A = Ao e^-kt?

Question

A = Ao e^-kt
(Where Ao and k are constants)

When the time t = 0, the amplitude A is 6cm. When t = 3, the amplitude has fallen to 1cm.

a) What are the values of Ao and k when t = 3?

b) What will the value of t be when A = 0.3?

Answer 1

a) Ao = 6 cm & k = 0.597

Using the fact that A = 6 when t = 0, plug these into the equation and solve for Ao:

A = Ao e^(-kt)
6 = Ao e^(-k*0)
6 = Ao e^0
6 = Ao * 1
Ao = 6 cm

Using that, you can solve for k by plugging in A = 1 when t = 3:

A = 6 e^(-kt)
1 = 6 e^(-k*3)
1/6 = e^(-3k)
ln(1/6) = -3k
-ln 6 = -3k
(1/3) ln 6 = k
k = (1/3) ln 6
k = 0.597253156

b) t = 1.161

Plug in A = .3 to the equation keeping the values of Ao and k we found in part a:

A = 6 e^(-((1/3) ln 6)t)
3 = 6 e^(-((1/3) ln 6)t)
1/2 = e^(-((1/3) ln 6)t)
ln(1/2) = -((1/3) ln 6)t
-ln 2 = -((1/3) ln 6)t
(3 ln 2)/(ln 6) = t
t = 1.16055842

Answer 2

A = Ao e^(-kt)

6 = Ao e^(0) => 6 = Ao
So A = 6 e^(-kt)

1 = 6 e^(-3k)
e^(-3k) = 1/6
-3k = ln(1/6)
k = [ln(1/6)]/(-3)
k = 0.597 (to 3 sig fig)

a) So Ao = 6 and k = 0.597

b)
A = 6 e^(-0.597t)
0.3 = 6 e^(-0.597t)
e^(-0.597t) = 0.3/6
-0.597t = ln(0.3/6)
t = ln(0.3/6) / (-0.597)
t = 5.02

Answer 3

A = Ao when t = 0
Ao = 6 cm

Question a)
1 = 6.e^(- 3k)
e^(3k) = 6
3k = ln 6
k = 0.597

Question b)
0.3 = 6.e^(-0.597).t
ln (0.05) = (- 0.597).t
t = 5.02

Answer 4

a) lnA=lnAoe^-kt> lnA=lnAo-kt> Ao=A=6cm when t=0
k=ln(Ao/A)/t= 0.5973 (using loga-logb=loga/b and -loga/b=logb/a)
b)ln0.3=ln6-0.5973t >t=ln0.05/-.5973=5sec approx

Answer 5

Thats the sort of stuff that makes my head hurt.