If P(t) is the amount of dollars in a savings bank account that pays a yearly interest rate of r% compounded continuously ,then dP/dt=(r/100)(P) , t in years . Assume the interest is 5% annually ,P(0)=$1000 ,and no monies are withdrawn a)how much will be in the account after 2 years ? b)when will the account reach $4000? c)if $1000 is added to the account every 12 months, how much will be in the account after 3.1/2 year ?

dP/dt = .05 P

dP/P = .05 dt
ln P = .05 t

P = Pi e^.05t
if Pi = 1000 and t = 2
P = 1000 e^(.1) = 1105.17

4000 = 1000 e^.05t
e^.05 t = 4
.05 t = ln 4
t = 27.72 years

1000 for 1 year = 1051
2051 for year 2 = 2156
3156 for year 3 = 3318
4318 * e^.025 =4427

a) After 2 years, let's calculate how much will be in the account using the formula:

P(t) = P(0)e^(rt)

where P(0) is the initial amount, r is the interest rate in decimal form, and t is the time in years.

Given that P(0) = $1000 and r = 5%, we have r = 0.05.

P(2) = 1000 * e^(0.05*2)
P(2) ≈ $1104.64

Therefore, there will be approximately $1104.64 in the account after 2 years.

b) To find when the account reaches $4000, we can rearrange the formula:

P(t) = P(0)e^(rt)

P(t)/P(0) = e^(rt)

Taking the natural logarithm of both sides, we get:

ln(P(t)/P(0)) = rt

Now, plug the given values into the equation:

ln(4000/1000) = 0.05t

ln(4) = 0.05t

Using a calculator, we find:

t ≈ 13.86 years

So, the account will reach $4000 in approximately 13.86 years.

c) If $1000 is added to the account every 12 months, the rate of change of the account balance will be affected. Taking this into consideration with the previous formula, we have:

dP/dt = (r/100)(P) + 1000

Given that r = 5%, which is 0.05 in decimal form, we can solve this differential equation using separation of variables.

dP/(rP+1000) = dt

Integrating both sides, we get:

ln(rP+1000) = rt + C

Applying initial condition P(0) = $1000, we find C = ln(1000).

ln(rP+1000) = rt + ln(1000)

Simplifying, we have:

ln(rP+1000) = rt + ln(1000)

Substituting t = 3.5 (3.1/2) into the equation, we can solve for P(3.5):

ln(0.05P(3.5)+1000) = 0.05 * 3.5 + ln(1000)

Using a calculator to compute the right-hand side, we find:

ln(0.05P(3.5)+1000) ≈ 0.175 + ln(1000)

Now, we can apply the natural exponential function to both sides:

0.05P(3.5) + 1000 ≈ e^(0.175 + ln(1000))

Simplifying the right-hand side:

0.05P(3.5) + 1000 ≈ 2.91813 * 1000

Subtracting 1000 from both sides:

0.05P(3.5) ≈ 1918.13

Solving for P(3.5):

P(3.5) ≈ 38362.60

Therefore, after 3.1/2 years, the account balance will be approximately $38,362.60.

a) To find the amount in the account after 2 years, we can use the formula:

P(t) = P(0) * e^(rt)

where P(t) is the amount in the account after time t, P(0) is the initial amount, r is the interest rate, and e is the base of the natural logarithm.

Given that P(0) = $1000 and r = 5% = 0.05, we can substitute these values into the formula:

P(2) = $1000 * e^(0.05 * 2)

To find the final amount, we need to evaluate the exponent and multiply it by the initial amount:

P(2) = $1000 * e^(0.1)
P(2) ≈ $1000 * 1.10517
P(2) ≈ $1105.17

Therefore, the amount in the account after 2 years will be approximately $1105.17.

b) To find when the account will reach $4000, we can rearrange the formula:

P(t) = P(0) * e^(rt)

to solve for t:

t = ln(P(t) / P(0)) / r

Given P(0) = $1000, P(t) = $4000, and r = 0.05, we can substitute these values into the formula:

t = ln($4000 / $1000) / 0.05
t = ln(4) / 0.05

Using a calculator, we can evaluate ln(4) / 0.05:

t ≈ 13.8629

Therefore, the account will reach $4000 after approximately 13.8629 years.

c) If $1000 is added to the account every 12 months, we can modify the formula to find the amount after 3.5 years:

P(t) = P(0) * e^(rt) + (1000 * (1+ r) * (t / 12))

Given P(0) = $1000, r = 0.05, and t = 3.5 years, we can substitute these values into the formula:

P(3.5) = $1000 * e^(0.05 * 3.5) + (1000 * (1 + 0.05) * (3.5 / 12))

Simplifying:

P(3.5) ≈ $1000 * e^(0.175) + (1000 * 1.05 * 0.2917)
P(3.5) ≈ $1000 * 1.1919 + 307.29
P(3.5) ≈ $1191.90 + 307.29
P(3.5) ≈ $1499.19

Therefore, the amount in the account after 3.5 years, with $1000 added every 12 months, will be approximately $1499.19.

To solve these problems, we will use the differential equation dP/dt = (r/100)P, where P(t) represents the amount of dollars in the savings bank account at time t.

Given: r = 5% = 0.05 and P(0) = $1000

a) To find the amount in the account after 2 years, we need to solve the differential equation. Taking separation of variables, we have:

dP/P = (r/100)dt

Integrating both sides, we get:

∫(1/P)dP = ∫(r/100)dt

ln|P| = (r/100)t + C1

Using the initial condition P(0) = $1000, we can substitute t = 0 and P = $1000 into the equation:

ln|1000| = (0.05/100)*0 + C1
C1 = ln|1000|

Therefore, the equation becomes:

ln|P| = (r/100)t + ln|1000|

Simplifying, we have:

ln|P| = (0.05/100)t + ln|1000|

Now, raising both sides as a power of "e" to get rid of the natural logarithm, we have:

e^(ln|P|) = e^[(0.05/100)t + ln|1000|]

P = e^[(0.05/100)t]*e^(ln|1000|)

Simplifying further, we get:

P = 1000 * e^(0.05/100*t)

Therefore, after 2 years (t = 2), the amount in the account will be:

P(2) = 1000 * e^(0.05/100*2)

Calculate this expression to find the value.

b) To find when the account will reach $4000, we set up the equation:

P(t) = 4000

1000 * e^(0.05/100*t) = 4000

Divide both sides by 1000:

e^(0.05/100*t) = 4

Take the natural logarithm of both sides:

ln(e^(0.05/100*t)) = ln(4)

Using the property ln(e^x) = x, we have:

(0.05/100)*t = ln(4)

Solve this equation for t to find when the account will reach $4000.

c) If $1000 is added to the account every 12 months, the rate of change of P per year will change. The new differential equation is:

dP/dt = (r/100)P + 1000

We can solve this new differential equation to find the amount in the account after 3.5 years. The process is similar to part (a) but with the modified differential equation.

Last line in answer has an error, it should say:

3318*e^.025=3402.27