Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2011)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2011

Part IV: A correct answer will receive 6 credits. Partial credit is available


39. The temperature, T, of a given cup of hot chocolate after it has been cooling for t minutes can best be modeled by the function below, where T₀ is the temperature of the room and k is a constant.

ln(T − T₀) = −kt + 4.718
A cup of hot chocolate is placed in a room that has a temperature of 68°. After 3 minutes, the temperature of the hot chocolate is 150°. Compute the value of k to the nearest thousandth. [Only an algebraic solution can receive full credit.]

Using this value of k, find the temperature, T, of this cup of hot chocolate if it has been sitting in this room for a total of 10 minutes. Express your answer to the nearest degree. [Only an algebraic solution can receive full credit.]

Answer:

For the first part, T is 150°, T₀ = 68°, and k = 3.

ln(T − T₀) = −kt + 4.718

ln(150 - 68) = −k(3) + 4.718

ln(82) = −3k + 4.718

4.4067 = −3k + 4.718

-0.3113 = −3k

0.1038 = k

k = 0.104

ln(T − T₀) = −kt + 4.718

ln(T − 68) = −(0.104)(10) + 4.718

ln(T − 68) = 3.678

T − 68 = e^3.678

T - 68 = 39.567

T = 107.567

T = 108 degrees

Note that if you got both numbers and showed no work, you received 1 credit. (One answer with no work is worth 0.)

If you solve for k, with work, and went no further, you received 2 credits, not 3.

If you used any method other than algebraic (such as graphing), you received half credit, which is 3 credits.

End of Exam.

More to come. Comments and questions welcome.

More Regents problems.

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