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Hiro is riding a carousel that is next to a wall.
His horizontal distance 
C(t) (in 
m ) away from the wall as a function of time 
t (in seconds) can be modeled by a sinusoidal expression of the form 
a*cos(b*t)+d.
At 
t=0, when he starts, he is closest to the wall, a distance of 
2m away. After 
7pi seconds he reaches his mid-way point from the wall, which is 
7m away.
Find 
C(t).

t should be in radians.

C(t)=

Hiro is riding a carousel that is next to a wall.\newlineHis horizontal distance C(t) C(t) (in m \mathrm{m} ) away from the wall as a function of time t t (in seconds) can be modeled by a sinusoidal expression of the form acos(bt)+d a \cdot \cos (b \cdot t)+d .\newlineAt t=0 t=0 , when he starts, he is closest to the wall, a distance of 2 m 2 \mathrm{~m} away. After 7π 7 \pi seconds he reaches his mid-way point from the wall, which is 7 m 7 \mathrm{~m} away.\newlineFind C(t) C(t) .\newlinet t should be in radians.\newlineC(t)= C(t)=

Full solution

Q. Hiro is riding a carousel that is next to a wall.\newlineHis horizontal distance C(t) C(t) (in m \mathrm{m} ) away from the wall as a function of time t t (in seconds) can be modeled by a sinusoidal expression of the form acos(bt)+d a \cdot \cos (b \cdot t)+d .\newlineAt t=0 t=0 , when he starts, he is closest to the wall, a distance of 2 m 2 \mathrm{~m} away. After 7π 7 \pi seconds he reaches his mid-way point from the wall, which is 7 m 7 \mathrm{~m} away.\newlineFind C(t) C(t) .\newlinet t should be in radians.\newlineC(t)= C(t)=
  1. Start at Maximum Value: Since Hiro is closest to the wall at t=0t=0, the cosine function starts at its maximum value. The amplitude aa is the maximum distance from the average value, which is 2m2\,\text{m} in this case.
  2. Average Value Calculation: The mid-way point after 7π7\pi seconds is 7m7m away from the wall. This is the average value of the function, so dd equals 7m7m.
  3. Finding Cycle Time: To find the value of bb, we use the fact that the cosine function completes a full cycle in 2πb\frac{2\pi}{b} seconds. Since Hiro reaches his mid-way point after 7π7\pi seconds, this must be half of the full cycle time. So, 2πb=2×7π\frac{2\pi}{b} = 2 \times 7\pi, which means b=17b = \frac{1}{7}.

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