document.write( "Question 170913: A person’s blood pressure, P(t), in millimeters of mercury (mm Hg), is modeled by the function P(t)=100=20cos(8π/3t), where t is the time in seconds.\r
\n" ); document.write( "\n" ); document.write( "a)What is the period of the function?
\n" ); document.write( "b)What does the value of the period mean in this situation?
\n" ); document.write( "c)Calculate the average rate of change in a person’s blood pressure on the interval [0.2,0.3]
\n" ); document.write( "d) Estimate the instantaneous rate of change in a person’s blood pressure at t=0.5.
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Algebra.Com's Answer #126181 by stanbon(75887)\"\" \"About 
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A person’s blood pressure, P(t), in millimeters of mercury (mm Hg), is modeled by the function P(t)=100+20cos(8π/3t), where t is the time in seconds.
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\n" ); document.write( "Form: y = acos(bx+c)+d
\n" ); document.write( "Period = b/(2pi)
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\n" ); document.write( "a)What is the period of the function?
\n" ); document.write( "Period = (8pi/3)/2pi = (4/3)pi
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\n" ); document.write( "b)What does the value of the period mean in this situation?
\n" ); document.write( "The time between highest pressure and the next occurrence of highest
\n" ); document.write( "pressure is (4/3)pi seconds
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\n" ); document.write( "c)Calculate the average rate of change in a person’s blood pressure on the interval [0.2,0.3]
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\n" ); document.write( "P(t)=100+20cos(8π/3t)
\n" ); document.write( "Average = [P(0.3)-P(0.2)]/[0.3-0.2]
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\n" ); document.write( "d) Estimate the instantaneous rate of change in a person’s blood pressure at t=0.5.
\n" ); document.write( "P(t)=100+20cos(8π/3t)
\n" ); document.write( "The instataneous rate of change at any time \"t\" is the derivative of P(t).
\n" ); document.write( "P'(t) = -20sin(8pi/3t)*(8pi/3)lnt
\n" ); document.write( "P'(0.5) = -20sin(16pi/3)*(8pi/3)ln(0.5)
\n" ); document.write( "etc.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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