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Consider the following problem:
The depth of the water in Harsha's bird bath is changing at a rate of 
r(t)=0.25 t-0.1 millimeters per hour (where 
t is the time in hours). At time 
t=0, the depth of the water is 35 millimeters. What is the depth of the water at 
t=3 hours?
Which expression can we use to solve the problem?
Choose 1 answer:
(A) 
35+int_(2)^(3)r(t)dt
(B) 
35+int_(3)^(3)r(t)dt
(C) 
35+int_(3)^(4)r(t)dt
(D) 
35+int_(0)^(3)r(t)dt

Consider the following problem:\newlineThe depth of the water in Harsha's bird bath is changing at a rate of r(t)=0.25t0.1 r(t)=0.25 t-0.1 millimeters per hour (where t t is the time in hours). At time t=0 t=0 , the depth of the water is 3535 millimeters. What is the depth of the water at t=3 t=3 hours?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) 35+23r(t)dt 35+\int_{2}^{3} r(t) d t \newline(B) 35+33r(t)dt 35+\int_{3}^{3} r(t) d t \newline(C) 35+34r(t)dt 35+\int_{3}^{4} r(t) d t \newline(D) 35+03r(t)dt 35+\int_{0}^{3} r(t) d t

Full solution

Q. Consider the following problem:\newlineThe depth of the water in Harsha's bird bath is changing at a rate of r(t)=0.25t0.1 r(t)=0.25 t-0.1 millimeters per hour (where t t is the time in hours). At time t=0 t=0 , the depth of the water is 3535 millimeters. What is the depth of the water at t=3 t=3 hours?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) 35+23r(t)dt 35+\int_{2}^{3} r(t) d t \newline(B) 35+33r(t)dt 35+\int_{3}^{3} r(t) d t \newline(C) 35+34r(t)dt 35+\int_{3}^{4} r(t) d t \newline(D) 35+03r(t)dt 35+\int_{0}^{3} r(t) d t
  1. Define Rate of Change: To find the depth of the water at t=3t=3 hours, we need to account for the initial depth and the change in depth over time. The rate of change of the depth is given by r(t)=0.25t0.1r(t) = 0.25t - 0.1 millimeters per hour. We need to integrate this rate from the starting time to t=3t=3 hours to find the total change in depth.
  2. Calculate Initial Depth: The initial depth of the water at t=0t=0 is given as 3535 millimeters. To find the depth at t=3t=3 hours, we add the initial depth to the integral of the rate of change from t=0t=0 to t=3t=3.
  3. Integrate Rate of Change: The correct expression to calculate the depth at t=3t=3 hours is the initial depth plus the integral of the rate of change from the start time (t=0t=0) to t=3t=3. This is represented by the expression 35+03r(t)dt35 + \int_{0}^{3} r(t) \, dt.
  4. Choose Correct Expression: Looking at the given choices, the expression that matches our description is (D) 35+03r(t)dt35 + \int_{0}^{3} r(t) \, dt. This is because it starts at t=0t=0, which is when the initial depth is given, and integrates up to t=3t=3, which is the time we want to find the depth for.

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