Answer:
s(15)= ![\frac{-3}{2} e^{15^{2} } +9.6](https://tex.z-dn.net/?f=%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7B15%5E%7B2%7D%20%7D%20%2B9.6)
Step-by-step explanation:
Before to getting started ,we have to consider the function which represents sleep decline rate S'(t) =
Since this function is interpreted in this case as a rate of change, we can deduce this funcion is the first derivative of the function of average amount of sleep (S(t)).
In order to find S(t), we will integrate S'(t)
![\int\limits^ {} -3t3e^{t^{2} } \, dx \\](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%20%7B%7D%20-3t3e%5E%7Bt%5E%7B2%7D%20%7D%20%5C%2C%20dx%20%5C%5C)
This case corresponds to intergration by substitution
Substituting u=![x^{2} \\](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%5C%5C)
Computting the differential of u
![\frac{du}{dt} =2tdt\\](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdt%7D%20%3D2tdt%5C%5C)
Now, let's go back to our integral and write it in terms of u, using both definition of u and its defferential.
![\int\limits^{}-3te^{u} \, \frac{1}{2t}du \\\\\int\limits^{}\frac{-3}{2} e^{u}du\\](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%7D-3te%5E%7Bu%7D%20%20%5C%2C%20%5Cfrac%7B1%7D%7B2t%7Ddu%20%20%5C%5C%5C%5C%5Cint%5Climits%5E%7B%7D%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7Bu%7Ddu%5C%5C)
The integral of a exponential function is itself. So the result is:
![\frac{-3}{2} e^{u} +c](https://tex.z-dn.net/?f=%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7Bu%7D%20%2Bc)
Now substitute u back in:
![\frac{-3}{2} e^{t^{2} } +c](https://tex.z-dn.net/?f=%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7Bt%5E%7B2%7D%20%7D%20%2Bc)
This answer correspond to function of average amount of sleep.
We know that when the semester begins (t=0) , math students sleep an average of 8.1 hours per day so we will use this info in order to find the value of c.
![s(0)=\frac{-3}{2} e^{0^{2} } +c=8.1\\\\\frac{-3}{2} +c=8.1](https://tex.z-dn.net/?f=s%280%29%3D%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7B0%5E%7B2%7D%20%7D%20%2Bc%3D8.1%5C%5C%5C%5C%5Cfrac%7B-3%7D%7B2%7D%20%2Bc%3D8.1)
Solve this equation
c=8.1+
c=9.6
Replacing the value of c on S(t)
S(t)= ![\frac{-3}{2} e^{t^{2} } +9.6](https://tex.z-dn.net/?f=%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7Bt%5E%7B2%7D%20%7D%20%2B9.6)
Now, let's to find the average of amount of sleep when 15 weeks has elapsed since the start date of the semester.
s(15)= ![\frac{-3}{2} e^{15^{2} } +9.6](https://tex.z-dn.net/?f=%5Cfrac%7B-3%7D%7B2%7D%20e%5E%7B15%5E%7B2%7D%20%7D%20%2B9.6)