(dL)/dt = KL^2 ln t, L(1)=-5
(dL)/dt = KL^2 ln t
(dL)/L^2 = K ln t dt
int(dL)/L^2 = K ln t dt
-1/L= K[tln(t)-t]+C
-1/-5= K[tln(1)-1]+C
1/5= -K+C
C= 1/5 + K
-1/L= K[tln(t)-t]+1/5 + K
color(orange)(L= -1/(K[tln(t)-t]+1/5 + K))
color(red)(a))
Due to Newton's law of cooling: (dT)/dt = k(T-75)
(dT)/dt = k(T-75)
y(t) = T(t) - 75
y(0) = T(0) - 75
y(0) = 185 - 75
y(0) = 110
(dy)/dt = ky
y(t) = 110e^(kt)
y(30) = 110e^(30t)
y(30) because it is going in the oven for half an hour
150-75 = 110e^(30t)
75 = 110e^(30t)
75/100 = e^(30t)
ln(75/110) = 30t
t= (ln(75/110)/30)
y(t) = 110e^(tln(75/110)/30)
y(45) = 110e^(45(1/30)ln(75/110)
= 62
y(45) = 62+ 75
color(orange)(y(45) = 137^@ F)
color(red)(b))
y(t) = T(t) - 75
y(110) = 110 - 75=35
35 = 110e^(tln(75/110)/30)
35/110 = e^(tln(75/110)/30)
ln(35/110) = tln(75/110)/30
30ln(35/110) = tln(75/110)
(30ln(35/110))/ln(75/110) = t
color(orange)(t=90 min
color(red)(a))
p(t)=100e^(kt)
290=100e^(k(1))
2.9=e^(k)
ln2.9=k
p(t)=100e^(ln(2.9)t)
color(orange)(p(t)=100(2.9)^t)
color(red)(b))
p(4)=100(2.9)^4
color(orange)(=7072.81) bacteria
color(red)(c))
p(t)=100e^(kt) take derivative of this
(d(p(t)))/dt=100((d)/dt)e^(kt)
p'(t)=100ke^(kt)
ln2.9=k
p'(t)=100(ln2.9)e^(ln2.9t)
p'(t)=100(ln2.9)(2.9^t)
p'(4)=100(ln2.9)(2.9^4)
color(orange)(p'(4)= 7530.50)
color(red)(d))
10000=100(2.9)^t
100=(2.9)^t
ln100=tln(2.9)
t= ln100/ln(2.9)
color(orange)(t= 4.3 hrs)
