Consider the following problem: $\newline$ 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 $r(t)=0.6 \sin \left(\frac{\pi t}{6}\right)$ 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? $\newline$ Which expression can we use to solve the problem? $\newline$ Choose $1$ answer: $\newline$ (A) $r^{\prime}(13)+62$ $\newline$ (B) $r^{\prime}(15)$ $\newline$ (C) $62+\int_{13}^{15} r(t) d t$ $\newline$ (D) $62+\int_{13}^{15} r^{\prime}(t) d t$

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Q. Consider the following problem:

\newline

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 \left(\frac{\pi t}{6}\right)

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?

\newline

Which expression can we use to solve the problem?

\newline

Choose

1

answer:

\newline

(A)

r^{\prime}(13)+62

\newline

(B)

r^{\prime}(15)

\newline

(C)

62+\int_{13}^{15} r(t) d t

\newline

(D)

62+\int_{13}^{15} r^{\prime}(t) d t

Understand the Problem: Understand the problem. $\newline$ 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. We need to find the air gap at a later time, $t=15$ hours.
Determine Approach: Determine the correct approach to find the air gap at $t=15$ hours. $\newline$ Since we are given the rate of change of the air gap, we need to integrate this rate from $t=13$ to $t=15$ to find the total change in the air gap over this time period. Then we add this change to the air gap at $t=13$ to find the air gap at $t=15$ .
Identify Expression: Identify the correct expression to use. $\newline$ The correct expression to use is the integral of the rate of change from $t=13$ to $t=15$ , added to the air gap at $t=13$ . This corresponds to option (C) $62 + \int_{13}^{15}r(t)\,dt$ .
Calculate Integral: Calculate the integral of the rate of change from $t=13$ to $t=15$ . We need to integrate $r(t) = 0.6 \sin\left(\frac{\pi t}{6}\right)$ from $t=13$ to $t=15$ . $\int_{13}^{15}r(t)\,dt = \int_{13}^{15}0.6 \sin\left(\frac{\pi t}{6}\right) \,dt$ Let's calculate this integral.
Perform Integration: Perform the integration. $\newline$ To integrate $0.6 \sin\left(\frac{\pi t}{6}\right)$ , we use the substitution method. Let $u = \frac{\pi t}{6}$ , then $du = \frac{\pi}{6} dt$ , and $dt = \frac{6}{\pi} du$ . $\newline$ The limits of integration change accordingly: when $t=13$ , $u=\frac{\pi\cdot13}{6}$ ; when $t=15$ , $u=\frac{\pi\cdot15}{6}$ . $\newline$ Now we integrate with respect to $u$ : $\newline$ $\int 0.6 \sin(u) \cdot \frac{6}{\pi} du$ from $u=\frac{\pi\cdot13}{6}$ to $u=\frac{\pi\cdot15}{6}$ .
Calculate New Integral: Calculate the new integral with the substitution. $\newline$ The integral becomes: $\newline$ $(0.6 \times \frac{6}{\pi}) \int \sin(u) \, du$ from $u=\frac{\pi\times13}{6}$ to $u=\frac{\pi\times15}{6}$ . $\newline$ This simplifies to: $\newline$ $(\frac{3.6}{\pi}) \times [-\cos(u)]$ from $u=\frac{\pi\times13}{6}$ to $u=\frac{\pi\times15}{6}$ .
Evaluate Integral: Evaluate the integral at the new limits. $\newline$ We substitute the limits into the antiderivative: $\newline$ $(\frac{3.6}{\pi}) * [-\cos(\frac{\pi*15}{6}) + \cos(\frac{\pi*13}{6})]$ .
Simplify Expression: Simplify the expression and calculate the values. $\newline$ We calculate the cosine values and simplify the expression: $\newline$ $(3.6/\pi) \cdot [-\cos((5\pi)/2) + \cos((13\pi)/6)]$ . $\newline$ Note that $\cos((5\pi)/2) = 0$ and $\cos((13\pi)/6)$ is equivalent to $\cos(\pi/6)$ due to periodicity of the cosine function. $\newline$ So, the expression simplifies to: $\newline$ $(3.6/\pi) \cdot [0 + \sqrt{3}/2]$ .
Multiply by Coefficient: Multiply the result by the coefficient. $\newline$ $(\frac{3.6}{\pi}) \times (\frac{\sqrt{3}}{2}) = \frac{3.6 \times \sqrt{3}}{2\pi}$ .
Add to Air Gap: Add the result to the air gap at $t=13$ to find the air gap at $t=15$ . The change in air gap from $t=13$ to $t=15$ is $\frac{3.6 \times \sqrt{3}}{2\pi}$ meters. We add this to the air gap at $t=13$ , which is $62$ meters. Air gap at $t=15$ = $62 + \frac{3.6 \times \sqrt{3}}{2\pi}$ .
Calculate Final Value: Calculate the final value. $\newline$ We calculate the numerical value of the expression: $\newline$ Air gap at $t=15 = 62 + \frac{(3.6 \times \sqrt{3})}{2\pi} \approx 62 + \frac{(3.6 \times 1.732)}{(2 \times 3.14159)} \approx 62 + \frac{(6.2352)}{(6.28318)} \approx 62 + 0.992 \approx 62.992$ meters.