Consider the following problem: The temperature of a cup of cocoa is decreasing at a rate of r(t)=-5.5e^(-0.09 t) degrees Celsius per minute (where t is the time in minutes). At time t=2, the temperature of the cocoa is 72 degrees Celsius. What is the temperature of the cocoa at t=10 minutes? Which expression can we use to solve the problem? Choose 1 answer: (A) int r(t)dt (B) int_(2)^(10)r(t)dt (C) int_(2)^(10)r(t)dt+72 (D) int r(t)dt+72

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Consider the following problem: The temperature of a cup of cocoa is decreasing at a rate of  r(t)=-5.5e^(-0.09 t) degrees Celsius per minute (where  t is the time in minutes). At time  t=2, the temperature of the cocoa is 72 degrees Celsius. What is the temperature of the cocoa at  t=10 minutes? Which expression can we use to solve the problem? Choose 1 answer: (A)  int r(t)dt (B)  int_(2)^(10)r(t)dt (C)  int_(2)^(10)r(t)dt+72 (D)  int r(t)dt+72

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Understand the problem: Understand the problem. We are given the rate of change of temperature of the cocoa as a function of time, r(t)=-5.5e^(-0.09 t), and the temperature at t=2 minutes, which is 72 degrees Celsius. We need to find the temperature at t=10 minutes.Determine expression: Determine the correct expression to use. To find the temperature at t=10 minutes, we need to integrate the rate of change from t=2 to t=10 and add the initial temperature at t=2. This is because the integral of the rate of change gives us the change in temperature, and adding the initial temperature gives us the final temperature.Choose correct expression: Choose the correct expression. The correct expression to use is the integral of the rate of change from t=2 to t=10 plus the initial temperature at t=2. This corresponds to choice (C) int_(2)^(10)r(t)dt+72.Calculate the integral: Calculate the integral. We need to calculate the integral of r(t) from t=2 to t=10 and then add 72 degrees Celsius to find the final temperature. The integral of r(t) from t=2 to t=10 is: int_(2)^(10)(-5.5e^(-0.09 t))dtPerform the integration: Perform the integration. To integrate r(t), we use the antiderivative of e^(-0.09 t), which is (-1/0.09)e^(-0.09 t). We then evaluate this from t=2 to t=10 and multiply by -5.5. So, the integral becomes: -5.5 * [(-1/0.09)e^(-0.09 t)] from t=2 to t=10Evaluate at bounds: Evaluate the integral at the bounds. First, evaluate the antiderivative at t=10: -5.5 * [(-1/0.09)e^(-0.09 * 10)] Then, evaluate the antiderivative at t=2: -5.5 * [(-1/0.09)e^(-0.09 * 2)]Subtract and add: Subtract the evaluations and add the initial temperature. Subtract the evaluation at t=2 from the evaluation at t=10 and then add 72 degrees Celsius to find the final temperature. Temperature at t=10 = -5.5 * [(-1/0.09)e^(-0.09 * 10) - (-1/0.09)e^(-0.09 * 2)] + 72Calculate final temperature: Calculate the final temperature. Now we perform the actual calculations: Temperature at t=10 = -5.5 * [(-1/0.09)(e^(-0.9) - e^(-0.18))] + 72Correct exponent and re-evaluate: Correct the calculation of the exponent and re-evaluate the integral. Correcting the exponents, we have: Temperature at t=10 = -5.5 * [(-1/0.09)(e^(-0.9) - e^(-0.18))] + 72 Now, let's calculate the values correctly: Temperature at t=10 = -5.5 * [(-1/0.09)(e^(-0.9) - e^(-0.18))] + 72Perform final calculation: Perform the final calculation. Now we calculate the numerical values: Temperature at t=10 = -5.5 * [(-1/0.09)(0.4066 - 0.8353)] + 72 Temperature at t=10 = -5.5 * [(-1/0.09)(-0.4287)] + 72 Temperature at t=10 = -5.5 * [47.6333] + 72 Temperature at t=10 = -261.98315 + 72 Temperature at t=10 = -189.98315 degrees CelsiusCorrect numerical values: Correct the numerical values and perform the correct arithmetic. Correcting the numerical values and performing the correct arithmetic, we have: Temperature at t=10 = -5.5 * [(-1/0.09)(e^(-0.9) - e^(-0.18))] + 72 First, calculate the values of the exponentials correctly: e^(-0.9) ≈ 0.4066 e^(-0.18) ≈ 0.8353 Now, calculate the change in temperature: Change in temperature = -5.5 * [(-1/0.09)(0.4066 - 0.8353)] Change in temperature = -5.5 * [(-1/0.09)(-0.4287)] Change in temperature = -5.5 * 47.6333 Change in temperature = -261.98315 Now, add the initial temperature: Temperature at t=10 = -261.98315 + 72 Temperature at t=10 = -189.98315 degrees Celsius This is still incorrect, as the temperature cannot be negative. Let's correct the arithmetic once more.Correct arithmetic: Correct the arithmetic and find the final temperature. Correcting the arithmetic, we have: Change in temperature = -5.5 * [(-1/0.09)(e^(-0.9) - e^(-0.18))] Change in temperature = -5.5 * [(-1/0.09)(-0.4287)] Change in temperature = -5.5 * 47.6333 Change in temperature = -261.98315 This value is the change in temperature from t=2 to t=10, not the final temperature. To find the final temperature, we need to add the initial temperature at t=2, which is 72 degrees Celsius: Temperature at t=10 = 72 - 261.98315 Temperature at t=10 = -189.98315 degrees Celsius This is still incorrect, as the temperature cannot be negative. The error lies in the sign of the change in temperature. Since the rate of change is negative, the temperature is decreasing, so we should be subtracting the change in temperature from the initial temperature, not adding it. Let's correct this and find the final temperature.Correct arithmetic and find: Correct the sign error and find the final temperature. Correcting the sign error, we have: Change in temperature = -5.5 * [(-1/0.09)(e^(-0.9) - e^(-0.18))] Change in temperature = -5.5 * [(-1/0.09)(-0.4287)] Change in temperature = -5.5 * 47.6333 Change in temperature = -261.98315 Now, subtract the change in temperature from the initial temperature at t=2, which is 72 degrees Celsius: Temperature at t=10 = 72 + 261.98315 Temperature at t=10 = 333.98315 degrees Celsius This is still incorrect, as the temperature is unrealistically high. The error lies in the calculation of the change in temperature. The correct calculation should be: Change in temperature = -5.5 * [(-1/0.09)(0.4066 - 0.8353)] Change in temperature = -5.5 * [(-1/0.09)(-0.4287)] Change in temperature = -5.5 * 4.76333 Change in temperature = -26.198315 Now, subtract the change in temperature from the initial temperature at t=2, which is 72 degrees Celsius: Temperature at t=10 = 72 - 26.198315 Temperature at t=10 = 45.801685 degrees Celsius This is the correct final temperature.

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