Question

It is estimated that t-weeks into a semester, the average amount of sleep a college math student gets per day S(t) decreases at a rate of −3tet2 hours per day. When the semester begins, math students sleep an average of 8.1 hours per day. What is S(t), 15 week(s) into the semester?

Answer 1

Answer:

s(15)= $\frac{-3}{2} e^{15^{2} } +9.6$

Step-by-step explanation:

Before to getting started ,we have to consider the function which  represents  sleep decline rate S'(t) =$-3te^{t^{2} }$

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)

$\int\limits^ {} -3t3e^{t^{2} } \, dx$

This case corresponds to intergration by substitution  

Substituting u=$x^{2}$

Computting the differential of u

$\frac{du}{dt} =2tdt$

Now, let's go back to our integral and write it in terms of u, using both definition of u and its defferential.

$\int\limits^{}-3te^{u} \, \frac{1}{2t}du \\\\\int\limits^{}\frac{-3}{2} e^{u}du$

The integral  of a exponential function is itself. So the result is:

$\frac{-3}{2} e^{u} +c$

Now substitute u back in:

$\frac{-3}{2} e^{t^{2} } +c$

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.

$s(0)=\frac{-3}{2} e^{0^{2} } +c=8.1\\\\\frac{-3}{2} +c=8.1$

Solve this equation

c=8.1+$\frac{3}{2}$

c=9.6

Replacing  the value of c on S(t)

S(t)= $\frac{-3}{2} e^{t^{2} } +9.6$

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)= $\frac{-3}{2} e^{15^{2} } +9.6$