022 Exam 2 Sample A, Problem 5

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Set up the equation to solve. You only need to plug in the numbers - not solve for particular values!

How much money would I have after 6 years if I invested \$3000 in a bank account that paid 4.5% interest,

(a) compounded monthly?
(b) compounded continuously?
Foundations:
The primary purpose of this problem is to demonstrate that you understand the difference between continuous compounding and compounding on an interval of time. When we compound on an interval, say monthly, the value in the account only changes at the end of each interval. In other words, there is no interest accrued for a week or a day. As a result, we use the formula
${\displaystyle A=P\left(1+{\frac {r}{n}}\right)^{nt},}$
where ${\displaystyle A}$ is the value of the account, ${\displaystyle P}$ is the principal (original amount invested), ${\displaystyle r}$ is the annual rate and ${\displaystyle n}$ is the number of compoundings per year. The value of ${\displaystyle n}$ is ${\displaystyle 365}$ for compounding daily, ${\displaystyle 52}$ for compounding weekly, and ${\displaystyle 12}$ for compounding monthly. As a result, the exponent ${\displaystyle nt}$, where ${\displaystyle t}$ is the time in years, is the number of compounding periods where we actually earn interest. Similarly, ${\displaystyle r/n}$ is the rate per compounding period (the annual rate divided by the number of compoundings per year).
For example, if we compound monthly for ${\displaystyle 7}$ years at a ${\displaystyle 6\%}$ rate, we would have ${\displaystyle nt\,=\,12(7)\,=\,84}$ months at a rate of ${\displaystyle 0.06/12\,=\,0.005}$, or ${\displaystyle 0.5\%}$), per monthly period. Notice that we always use the decimal version for interest rates in these equations.
Step 3:
We can now use the chain rule to find
${\displaystyle {\begin{array}{rcl}y'&=&\left(g\circ f\right)'(x)\\\\&=&g'\left(f(x)\right)\cdot f'(x)\\\\&=&\displaystyle {{\frac {x}{(x+5)(x-1)}}\cdot {\frac {x^{2}-9x+5}{x^{2}}}}\\\\&=&\displaystyle {{\frac {x^{2}-9x+5}{x^{3}+4x^{2}-5x}}.}\end{array}}}$

Note that many teachers do not prefer a cleaned up answer, and may request that you do not simplify. In this case, we could write the answer as

${\displaystyle y'=\displaystyle {{\frac {x}{(x+5)(x-1)}}\cdot {\frac {(2x+5)x-(x^{2}+4x-5)(1)}{x^{2}}}.}}$
${\displaystyle y'\,=\,\displaystyle {{\frac {x^{2}-9x+5}{x^{3}+4x^{2}-5x}}.}}$