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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.\newlineThe dog's weight W(t) W(t) (in kg \mathrm{kg} ) as a function of time t t (in days) over the course of a year 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 , the start of the year, he is at his maximum weight of 9.1 kg 9.1 \mathrm{~kg} . One-quarter of the year later, when t=91.25 t=91.25 , he is at his average weight of 8.2 kg 8.2 \mathrm{~kg} .\newlineFind W(t) W(t) .\newlinet t 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.\newlineThe dog's weight W(t) W(t) (in kg \mathrm{kg} ) as a function of time t t (in days) over the course of a year 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 , the start of the year, he is at his maximum weight of 9.1 kg 9.1 \mathrm{~kg} . One-quarter of the year later, when t=91.25 t=91.25 , he is at his average weight of 8.2 kg 8.2 \mathrm{~kg} .\newlineFind W(t) W(t) .\newlinet t 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, and the average weight is given as 8.2kg8.2\,\text{kg} at t=91.25days.t=91.25\,\text{days}.
  2. Determine Amplitude: Determine the amplitude of the sinusoidal function.\newlineThe amplitude aa is the difference between the maximum weight and the average weight.\newlinea=9.1kg8.2kg=0.9kga = 9.1\,\text{kg} - 8.2\,\text{kg} = 0.9\,\text{kg}
  3. Determine Vertical Shift: Determine the vertical shift dd of the sinusoidal function.\newlineThe vertical shift is the average weight, which is given as 8.28.2 kg.\newlined=8.2d = 8.2 kg
  4. Determine Period: Determine the period of the sinusoidal function.\newlineSince the weight fluctuates over the course of a year, the period is one year. In days, this is 365365 days. However, we need to convert this period into radians since the problem asks for tt in radians.\newlineThe period (T)(T) in radians is given by T=2π/bT = 2\pi/b, where bb is the frequency.
  5. Calculate Frequency: Calculate the frequency bb using the period in days.\newlineWe know that one complete cycle occurs in 365365 days, so we can find bb by rearranging the period formula:\newlineT=2πbT = \frac{2\pi}{b}\newline365=2πb365 = \frac{2\pi}{b}\newlineb=2π365b = \frac{2\pi}{365}
  6. Write Sinusoidal Function: Write the sinusoidal function using the values for amplitude aa, frequency bb, and vertical shift dd.W(t)=acos(bt)+dW(t) = a\cos(bt) + dW(t)=0.9cos(2π365t)+8.2W(t) = 0.9\cos\left(\frac{2\pi}{365}t\right) + 8.2
  7. Verify Conditions: Verify that the function meets the given conditions.\newlineAt t=0t=0, W(0)W(0) should be the maximum weight:\newlineW(0)=0.9cos(2π3650)+8.2=0.91+8.2=9.1kgW(0) = 0.9\cdot\cos\left(\frac{2\pi}{365}\cdot0\right) + 8.2 = 0.9\cdot1 + 8.2 = 9.1\,\text{kg}\newlineAt t=91.25t=91.25, W(91.25)W(91.25) should be the average weight:\newlineW(91.25)=0.9cos(2π36591.25)+8.2W(91.25) = 0.9\cdot\cos\left(\frac{2\pi}{365}\cdot91.25\right) + 8.2\newlineSince cos(2π36591.25)=cos(π2)=0\cos\left(\frac{2\pi}{365}\cdot91.25\right) = \cos\left(\frac{\pi}{2}\right) = 0, we get:\newlineW(91.25)=0.90+8.2=8.2kgW(91.25) = 0.9\cdot0 + 8.2 = 8.2\,\text{kg}\newlineBoth conditions are satisfied.

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