document.write( "Question 1149301: Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 53 degrees occurs at 4 PM and the average temperature for the day is 45 degrees. Find the temperature, to the nearest degree, at 8 AM. \n" ); document.write( "
Algebra.Com's Answer #770688 by greenestamps(13203)\"\" \"About 
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\n" ); document.write( "We have an average temperature of 45 degrees and a high of 53 degrees. That makes the midline 45 and the amplitude 8.

\n" ); document.write( "We could spend some time trying to figure out how to get a graph in which x=0 corresponds to midnight. However; that is not necessary. We can use any sinusoidal function we want, as long as we understand which part of the graph corresponds to what hour of the day.

\n" ); document.write( "So with midline 45 and amplitude 8, let's choose a simple function:

\n" ); document.write( "\"f%28x%29+=+45%2B8sin%28x%29\"

\n" ); document.write( "A graph (showing 1.5 periods, from -pi to 2pi)....

\n" ); document.write( "\"graph%28400%2C200%2C-pi%2C2pi%2C36%2C54%2C45%2B8sin%28x%29%29\"

\n" ); document.write( "The time at which we are to find the temperature, 8 AM, is 8 hours before 4 PM, which is the time when the temperature is maximum. 8 hours is 1/3 of a day, so we want to find the temperature 1/3 of a cycle before 4 PM.

\n" ); document.write( "Since we are using a sine function with no horizontal shift, the time of maximum temperature, 4 PM, corresponds to 1/4 of the way through a cycle, at pi/2.

\n" ); document.write( "We want to go back 1/3 of a cycle, or (2/3)pi, from pi/2; that puts us at -pi/6.

\n" ); document.write( "That makes it easy to find the temperature at 4 AM without a calculator:

\n" ); document.write( "\"45%2B8%2Asin%28-pi%2F6%29+=+45%2B8%28-.5%29+=+45-4+=+41\"

\n" ); document.write( "The temperature at 4 AM is 41 degrees.
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