Consider the following problem:The temperature of a soup is increasing at a rate of r(t)=30e−0.3t 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) ∫08r(t)dt(C) ∫88r(t)dt(D) r′(8)−r′(0)
Q. Consider the following problem:The temperature of a soup is increasing at a rate of r(t)=30e−0.3t 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) ∫08r(t)dt(C) ∫88r(t)dt(D) r′(8)−r′(0)
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)=30e−0.3t. We are interested in the interval from t=0 to t=8 minutes.
Calculate Total Increase: The correct expression to calculate the total increase in temperature from t=0 to t=8 is the integral of r(t) from 0 to 8. This is because integration will give us the total accumulation of the temperature change over the time interval.
Mathematical Representation: The integral of r(t) from 0 to 8 is represented mathematically as ∫08r(t)dt. This corresponds to choice (B) ∫08r(t)dt.
Calculate Integral: We can now calculate the integral to find the total temperature increase. The integral of 30e−0.3t from 0 to 8 is:∫0830e−0.3tdt=[−100e−0.3t] (from 0 to 8)
Evaluate Integral: Evaluating the integral at the bounds gives us:\(-100e^{(−0.3\times 8)} - (−100e^{(−0.3\times 0)}) = −100e^{(−2.4)} - (−100e^{(0)})
Simplify Expression: Simplifying the expression gives us:−100/e2.4−(−100)=100−100/e2.4
Calculate Value of e2.4: Using a calculator to find the value of e2.4, we get approximately:100−100/11.02317638≈100−9.076923
Subtract Values: Subtracting the two values gives us the total temperature increase: 100−9.076923≈90.923077
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