Science:Math Exam Resources/Courses/MATH105/April 2010/Question 01 (k)

MATH105 April 2010

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Question 01 (k)
Suppose that money is deposited steadily into a savings account at the rate of $3000 per year. Determine the balance at the end of 5 years if the account pays 6% interest compounded continuously.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Can you describe the instantaneous rate of change of capital as a differential equation?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The instantaneous rate of change of capital has two components: We continuously add money at a rate of $3,000 per year; We earn 6% interest compounded continuously. If we denote by A(t) the balance of the account at time t, then we can write the differential equation ${\begin{aligned}{\frac {dA}{dt}}&=3000+6\%\,A(t)\\&=3000+0.06\,A(t)\end{aligned}}$ We solve this equation using separation of variables: $\int {\frac {dA}{3000+0.06A}}=\int \,dt$ Solving both sides separately yields ${\frac {1}{0.06}}\ln |3000+0.06A|+C_{1}=t+C_{2}$ Since A is positive, as the amount of money in the account, we can drop the absolute value sign. This yields $\displaystyle \ln(3000+0.06A)=0.06t+C_{3}$ or, equivalently $3000+0.06A=e^{0.06t+C_{3}}=e^{0.06t}e^{C_{3}}=C_{4}e^{0.06t}.$ C4 is another arbitrary, but positive constant. Hence we obtain for A(t): $A(t)={\frac {1}{0.06}}\left(C_{4}e^{0.06t}-3000\right).$ Now it's time to think about the initial condition. Since the account does not start with any balance initially, we know that A(0) = 0. This yields for C4 that $0={\frac {1}{0.06}}\left(C_{4}e^{0}-3000\right),\quad {\text{which implies}}\quad C_{4}=3000.$ With this we finally find the expression for the balance in dollars at t years: $A(t)={\frac {3000}{0.06}}\left(e^{0.06t}-1\right).$ This implies that at t = 5 years the balance is $A(5)={\frac {3000}{0.06}}\left(e^{0.06\cdot 5}-1\right)=17492.94$ Reality check: Without interest we would expect to see 5 x 3000 = 15 000 dollars in the account. The amount with interest should be in the same order of magnitude and a little larger.

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