f $20,000 is invested in a savings account offering 3.5% per year, compounded continuously, how fast is the balance growing after 4 years? (Round your answer to the nearest cent.)
$ 1 per yr
The effective interest rate with continuous componding is
i = e^(.035)-1 = 3.562%
After four years the balance is
(with continuous compounding)
20,000*(1.0362)^4 = $23,005.48.
The balance is growing at an instantaneous rate of
0.035*23,005.48 = $805.19 /year
but over the year to follow, the earnings will be
$832.80 /yr
Sorry, how did you get the 1.0362?
thanks
Well, buckle up because we're taking a wild ride on the compounding train!
To calculate the growth rate, we can use the continuous compounding formula:
A = P * e^(rt)
Where:
A = final balance
P = initial investment
e = mathematical constant approximately equal to 2.71828
r = annual interest rate (as a decimal)
t = time in years
Plugging in the values:
A = 20000 * e^(0.035 * 4)
So, let's grab our math clown shoes and calculate:
A ≈ 20000 * 2.71828^(0.14)
After doing the math magic, we find that:
A ≈ 20000 * 1.150268
Drumroll, please...
A ≈ $23,005.36
Hold on to your hat! The balance is growing at an approximate rate of $1 per year. So, after 4 years, the balance would have increased by a total of $4.
Remember though, this is just an approximation, so don't go planning your retirement around my clown calculations!
To find out how fast the balance is growing after 4 years, we need to use the formula for continuous compound interest:
A = P * e^(rt)
Where:
A = the final balance
P = the initial principal (amount invested)
e = the mathematical constant approximately equal to 2.71828
r = the annual interest rate (in decimal form)
t = the time in years
In this case:
P = $20,000
r = 3.5% = 0.035 (converted to decimal form)
t = 4 years
Plugging in the values into the formula, we get:
A = 20000 * e^(0.035 * 4)
Now we need to find the derivative of A with respect to time (t) to determine how fast the balance is growing.
dA/dt = 20000 * 0.035 * e^(0.035 * 4)
Evaluating this expression will give us the rate at which the balance is growing after 4 years.
Please note that this is a continuous compound interest problem, so the growth rate is continuously changing.