Answer:

a. compounding by the day: A = $5967.17

b. compounding by the hour: A = $5967.29

c. compounding by the minute: A = $5967.30

d. compounding by the second: A = $5967.30

Step-by-step explanation:

\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}
$ A=P\left(1+\frac{r}{n}\right)^{nt}$
where:
\phantom{ww}$\bullet$ $A =$ final amount
\phantom{ww}$\bullet$ $P =$ principal amount
\phantom{ww}$\bullet$ $r =$ interest rate (in decimal form)
\phantom{ww}$\bullet$ $n =$ number of times interest is applied per year
\phantom{ww}$\bullet$ $t =$ time (in years)
\end{minipage}}

Part (a)

If the interest is compounding by the day then n = 365.

Given:

• P = $4000
• r = 4% = 0.04
• n = 365
• t = 10 years                  

Substitute the values into the compound interest formula and solve for A:

\implies A=4000\left(1+\dfrac{0.04}{365}\right)^{365 \times 10}

\implies A=4000\left(1.00010958...\right)^{3650}

\implies A=4000(1.49179200...)

\implies A=5967.16801...

\implies A=\$5967.17

Part (b)

If the interest is compounding by the hour then:

• n = 365 × 24 = 8760

Given:

• P = $4000
• r = 4% = 0.04
• n = 8760
• t = 10 years                  

Substitute the values into the compound interest formula and solve for A:

\implies A=4000\left(1+\dfrac{0.04}{8760}\right)^{8760 \times 10}

\implies A=4000\left(1.00000456...\right)^{87600}

\implies A=4000\left(1.49182333...\right)

\implies A=5967.29333...

\implies A=\$5967.29

Part (c)

If the interest is compounding by the minute then:

• n = 365 × 24 × 60 = 525600

Given:

• P = $4000
• r = 4% = 0.04
• n = 525600
• t = 10 years          

Substitute the values into the compound interest formula and solve for A:

\implies A=4000\left(1+\dfrac{0.04}{525600}\right)^{525600 \times 10}

\implies A=4000\left(1.00000007...\right)^{5256000}

\implies A=4000(1.49182466...)

\implies A=5967.29867...

\implies A=\$5967.30

Part (d)

If the interest is compounding by the second then:

• n = 365 × 24 × 60 × 60 = 31536000

Given:

• P = $4000
• r = 4% = 0.04
• n = 31536000
• t = 10 years                  

Substitute the values into the compound interest formula and solve for A:

\implies A=4000\left(1+\dfrac{0.04}{31536000}\right)^{31536000 \times 10}

\implies A=4000\left(1.00000000...\right)^{315360000}

\implies A=4000\left(1.49182390...\right)

\implies A=5967.29562...

\implies A=\$5967.30

The more compounding periods throughout the year, the higher the future value of the investment.  However, the difference between compounding by the day and compounding by the second results in a difference of 13 cents over the year, which is negligible comparatively.