If $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
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To find out how fast the balance is growing after 4 years, we can calculate the derivative of the balance function with respect to time.
The balance function for continuous compounding is given by:
B(t) = P * e^(rt)
Where:
B(t) = Balance at time t
P = Principal amount (initial investment)
r = Annual interest rate
t = Time in years
e = Euler's number (approximately equal to 2.71828)
In this case:
P = $20,000
r = 3.5% = 0.035
t = 4
So, the balance function becomes:
B(t) = 20000 * e^(0.035t)
To find how fast the balance is growing after 4 years, we need to find the derivative of B(t) with respect to t.
dB(t)/dt = 20000 * 0.035 * e^(0.035t)
Now we can evaluate this expression at t = 4 to find the rate at which the balance is growing:
dB(t)/dt = 20000 * 0.035 * e^(0.035*4)
Calculate the value of e^(0.035*4) using a calculator:
e^(0.035*4) ≈ 1.14986
Now substitute this value back into the derivative expression:
dB(t)/dt ≈ 20000 * 0.035 * 1.14986
Evaluate this expression to find the rate at which the balance is growing after 4 years:
dB(t)/dt ≈ $805.90
Therefore, the balance is growing at a rate of approximately $805.90 per year after 4 years of continuous compounding.