Let $$ P(t)=11,700 e^{0.10 t}-11,000 $$ be the balance after $$ t $$ years in a savings account with a continuous stream of income. (a) How much money was originally deposited? (b) What is the interest rate? (c) How much money will be in the account after 8 years? (d) What differential equation and initial condition are satisfied by $$ y=P(t) $$ ? (e) Use the differential equation in (d) to determine how fast the balance is growing when $$ \$ 1000 $$ is in the account?

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(a) The original deposit is the value at $$ t=0 $$. Substitute $$ t=0 $$ into 
$$
P(t)=11\,700\, e^{0.10\,t}-11\,000:
$$
$$
P(0)=11\,700\, e^{0.10\cdot 0}-11\,000=11\,700\cdot 1-11\,000=700.
$$
Thus, the original deposit was $$\$700$$.

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(b) The continuous interest rate appears in the exponent of the exponential function. Since the given function is 
$$
P(t)=11\,700\, e^{0.10\,t}-11\,000,
$$
the interest rate is $$0.10$$, i.e., $$10\%$$.

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(c) To find the balance after 8 years, substitute $$ t=8 $$ into $$ P(t) $$:
$$
P(8)=11\,700\, e^{0.10\cdot8}-11\,000=11\,700\, e^{0.8}-11\,000.
$$
Approximating $$ e^{0.8}\approx 2.22554 $$:
$$
P(8)\approx 11\,700\cdot 2.22554-11\,000\approx 26\,040.8-11\,000\approx 15\,040.8.
$$
Thus, after 8 years there will be approximately $$\$15\,041$$ in the account.

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(d) We wish to find a differential equation satisfied by $$ y=P(t) $$. Notice that the given solution can be rewritten as
$$
y+11\,000=11\,700\, e^{0.10t}.
$$
Differentiating both sides with respect to $$ t $$:
$$
\frac{d}{dt}\Bigl(y+11\,000\Bigr)=11\,700\cdot 0.10\, e^{0.10t},
$$
$$
y'(t)=0.10\,(11\,700\, e^{0.10t}).
$$
But since $$ 11\,700\, e^{0.10t}=y+11\,000 $$, we have
$$
y'(t)=0.10\,(y+11\,000).
$$
Thus, the differential equation is
$$
\frac{dy}{dt}=0.10\,y+1\,100,
$$
with the initial condition $$ y(0)=700 $$.

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(e) To determine how fast the balance is growing when there is $$\$1000$$ in the account, we use the differential equation:
$$
\frac{dy}{dt}=0.10\,(y+11\,000).
$$
When $$ y=1000 $$:
$$
\frac{dy}{dt}=0.10\,(1000+11\,000)=0.10\,(12\,000)=1\,200.
$$
Thus, when the account balance is $$\$1000$$, it is growing at a rate of $$\$1\,200$$ per year.
(a) The original deposit was \$700. (b) The interest rate is 10%. (c) After 8 years, there will be approximately \$15,041 in the account. (d) The differential equation satisfied by $$ y = P(t) $$ is $$ \frac{dy}{dt} = 0.10y + 1100 $$ with the initial condition $$ y(0) = 700 $$. (e) When the account balance is \$1,000, it is growing at a rate of \$1,200 per year.

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