correct
A = 25000 e^(.084t)
dA/dt = Pr e^(rt) we know r = .084
dA/dt = .084(25000) e^(.084t)
= 2100 e^(.084t)
when A = 30,000
30,000 = 25,000 e^(.084t)
1.2 = e^(.084t)
ln 1.2 = .084t ---> I assume you know log rules
t = ln 1.2 / .084
= appr 2.17
Plug that into 2100 e^(.084t)
An investment of 25,000 earns interest at an annual rate of 8.4% compounded continuosly
a. find the instantaneous rate of change of the amount in the account after years
b. find the instantaneous rate of change of the amount in the account at the time the amount is eqaul to 30,000
So i know that the equation is A=Pe^rt. So to find the instanteous rate of change I need to find the derivative right?
2 answers
Which point is located at (0.25,-0.5)