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Consider the following problem:
The temperature of a soup is increasing at a rate of 
r(t)=30e^(-0.3 t) degrees Celsius per minute (where 
t is the time in minutes). At time 
t=0, the temperature of the soup is 21 degrees Celsius. By how much does the temperature increase between 
t=0 and 
t=8 minutes?
Which expression can we use to solve the problem?
Choose 1 answer:
(A) 
r(8)-r(0)
(B) 
int_(0)^(8)r(t)dt
(C) 
int_(8)^(8)r(t)dt
(D) 
r^(')(8)-r^(')(0)

Consider the following problem:\newlineThe temperature of a soup is increasing at a rate of r(t)=30e0.3t r(t)=30 e^{-0.3 t} degrees Celsius per minute (where t t is the time in minutes). At time t=0 t=0 , the temperature of the soup is 2121 degrees Celsius. By how much does the temperature increase between t=0 t=0 and t=8 t=8 minutes?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) r(8)r(0) r(8)-r(0) \newline(B) 08r(t)dt \int_{0}^{8} r(t) d t \newline(C) 88r(t)dt \int_{8}^{8} r(t) d t \newline(D) r(8)r(0) r^{\prime}(8)-r^{\prime}(0)

Full solution

Q. Consider the following problem:\newlineThe temperature of a soup is increasing at a rate of r(t)=30e0.3t r(t)=30 e^{-0.3 t} degrees Celsius per minute (where t t is the time in minutes). At time t=0 t=0 , the temperature of the soup is 2121 degrees Celsius. By how much does the temperature increase between t=0 t=0 and t=8 t=8 minutes?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) r(8)r(0) r(8)-r(0) \newline(B) 08r(t)dt \int_{0}^{8} r(t) d t \newline(C) 88r(t)dt \int_{8}^{8} r(t) d t \newline(D) r(8)r(0) r^{\prime}(8)-r^{\prime}(0)
  1. Rate of Temperature Change Function: To find the total increase in temperature over a period of time, we need to integrate the rate of temperature change over that time interval. The rate of temperature change is given by the function r(t)=30e0.3tr(t) = 30e^{-0.3t}. We are interested in the interval from t=0t=0 to t=8t=8 minutes.
  2. Calculate Total Increase: The correct expression to calculate the total increase in temperature from t=0t=0 to t=8t=8 is the integral of r(t)r(t) from 00 to 88. This is because integration will give us the total accumulation of the temperature change over the time interval.
  3. Mathematical Representation: The integral of r(t)r(t) from 00 to 88 is represented mathematically as 08r(t)dt\int_{0}^{8} r(t) \, dt. This corresponds to choice (B) 08r(t)dt\int_{0}^{8}r(t)\,dt.
  4. Calculate Integral: We can now calculate the integral to find the total temperature increase. The integral of 30e0.3t30e^{-0.3t} from 00 to 88 is:\newline0830e0.3tdt=[100e0.3t] (from 0 to 8)\int_{0}^{8} 30e^{-0.3t} dt = [-100e^{-0.3t}] \text{ (from } 0 \text{ to } 8)
  5. Evaluate Integral: Evaluating the integral at the bounds gives us:\newline\(-100e^{(0-0.33\times 88)} - (100-100e^{(0-0.33\times 00)}) = 100-100e^{(2-2.44)} - (100-100e^{(00)})
  6. Simplify Expression: Simplifying the expression gives us:\newline100/e2.4(100)=100100/e2.4-100/e^{2.4} - (-100) = 100 - 100/e^{2.4}
  7. Calculate Value of e2.4e^{2.4}: Using a calculator to find the value of e2.4e^{2.4}, we get approximately:\newline100100/11.023176381009.076923100 - 100/11.02317638 \approx 100 - 9.076923
  8. Subtract Values: Subtracting the two values gives us the total temperature increase: 1009.07692390.923077100 - 9.076923 \approx 90.923077

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