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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) r^(')(13)+62 (B) r^(')(15) (C) 62+int_(13)^(15)r(t)dt (D) 62+int_(13)^(15)r^(')(t)dt

Consider the following problem:\newlineThe 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)=0.6sin⁡(πt6) r(t)=0.6 \sin \left(\frac{\pi t}{6}\right) meters per hour (where t t is the time in hours). At time t=13 t=13 , the air gap is 6262 meters. What is the air gap when t=15 t=15 hours?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) r′(13)+62 r^{\prime}(13)+62 \newline(B) r′(15) r^{\prime}(15) \newline(C) 62+∫1315r(t)dt 62+\int_{13}^{15} r(t) d t \newline(D) 62+∫1315r′(t)dt 62+\int_{13}^{15} r^{\prime}(t) d t

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Q. Consider the following problem:\newlineThe 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)=0.6sin⁡(πt6) r(t)=0.6 \sin \left(\frac{\pi t}{6}\right) meters per hour (where t t is the time in hours). At time t=13 t=13 , the air gap is 6262 meters. What is the air gap when t=15 t=15 hours?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) r′(13)+62 r^{\prime}(13)+62 \newline(B) r′(15) r^{\prime}(15) \newline(C) 62+∫1315r(t)dt 62+\int_{13}^{15} r(t) d t \newline(D) 62+∫1315r′(t)dt 62+\int_{13}^{15} r^{\prime}(t) d t
  1. Understand the Problem: Understand the problem.\newlineWe are given the rate of change of the air gap as a function of time, r(t)r(t), and the air gap at a specific time, t=13t=13 hours. We need to find the air gap at a later time, t=15t=15 hours.
  2. Determine Approach: Determine the correct approach to find the air gap at t=15t=15 hours.\newlineSince we are given the rate of change of the air gap, we need to integrate this rate from t=13t=13 to t=15t=15 to find the total change in the air gap over this time period. Then we add this change to the air gap at t=13t=13 to find the air gap at t=15t=15.
  3. Identify Expression: Identify the correct expression to use.\newlineThe correct expression to use is the integral of the rate of change from t=13t=13 to t=15t=15, added to the air gap at t=13t=13. This corresponds to option (C) 62+∫1315r(t) dt62 + \int_{13}^{15}r(t)\,dt.
  4. Calculate Integral: Calculate the integral of the rate of change from t=13t=13 to t=15t=15. We need to integrate r(t)=0.6sin⁡(πt6)r(t) = 0.6 \sin\left(\frac{\pi t}{6}\right) from t=13t=13 to t=15t=15. ∫1315r(t) dt=∫13150.6sin⁡(πt6) dt\int_{13}^{15}r(t)\,dt = \int_{13}^{15}0.6 \sin\left(\frac{\pi t}{6}\right) \,dt Let's calculate this integral.
  5. Perform Integration: Perform the integration.\newlineTo integrate 0.6sin⁡(πt6)0.6 \sin\left(\frac{\pi t}{6}\right), we use the substitution method. Let u=πt6u = \frac{\pi t}{6}, then du=π6dtdu = \frac{\pi}{6} dt, and dt=6πdudt = \frac{6}{\pi} du.\newlineThe limits of integration change accordingly: when t=13t=13, u=π⋅136u=\frac{\pi\cdot13}{6}; when t=15t=15, u=π⋅156u=\frac{\pi\cdot15}{6}.\newlineNow we integrate with respect to uu:\newline∫0.6sin⁡(u)⋅6πdu\int 0.6 \sin(u) \cdot \frac{6}{\pi} du from u=π⋅136u=\frac{\pi\cdot13}{6} to u=π⋅156u=\frac{\pi\cdot15}{6}.
  6. Calculate New Integral: Calculate the new integral with the substitution.\newlineThe integral becomes:\newline(0.6×6π)∫sin⁡(u) du(0.6 \times \frac{6}{\pi}) \int \sin(u) \, du from u=π×136u=\frac{\pi\times13}{6} to u=π×156u=\frac{\pi\times15}{6}.\newlineThis simplifies to:\newline(3.6π)×[−cos⁡(u)](\frac{3.6}{\pi}) \times [-\cos(u)] from u=π×136u=\frac{\pi\times13}{6} to u=π×156u=\frac{\pi\times15}{6}.
  7. Evaluate Integral: Evaluate the integral at the new limits.\newlineWe substitute the limits into the antiderivative:\newline(3.6π)∗[−cos⁡(π∗156)+cos⁡(π∗136)](\frac{3.6}{\pi}) * [-\cos(\frac{\pi*15}{6}) + \cos(\frac{\pi*13}{6})].
  8. Simplify Expression: Simplify the expression and calculate the values.\newlineWe calculate the cosine values and simplify the expression:\newline(3.6/π)⋅[−cos⁡((5π)/2)+cos⁡((13π)/6)](3.6/\pi) \cdot [-\cos((5\pi)/2) + \cos((13\pi)/6)].\newlineNote that cos⁡((5π)/2)=0\cos((5\pi)/2) = 0 and cos⁡((13π)/6)\cos((13\pi)/6) is equivalent to cos⁡(π/6)\cos(\pi/6) due to periodicity of the cosine function.\newlineSo, the expression simplifies to:\newline(3.6/π)⋅[0+3/2](3.6/\pi) \cdot [0 + \sqrt{3}/2].
  9. Multiply by Coefficient: Multiply the result by the coefficient.\newline(3.6π)×(32)=3.6×32π(\frac{3.6}{\pi}) \times (\frac{\sqrt{3}}{2}) = \frac{3.6 \times \sqrt{3}}{2\pi}.
  10. Add to Air Gap: Add the result to the air gap at t=13t=13 to find the air gap at t=15t=15. The change in air gap from t=13t=13 to t=15t=15 is 3.6×32π\frac{3.6 \times \sqrt{3}}{2\pi} meters. We add this to the air gap at t=13t=13, which is 6262 meters. Air gap at t=15t=15 = 62+3.6×32π62 + \frac{3.6 \times \sqrt{3}}{2\pi}.
  11. Calculate Final Value: Calculate the final value.\newlineWe calculate the numerical value of the expression:\newlineAir gap at t=15=62+(3.6×3)2π≈62+(3.6×1.732)(2×3.14159)≈62+(6.2352)(6.28318)≈62+0.992≈62.992t=15 = 62 + \frac{(3.6 \times \sqrt{3})}{2\pi} \approx 62 + \frac{(3.6 \times 1.732)}{(2 \times 3.14159)} \approx 62 + \frac{(6.2352)}{(6.28318)} \approx 62 + 0.992 \approx 62.992 meters.

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