1. Thermometer reading 75oF is taken out where temperature is 20oF. The reading is 30oF 4 minutes later. Find:
a.)Thermometer reading 7 minutes after the thermometer was brought outside. b.)The time taken for the reading to drop from 75oF to within a half degree of the air temperature
2. Proved the identity of trigonometry
3. Determine the differential equation in behind
, find when t=√2
5. Find the solution

Solution:
1. Known :
75oF 30oF
u is the thermometer temperature (oF)
t is time (minutes)
when t=0 temperature is 75o (u=75o)
when t=4 temperature is 30o (u=30o)
-k = decrease of temperature in thermometer
(u-20)=difference of temperature
Differential equation
Substitution t=0 and u=75 in formula where we get it.
u=20+ce-kt
75=20+ce-k.0
75-20=ce0
55=c.1
c=55
the equation become u=20+55e-kt
to find k, substitution t=4 and u=30o
30=20+55e-kt
10=55 e-kt
e4t=5,5
ln e4k=ln5,5
4k lne=ln 5,5
k=
so u=20+55exp(-0,43t)
a) If t=7
And you input t in the formula so
u=20+55exp(-0,43.7)
=20+55.0,5
=22,7
So the thermometer reading 22,7oF
b) If u=75/2=37,5
And you input u in the formula so
u=20+55exp(-0,43t)
37,5=20+55exp(-0,43t)
17,5=55exp(-0,43t)
17,5:55=exp(-0,43t)
So the time taken for the reading to drop from 75oF whiting a half degree of the air temperature is 2,66 minutes
2.

Answer:
We choose left segment to prove it
So the equation proven
3.
we can write the differential equation
From that:

Integration factor :
The general solution differential equation is:
So the solution is
and
find
when t=√2
Answer:
If t=√2, x=√5 and

So the solution dy/dx is
5. Find the solution
Answer: Choose u=secx, dv=sec2xdx, so dv=secxtanxdx, v=tanx
so
So the solution











