I am stuck on this question. Could someone work it out. Id appreciate it.
Find the amount of A in an account after t years given the following conditions.
dA/dt = 0.04A
A(0) = 7000
A(t) =
dA/dt = 0.04A
Separate the variables:
dA/A = 0.04dt
Integrate both sides
ln(A)=0.04t + C'
A=Ce^(0.04t)
Substitute A(0)=7000 [i.e. t=0,A=7000]
7000=C*1
or
C=7000
Therefore
A = 7000e^(0.04t)
What is happening?
To solve this question, we need to find the function A(t) that represents the amount of A in the account after t years.
The given condition is:
dA/dt = 0.04A
This is a differential equation, which relates the rate of change of A with respect to time (dA/dt) to the current value of A (A). To solve this differential equation, we can use separation of variables.
First, let's rearrange the equation:
dA/A = 0.04dt
Now, let's integrate both sides of the equation with respect to their respective variables:
∫(1/A) dA = ∫0.04 dt
The integral of (1/A) with respect to A is ln|A|, and the integral of 0.04 with respect to t is 0.04t:
ln|A| = 0.04t + C
Here, C is the constant of integration.
Now, we can solve for A. Taking the exponent of both sides of the equation, we have:
|A| = e^(0.04t + C)
Since the exponential function is always positive, we can drop the absolute value signs:
A = e^(0.04t + C)
To find the particular solution for the given initial condition A(0) = 7000, plug in t = 0 and A = 7000 into the equation:
7000 = e^(0.04(0) + C)
Since e^0 = 1, the equation simplifies to:
7000 = e^C
Taking the natural logarithm of both sides, we can solve for C:
ln(7000) = C
Now we have the specific solution for C, which is:
C = ln(7000)
Finally, substitute this value of C back into the equation A = e^(0.04t + C) to get the function A(t):
A(t) = e^(0.04t + ln(7000))
A(t) = e^(ln(7000)) * e^(0.04t)
A(t) = 7000 * e^(0.04t)
Therefore, the amount of A in the account after t years is given by the function A(t) = 7000 * e^(0.04t).