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Consider the following problem: The water level under a bridge is changing at a rate of r(t)=40 sin((pi t)/(6)) centimeters per hour (where t is the time in hours). At time t=3, the water level is 91 centimeters. By how much does the water level change during the 4^("th ") hour? Which expression can we use to solve the problem? Choose 1 answer: (A) int_(0)^(4)r(t)dt (B) int_(4)^(5)r(t)dt (C) int_(3)^(4)r(t)dt (D) int_(4)^(4)r(t)dt

Consider the following problem:\newlineThe water level under a bridge is changing at a rate of r(t)=40sin⁡(πt6) r(t)=40 \sin \left(\frac{\pi t}{6}\right) centimeters per hour (where t t is the time in hours). At time t=3 t=3 , the water level is 9191 centimeters. By how much does the water level change during the 4th  4^{\text {th }} hour?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫04r(t)dt \int_{0}^{4} r(t) d t \newline(B) ∫45r(t)dt \int_{4}^{5} r(t) d t \newline(C) ∫34r(t)dt \int_{3}^{4} r(t) d t \newline(D) ∫44r(t)dt \int_{4}^{4} r(t) d t

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Full solution

Q. Consider the following problem:\newlineThe water level under a bridge is changing at a rate of r(t)=40sin⁡(πt6) r(t)=40 \sin \left(\frac{\pi t}{6}\right) centimeters per hour (where t t is the time in hours). At time t=3 t=3 , the water level is 9191 centimeters. By how much does the water level change during the 4th  4^{\text {th }} hour?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫04r(t)dt \int_{0}^{4} r(t) d t \newline(B) ∫45r(t)dt \int_{4}^{5} r(t) d t \newline(C) ∫34r(t)dt \int_{3}^{4} r(t) d t \newline(D) ∫44r(t)dt \int_{4}^{4} r(t) d t
  1. Define Time Interval: To find the change in water level during the 4th4^{\text{th}} hour, we need to evaluate the rate of change function r(t)r(t) from the start to the end of the 4th4^{\text{th}} hour. The 4th4^{\text{th}} hour starts at t=3t=3 and ends at t=4t=4.
  2. Use Integral Expression: The correct expression to use for this problem is the integral of r(t)r(t) from t=3t=3 to t=4t=4, which represents the total change in water level during the 4th4^{\text{th}} hour. This corresponds to option (C) ∫34r(t) dt\int_{3}^{4}r(t)\,dt.
  3. Calculate Integral: Now we will calculate the integral of r(t)r(t) from t=3t=3 to t=4t=4. ∫3440sin⁡(πt6)dt\int_{3}^{4}40 \sin\left(\frac{\pi t}{6}\right) dt
  4. Apply Antiderivative Rule: To integrate, we use the antiderivative of sin⁡(kx)\sin(kx), which is −1kcos⁡(kx)-\frac{1}{k} \cos(kx), where kk is a constant. In this case, k=π6k = \frac{\pi}{6}.∫3440sin⁡(πt6)dt=−40π/6[cos⁡(πt6)]\int_{3}^{4}40 \sin\left(\frac{\pi t}{6}\right) dt = -\frac{40}{\pi/6} \left[\cos\left(\frac{\pi t}{6}\right)\right] | from t=3t=3 to t=4t=4
  5. Simplify Constant Factor: Simplify the constant factor in front of the integral. \newline−40π/6=−240π-\frac{40}{\pi/6} = -\frac{240}{\pi}\newlineNow we have:\newline−240π[cos⁡(πt6)]-\frac{240}{\pi} [\cos(\frac{\pi t}{6})] | from t=3t=3 to t=4t=4
  6. Evaluate Antiderivative: Evaluate the antiderivative at the upper and lower limits of the integral. \newline−240π[cos⁡(π⋅46)−cos⁡(π⋅36)]-\frac{240}{\pi} [\cos(\frac{\pi \cdot 4}{6}) - \cos(\frac{\pi \cdot 3}{6})]
  7. Calculate Cosine Values: Calculate the cosine values.\newlinecos⁡(π⋅46)=cos⁡(2π3)=−12\cos\left(\frac{\pi \cdot 4}{6}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} (since cos⁡(2π3)\cos\left(\frac{2\pi}{3}\right) is in the second quadrant where cosine is negative)\newlinecos⁡(π⋅36)=cos⁡(π2)=0\cos\left(\frac{\pi \cdot 3}{6}\right) = \cos\left(\frac{\pi}{2}\right) = 0 (since cos⁡(π2)\cos\left(\frac{\pi}{2}\right) is the cosine of 9090 degrees, which is 00)
  8. Substitute Values: Substitute the cosine values into the expression.\newline−240π[−12−0]=240π×12-\frac{240}{\pi} \left[-\frac{1}{2} - 0\right] = \frac{240}{\pi} \times \frac{1}{2}
  9. Simplify Expression: Simplify the expression to find the change in water level during the 4th4^{\text{th}} hour.240π×12=120π\frac{240}{\pi} \times \frac{1}{2} = \frac{120}{\pi} centimeters

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