Consider the following problem:The depth of the water in Harsha's bird bath is changing at a rate of r(t)=0.25t−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+∫23r(t)dt(B) 35+∫33r(t)dt(C) 35+∫34r(t)dt(D) 35+∫03r(t)dt
Q. Consider the following problem:The depth of the water in Harsha's bird bath is changing at a rate of r(t)=0.25t−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+∫23r(t)dt(B) 35+∫33r(t)dt(C) 35+∫34r(t)dt(D) 35+∫03r(t)dt
Define Rate of Change: To find the depth of the water at t=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)=0.25t−0.1 millimeters per hour. We need to integrate this rate from the starting time to t=3 hours to find the total change in depth.
Calculate Initial Depth: The initial depth of the water at t=0 is given as 35 millimeters. To find the depth at t=3 hours, we add the initial depth to the integral of the rate of change from t=0 to t=3.
Integrate Rate of Change: The correct expression to calculate the depth at t=3 hours is the initial depth plus the integral of the rate of change from the start time (t=0) to t=3. This is represented by the expression 35+∫03r(t)dt.
Choose Correct Expression: Looking at the given choices, the expression that matches our description is (D) 35+∫03r(t)dt. This is because it starts at t=0, which is when the initial depth is given, and integrates up to t=3, which is the time we want to find the depth for.
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