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Joan brings a cup of hot tea into a room and places the cup on a table - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 4
Question 5
Joan brings a cup of hot tea into a room and places the cup on a table. At time t minutes after Joan places the cup on the table, the temperature, θ°C, of the tea is... show full transcript
Worked Solution & Example Answer:Joan brings a cup of hot tea into a room and places the cup on a table - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 4
Step 1
Step 2
(b) Show that k = \frac{1}{5} \ln 2.
Answer
For this part, we substitute ( t = 5 ) and ( \theta = 55 )°C into the equation:
[ \theta = 20 + 70e^{-5k} ] [ 55 = 20 + 70e^{-5k} ] [ 35 = 70e^{-5k} ] [ e^{-5k} = \frac{35}{70} = \frac{1}{2} ]
Taking the natural logarithm:
[ -5k = \ln\left(\frac{1}{2}\right) = -\ln(2) ] [ k = \frac{1}{5} \ln(2) ]
Therefore, we have shown that ( k = \frac{1}{5} \ln 2 ).
Step 3
(c) Find the rate at which the temperature of the tea is decreasing at the instant when t = 10.
Answer
To find the rate of change of temperature, we first differentiate the equation:
[ \theta = 20 + 70e^{-kt} ] [ \frac{d\theta}{dt} = -70ke^{-kt} ]
Now we substitute ( k = \frac{1}{5} \ln 2 ) into the derivative:
[ \frac{d\theta}{dt} = -70 \cdot \frac{1}{5} \ln(2) e^{-\frac{1}{5}\ln(2) \cdot 10} ] [ = -14 \ln(2) e^{-2\ln(2)} ] [ = -14 \ln(2) \cdot \left( \frac{1}{2} \right)^2 ] [ = -14 \ln(2) \cdot \frac{1}{4} ] [ = -\frac{14}{4} \ln(2) \approx -2.426 \text{°C/min} ]
Thus, the rate of decrease of ( \theta ) at ( t = 10 ) minutes is approximately ( -2.426 ) °C per minute.