Question

q11. the body temperature of a patient with a nine - day infection is approximated by

$t(t)=100^{circ}+1.4sin(30^{circ}t)$ where t is in days.

a. what is the temperature of this patient at the beginning of their illness?
b. what is their temperature after 3 days? after 6 days? after 9 days?
c. graph one cycle of the function t.

Explanation:

Part (a)

Step1: Identify \( t \) at start

At the beginning of illness, \( t = 0 \) days.

Step2: Substitute \( t = 0 \) into \( T(t) \)

\( T(0)=100^{\circ}+1.4\sin(30^{\circ}\times0) \)
Since \( \sin(0)=0 \), we have \( T(0)=100^{\circ}+1.4\times0 = 100^{\circ} \)

Part (b)

Step1: Temperature after 3 days (\( t = 3 \))

Substitute \( t = 3 \) into \( T(t) \):
\( T(3)=100^{\circ}+1.4\sin(30^{\circ}\times3) \)
\( 30^{\circ}\times3 = 90^{\circ} \), and \( \sin(90^{\circ}) = 1 \)
So \( T(3)=100^{\circ}+1.4\times1 = 101.4^{\circ} \)

Step2: Temperature after 6 days (\( t = 6 \))

Substitute \( t = 6 \) into \( T(t) \):
\( T(6)=100^{\circ}+1.4\sin(30^{\circ}\times6) \)
\( 30^{\circ}\times6 = 180^{\circ} \), and \( \sin(180^{\circ}) = 0 \)
So \( T(6)=100^{\circ}+1.4\times0 = 100^{\circ} \)

Step3: Temperature after 9 days (\( t = 9 \))

Substitute \( t = 9 \) into \( T(t) \):
\( T(9)=100^{\circ}+1.4\sin(30^{\circ}\times9) \)
\( 30^{\circ}\times9 = 270^{\circ} \), and \( \sin(270^{\circ})=- 1 \)
So \( T(9)=100^{\circ}+1.4\times(-1)=98.6^{\circ} \)

Part (c)

To graph \( T(t)=100^{\circ}+1.4\sin(30^{\circ}t) \), we first analyze the sine function:

• The general form of a sine function is \( y = A\sin(Bx)+C \), where:
• Amplitude \( A = 1.4 \)
• Period \( P=\frac{360^{\circ}}{|B|}=\frac{360^{\circ}}{30^{\circ}} = 12 \) days (but since the infection is 9 - day, we can consider one cycle or part of it)
• Vertical shift \( C = 100^{\circ} \)

1. Key Points for Graphing:

• When \( t = 0 \), \( T(0)=100^{\circ} \) (from part a)
• When \( t = 3 \), \( T(3)=101.4^{\circ} \) (from part b)
• When \( t = 6 \), \( T(6)=100^{\circ} \) (from part b)
• When \( t = 9 \), \( T(9)=98.6^{\circ} \) (from part b)
• When \( t = 12 \), \( T(12)=100^{\circ}+1.4\sin(30^{\circ}\times12)=100^{\circ}+1.4\sin(360^{\circ}) = 100^{\circ} \) (completing one cycle)

1. Plotting the Graph:

• The vertical axis (y - axis) represents temperature \( T(t) \) (in degrees) and the horizontal axis (x - axis) represents time \( t \) (in days).
• Plot the points \((0,100^{\circ})\), \((3,101.4^{\circ})\), \((6,100^{\circ})\), \((9,98.6^{\circ})\), \((12,100^{\circ})\) and draw a smooth sine - curve connecting these points, centered around the horizontal line \( T = 100^{\circ} \) with amplitude 1.4.

Answer:

s:
a. \(\boldsymbol{100^{\circ}}\)

b. After 3 days: \(\boldsymbol{101.4^{\circ}}\); After 6 days: \(\boldsymbol{100^{\circ}}\); After 9 days: \(\boldsymbol{98.6^{\circ}}\)

c. (Graph description as above, with key points \((0,100)\), \((3,101.4)\), \((6,100)\), \((9,98.6)\), \((12,100)\) plotted and a sine - curve drawn through them, centered at \( y = 100 \) with amplitude 1.4)