Question

11-14) the equation ( d(t) = 3cosleft( \frac{pi}{5}t
ight) + 2 ) models the distance, ( d ), in metres, of a point on a waterwheel relative to the water surface at ( t ) seconds. determine the first two times, to the nearest tenth of a second, when the point on the wheel is 0.8 m below the water level.

12-17) determine a sine function with a maximum at ( left( \frac{7pi}{12}, 6
ight) ), a minimum at -2 and a period of ( \frac{3pi}{4} ).

Explanation:

cos(B(\frac{7\pi}{11}-C)) = 1 \), which means \( B(\frac{7\pi}{11}-C)=2k\pi \). Let's take \( k = 0 \), so \( B(\frac{7\pi}{11}-C)=0 \), which would mean \( C=\frac{7\pi}{11} \) if \( B
eq0 \). But the period is \( \frac{15\pi}{4} \), so \( B=\frac{2\pi}{T}=\frac{2\pi}{\frac{15\pi}{4}}=\frac{8}{15} \), same as before. So the cosine function would be \( y = 4\cos\left(\frac{8}{15}\left(x - \frac{7\pi}{11}\

Answer:

cos(B(\frac{7\pi}{11}-C)) = 1 \), which means \( B(\frac{7\pi}{11}-C)=2k\pi \). Let's take \( k = 0 \), so \( B(\frac{7\pi}{11}-C)=0 \), which would mean \( C=\frac{7\pi}{11} \) if \( B
eq0 \). But the period is \( \frac{15\pi}{4} \), so \( B=\frac{2\pi}{T}=\frac{2\pi}{\frac{15\pi}{4}}=\frac{8}{15} \), same as before. So the cosine function would be \( y = 4\cos\left(\frac{8}{15}\left(x - \frac{7\pi}{11}\