Consider the following problem: The air gap (the distance from the bottom of the bridge to the surface of the water) under a bridge is changing at a rate of r(t)=0.6 sin((pi t)/(6)) meters per hour (where t is the time in hours). At time t=13, the air gap is 62 meters. What is the air gap when t=15 hours? Which expression can we use to solve the problem? Choose 1 answer: (A) 62+int_(13)^(15)r^(')(t)dt (B) r^(')(15) (C) r^(')(13)+62 (D) 62+int_(13)^(15)r(t)dt

Question

Consider the following problem: The air gap (the distance from the bottom of the bridge to the surface of the water) under a bridge is changing at a rate of  r(t)=0.6 sin((pi t)/(6)) meters per hour (where  t is the time in hours). At time  t=13, the air gap is 62 meters. What is the air gap when  t=15 hours? Which expression can we use to solve the problem? Choose 1 answer: (A)  62+int_(13)^(15)r^(')(t)dt (B)  r^(')(15) (C)  r^(')(13)+62 (D)  62+int_(13)^(15)r(t)dt

Bytelearn · Accepted Answer

Understand and Determine Approach: Understand the problem and determine the correct approach. We are given the rate of change of the air gap as a function of time, r(t), and the air gap at a specific time, t=13 hours. To find the air gap at t=15 hours, we need to integrate the rate of change from t=13 to t=15 and add it to the air gap at t=13.Identify Correct Expression: Identify the correct expression to use. The correct expression to use is the one that adds the integral of the rate of change from t=13 to t=15 to the air gap at t=13. This is because the integral of a rate of change gives us the total change over the interval. The correct expression is therefore: 62 + ∫_(13)^(15) r(t) dt This corresponds to option (D).Calculate Integral from t=13 to t=15: Calculate the integral of the rate of change from t=13 to t=15. We need to integrate r(t) = 0.6 sin((pi t)/(6)) from t=13 to t=15. ∫_(13)^(15) 0.6 sin((pi t)/(6)) dtPerform Integration with Substitution: Perform the integration. To integrate 0.6 sin((pi t)/(6)), we use the substitution method. Let u = (pi t)/(6), then du = (pi/6) dt, and dt = (6/pi) du. The limits of integration also change accordingly: when t=13, u=(pi*13)/6; when t=15, u=(pi*15)/6. ∫ 0.6 sin(u) * (6/pi) duEvaluate Integral at New Limits: Calculate the new integral with the substitution. The integral becomes: (0.6 * 6/pi) ∫ sin(u) du from u=(pi*13)/6 to u=(pi*15)/6 This simplifies to: (3.6/pi) * [-cos(u)] from u=(pi*13)/6 to u=(pi*15)/6Simplify Expression: Evaluate the integral at the new limits. We substitute the limits into the antiderivative: (3.6/pi) * [-cos((pi*15)/6) + cos((pi*13)/6)]Calculate Cosine Value: Simplify the expression. We calculate the cosine values and multiply by the constant: (3.6/pi) * [-cos((5pi)/2) + cos((13pi)/6)] Note that cos((5pi)/2) = 0 and cos((13pi)/6) is a value we need to calculate.Complete Calculation: Calculate the cosine value and complete the calculation. cos((13pi)/6) is equivalent to cos((pi)/6) because cosine is periodic with period 2pi. Therefore, cos((13pi)/6) = cos((pi)/6) = sqrt(3)/2. Now we can complete the calculation: (3.6/pi) * [0 + sqrt(3)/2] This simplifies to: (3.6/pi) * (sqrt(3)/2)Calculate Numerical Value: Calculate the numerical value of the integral. (3.6/pi) * (sqrt(3)/2) ≈ (3.6/pi) * (0.866) ≈ 0.9956Add Result to Air Gap: Add the result of the integral to the air gap at t=13 to find the air gap at t=15. 62 + 0.9956 ≈ 62.9956 meters

Full solution

More problems from Find derivatives of sine and cosine functions

Related topics