ⋅ v ' ( x ) , we should consider using it when the integrand is a product of two functions. Important Trigonometric Identities Identity 2 :Half-angle Identity for sin 2 x sin 2 x = 1 − cos 2 x 2 Here in this problem, it is given that a 100 − w a t t ( W ) bulb has resistance R = 144 o h m s ( Ω ) when attached to household current, where the voltage varies as V = V 0 ⋅ sin ( 2 ⋅ π ⋅ f ⋅ t ) in which V 0 = 110 V , and f = 60 H z . The energy (in joules) expended by the bulb over a period of T seconds is U = ∫ 0 T P ( t ) ⋅ d t . Power is derived by the formula P = V 2 R ( J / s ) . We have to find the value of energy if the bulb remains of for 5 hours. We have to substitute the values of voltage and resistance in expression of power then we have to substitute the expression of power in the given integral to find the required energy. To evaluate this integral we initially use substitution process and trigonometric identity. Calculation: Description: ∫ u ( x ) ⋅ v ' ( x ) ⋅ d x = u ( x ) ⋅ v ( x ) − ∫ u ' ( x ) ⋅ v ( x ) ⋅ d x Step 1 : Write the given integral. U = ∫ 0 T P ( t ) ⋅ d t Step 2 : Substitute the value of V a n d R in the formula of power to find P ( t ) . P = V 2 R V = V 0 ⋅ sin ( 2 ⋅ π ⋅ f ⋅ t ) Here it is given that V 0 = 110 V and f = 60 H z . Substitute these values in the above equation of voltage, V = 110 ⋅ sin ( 2 ⋅ π ⋅ 60 ⋅ t ) = 110 sin 120 π t Substitute the above value of V and the given value of R = 144 in the equation of power, P = V 2 R = ( 110 sin 120 π t ) 2 144 = ( 110 ) 2 ⋅ sin 2 ( 120 π ⋅ t ) 144 P = ( 110 ) 2 144 ⋅ sin 2 ( 120 π ⋅ t ) Step 3 : Substitute the above obtained value of P and the given time T . P = ( 110 ) 2 144 ⋅ sin 2 ( 120 π ⋅ t ) T = 5 h = 5 × 3600 s = 18000 s Substitute these values in the given integral, U = ∫ 0 18000 ( 110 ) 2 144 ⋅ sin 2 ( 120 π ⋅ t ) ⋅ d t U = ( 110 ) 2 144 ∫ 0 18000 sin 2 ( 120 π ⋅ t ) ⋅ d t Step 4 : To find the value of the integral ∫ 0 18000 sin 2 ( 120 π ⋅ t ) ⋅ d t , initially we use substitution process, and then we use trigonometric identity...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.2, Problem 76E

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$U$ if bulb remains for the given time

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U if bulb remains for the given time

Solution Summary: The author explains that many trigonometric functions can be integrated by combining substitution and Integration by parts with the appropriate trigonemetric identities.

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