Answer:
s(15)= [tex]\frac{-3}{2} e^{15^{2} } +9.6[/tex]
Step-by-step explanation:
Before to getting started ,we have to consider the function which represents sleep decline rate S'(t) =[tex]-3te^{t^{2} } \\[/tex]
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)
[tex]\int\limits^ {} -3t3e^{t^{2} } \, dx \\[/tex]
This case corresponds to intergration by substitution
Substituting u=[tex]x^{2} \\[/tex]
Computting the differential of u
[tex]\frac{du}{dt} =2tdt\\[/tex]
Now, let's go back to our integral and write it in terms of u, using both definition of u and its defferential.
[tex]\int\limits^{}-3te^{u} \, \frac{1}{2t}du \\\\\int\limits^{}\frac{-3}{2} e^{u}du\\[/tex]
The integral of a exponential function is itself. So the result is:
[tex]\frac{-3}{2} e^{u} +c[/tex]
Now substitute u back in:
[tex]\frac{-3}{2} e^{t^{2} } +c[/tex]
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.
[tex]s(0)=\frac{-3}{2} e^{0^{2} } +c=8.1\\\\\frac{-3}{2} +c=8.1[/tex]
Solve this equation
c=8.1+[tex]\frac{3}{2}[/tex]
c=9.6
Replacing the value of c on S(t)
S(t)= [tex]\frac{-3}{2} e^{t^{2} } +9.6[/tex]
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)= [tex]\frac{-3}{2} e^{15^{2} } +9.6[/tex]
