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Math Help - Problem Set

  1. #1
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    Problem Set

    The Variation in body temperature is an example of a circadian rhythm, a biological process that repeats itself approximately every 24 hours. Body temperature is highest about 5 p.m. and lowest at 5 a.m. Let y denote the body temperature (in F degrees) and let t=0 correspond to midnight. If the low and high body temperature are 98.3 and 98.9 respectively (normal is 98.6) find an equation of the for y=98.6+asin(bt+c)

    How would i do a problem like this? THanks for the help
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  2. #2
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    Quote Originally Posted by aussiekid90
    The Variation in body temperature is an example of a circadian rhythm, a biological process that repeats itself approximately every 24 hours. Body temperature is highest about 5 p.m. and lowest at 5 a.m. Let y denote the body temperature (in F degrees) and let t=0 correspond to midnight. If the low and high body temperature are 98.3 and 98.9 respectively (normal is 98.6) find an equation of the for y=98.6+asin(bt+c)

    How would i do a problem like this? THanks for the help
    The difference between the high and low is .6, now the amplitude is the half distance between the max and min. Thus, the amplidute a=.3. Notice that at 5 pm which is t=17 we have a minimum. Thus, \sin(bt+c)=-1 because that is the smallest it every gets. Thus, 17b+c=\frac{3\pi}{2} and the maximum is at 5 am which is t=5 thus, \sin(bt+c)=1 because that is the maximum sine ever gets. Thus, 5b+c=\frac{\pi}{2}. Now we have two systems,
    \left\{ \begin{array}{cc}17b+c&=3\pi/2\\5b+c&=\pi/2\end{array} \right
    Solving them we get, (b,c)=(\pi/12,\pi/12)
    Thus, y=98.6+.3\sin \left( \frac{\pi}{12}t+\frac{\pi}{12} \right)
    Q.E.D.
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  3. #3
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    Quote Originally Posted by aussiekid90
    The Variation in body temperature is an example of a circadian rhythm, a biological process that repeats itself approximately every 24 hours. Body temperature is highest about 5 p.m. and lowest at 5 a.m. Let y denote the body temperature (in F degrees) and let t=0 correspond to midnight. If the low and high body temperature are 98.3 and 98.9 respectively (normal is 98.6) find an equation of the for y=98.6+asin(bt+c)

    How would i do a problem like this? THanks for the help
    So the cyrcadian rhythm of the body temperature follows a sine curve
    y = 98.6 +A*sin(Bt +C) ----------(i)

    Given:
    ---y-axis is vertical, while t-axis is horizontal.
    ---t=0 at midnight or 0000 h0ur.
    ---period or one cycle is 24 hours.
    ---minimum y, (y=98.3), is at 5AM or 0500 hr.
    ---maximum y, (y=98.9), is at 5PM or 1700 hr.
    ---mean or normal y is 98.6 ----that sure follows a sine curve because it is midway between max and min y's.

    I always use the zero-value of the basic sine curve to find the equation of the sine curve. The basic sine curve here is that one based on the y=98.6 horizontal line. The basic sine curve was shifted vertically up from y=0 to y=98.6. That is why the Eq.(i) shows a 98.6 vertical shift.

    The A or amplitude of the sine curve is the vertical absolute distance of the maximum or minimum points from the shifted horizontal axis, which is the y=98.6 horizontal axis.
    So, A = 98.6 -98.3 = 0.3, or,
    A = 98.9 -98.6 = 0.3 ----------------***

    The B is the frequency of the cycle, or, using degrees as the units of the angles,
    period = 360/B ----***
    So, since the period is given as 24hrs,
    24 = 360/B
    B = 360/24 = 60/4 = 15 -----***

    The C represents the horizontal shift, (or phase-shift), that is done on the basic sine curve.
    Initially, the unshifted sine curve starts at y=0 when t=0, [y = sin(0) = 0]. Now the y=0 is not at the t=0 anymore---it was shifted to midway the t=5 and t=17.
    In any sine curve, the curve passes or intersects the horizontal axis at midway the highest and lowest points of the curve.
    So, the y=0 here is at (5+17)/2 = 22/2 = 11 AM.
    This t=11 is to the right of the (0,0) origin, so in the equation, this t=11 will have a negative sign.
    The C is not -11. The C carries the B, or the C is a multiple of the B, so,
    C = -11*15 = -165 -------------***

    Therefore, the equation is
    y = 98.6 +A*sin(Bt +C) ------(i)
    y = 98.6 +(0.3)sin(15t -165) --------------answer.

    Check when t=5 and y=98.3,
    98.3 =? 98.6 +(0.3)sin(15*5 -165)
    98.3 =? 98.6 +(0.3)sin(-90deg)
    98.3 =? 98.6 +(0.3)(-1.000)
    98.3 =? 98.3
    Yes, so, OK.

    Check when t=17 and y=98.9,
    98.9 =? 98.6 +(0.3)sin(15*17 -165)
    98.9 =? 98.6 +(0.3)sin(90deg)
    98.9 =? 98.6 +(0.3)(1.000)
    98.9 =? 98.9
    Yes, so, OK.
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  4. #4
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    Quote Originally Posted by Rich B.
    aussiekid90:

    Note that solutions found by ThePerfectHacker and ticbol are equivalent. The latter solution, i.e., y = 98.6 +(0.3)sin(15t -165), simplifies to y = 98.6 +(0.3)sin(15t + 15), as sin(u)=sin(u+180) for all real u. The argument (15t + 15) being expressed in degrees converts to ((pi/12)(t) + pi/12) in radian measure (15 = pi/12 radians). Hence the equation is equivalent to y = 98.6 +(0.3)sin((pi/12)(t) + pi/12), which is identical to the solution presented by P.Hckr..

    One need only remember to set one's calculator to radian mode if evaluating sin[(pi/12)(t)+pi/12], and to degree mode when working sin(15t+15).

    I just didn't want you to go nuts trying to figure out which of the two presentations is correct and, consequently, the one you should focus upon. They are both correct.

    Regards,

    Rich B.
    I don't want to comment on anybody's answer, but I cannot help it here.

    How can my y = 98.6 +(0.3)sin(15t -165) be the same as y = 98.6 +(0.3)sin(15t +15)?

    Let us check the latter when t=5 and y=98.3,
    98.3 =? 98.6 +(0.3)sin(15*5 +15)
    98.3 =? 98.6 +(0.3)sin(90deg)
    98.3 =? 98.6 +(0.3)(1.000)
    98.3 =? 98.9
    No.

    So, how can they be the same?
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  5. #5
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    Dear Ticbol:

    You are absolutely right. I don't know what I was thinking -- certainly not correctly.

    Enjoy the day,

    Rich B.
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