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Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates throughout the year.
The dog's weight W(t) (in kg ) as a function of time t (in days) over the course of a year can be modeled by a sinusoidal expression of the form a*cos(b*t)+d.
At t=0, the start of the year, he is at his maximum weight of 9.1kg. One-quarter of the year later, when 
t=91.25, he is at his average weight of 8.2kg.
Find W(t).
t should be in radians.

W(t)=◻

Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates throughout the year. The dog's weight W(t)W(t) (in kg\text{kg}) as a function of time tt (in days) over the course of a year can be modeled by a sinusoidal expression of the form aa\cdotcos(b\cos(b\cdottt))+d+d. \newlineAt t=0t=0, the start of the year, he is at his maximum weight of 9.1kg9.1\,\text{kg}. One-quarter of the year later, when t=91.25t=91.25, he is at his average weight of 8.2kg8.2\,\text{kg}.\newline Find W(t)W(t). \newlinett should be in radians. \newlineW(t)W(t) ==

Full solution

Q. Aditya's dog routinely eats Aditya's leftovers, which vary seasonally. As a result, his weight fluctuates throughout the year. The dog's weight W(t)W(t) (in kg\text{kg}) as a function of time tt (in days) over the course of a year can be modeled by a sinusoidal expression of the form aa\cdotcos(b\cos(b\cdottt))+d+d. \newlineAt t=0t=0, the start of the year, he is at his maximum weight of 9.1kg9.1\,\text{kg}. One-quarter of the year later, when t=91.25t=91.25, he is at his average weight of 8.2kg8.2\,\text{kg}.\newline Find W(t)W(t). \newlinett should be in radians. \newlineW(t)W(t) ==
  1. Identify Maximum and Average Weight: Identify the maximum weight and the average weight from the given information.\newlineThe maximum weight is given as 9.1kg9.1\,\text{kg} at t=0t=0, which is the start of the year. This means that the value of dd (the vertical shift) is the average weight, and the amplitude aa is the difference between the maximum weight and the average weight.
  2. Calculate Amplitude: Calculate the amplitude ( extit{a}) of the sinusoidal function.\newlineThe amplitude is the difference between the maximum weight and the average weight.\newlinea=maximum weightaverage weighta = \text{maximum weight} - \text{average weight}\newlinea=9.1kg8.2kga = 9.1\,\text{kg} - 8.2\,\text{kg}\newlinea=0.9kga = 0.9\,\text{kg}
  3. Determine Value of \newlinedd: Determine the value of \newlinedd, which is the average weight.\newlineSince the dog's weight is at its average at \newlinet=91.25t=91.25 days, and this is one-quarter of the year, the average weight is also the vertical shift of the sinusoidal function.\newline\newlined=average weightd = \text{average weight}\newline\newlined=8.2kgd = 8.2 \, \text{kg}
  4. Calculate Value of b: Calculate the value of bb, which is related to the period of the sinusoidal function.\newlineSince the dog's weight fluctuates over the course of a year, the period of the sinusoidal function is one year. In days, this is 365365 days. Since t=91.25t=91.25 corresponds to one-quarter of the year, the cosine function should complete one-quarter of its cycle by t=91.25t=91.25.\newlineThe period TT of a cosine function is given by T=2πbT = \frac{2\pi}{b}.\newlineTherefore, b=2πTb = \frac{2\pi}{T}.\newlineSince one-quarter of the period corresponds to t=91.25t=91.25, we have:\newline2π4b=91.25\frac{2\pi}{4b} = 91.25\newlineb=2π4×91.25b = \frac{2\pi}{4 \times 91.25}\newline36536500
  5. Write Sinusoidal Function: Write the equation of the sinusoidal function W(t)W(t).\newlineWe have determined that a=0.9a = 0.9, b=π182.5b = \frac{\pi}{182.5}, and d=8.2d = 8.2. The function is a cosine function, which starts at its maximum value at t=0t=0. Therefore, the phase shift is 00.\newlineThe equation of the sinusoidal function is:\newlineW(t)=acos(bt)+dW(t) = a\cos(b\cdot t) + d\newlineW(t)=0.9cos(π182.5t)+8.2W(t) = 0.9\cos\left(\frac{\pi}{182.5}\cdot t\right) + 8.2

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