Respuesta :
Answer:
a) Rate of change of amount
[tex]A'(t) =110e^{0.055t}[/tex]
b) &170.79
c) 0.21
Step-by-step explanation:
We are given the following in the question:
The balance is given by the equation:
[tex]A(t) =- 2000e^{0.055t}[/tex]
where t is the time in years and the initial investment is $2000 when compounded continuously.
a) Rate of change of amount
[tex]\dfrac{d(A(t))}{dt} = \dfrac{d}{dt}(2000e^{0.055t})\\\\\dfrac{d(A(t))}{dt} = 2000e^{0.055t}\times 0.055\\\\\dfrac{d(A(t))}{dt} =110e^{0.055t}[/tex]
b) We have to find the value of A'(8)
[tex]A'(t) =110e^{0.055t}\\A'(8) = 110e^{0.055(8)} = 170.79[/tex]
Interpretation:
The future value of 9 year investment of $2000 will be $170.79 more than the future value of 8 year investment.
c) Comparison
Approximation = $171
Actual change = $170.79
Difference =
[tex]\text{Approximation - Actual change}\\=171 - 170.79\\=0.21[/tex]
Thus, the error is 0.21
