The average student enrolled in the 20-wk Court Reporting I course at the American Institute of Court Reporting progresses according to the rule below where 0 t 20, and N'(t) measures the rate of change in the number of words/minute dictation the student takes in machine shorthand after t wk in the course.
N '(t) = 2e-0.02t
Assuming that the average student enrolled in the course begins with a dictation speed of 45 words/minute, find an expression N(t) that gives the dictation speed of the student after t weeks in the course.
N '(t) = 2e-0.02t
N = (2/-.02) e^-.02t + c
N = -100 e^-.02t + c
when t=0 , N = 45
45 = -100(1) + c
c = 145
N = -100 e^-.02t + 145
testing:
if t = 0
N = -100 e^0 +145
= 45 , good
if t = 5
N = -100 e^-.1 + 145
= -90.5 + 145
= 54.5 , improving
if t = 20
N = 78 , looks logical, and my answer is reasonable.
To find an expression N(t) that gives the dictation speed of the student after t weeks in the course, we need to integrate the given rate of change function N'(t) = 2e^(-0.02t).
The initial dictation speed is given as 45 words/minute, which will serve as the integration constant.
So, let's integrate N'(t) with respect to t with an initial condition:
∫ N'(t) dt = ∫ 2e^(-0.02t) dt
Integrating 2e^(-0.02t) with respect to t, we get:
N(t) = -100e^(-0.02t) + C
Applying the initial condition where N(0) = 45, we can solve for the integration constant C:
45 = -100e^0 + C
C = 45 + 100
C = 145
Therefore, the expression N(t) that gives the dictation speed of the student after t weeks in the course is:
N(t) = -100e^(-0.02t) + 145
To find the expression N(t) that gives the dictation speed of the student after t weeks in the course, we need to integrate the given rate function N'(t) with respect to t.
Given: N'(t) = 2e^(-0.02t)
To integrate N'(t), we can use the formula:
∫ e^(kt) dt = (1/k) * e^(kt) + C
Where C is the constant of integration.
Integrating N'(t) = 2e^(-0.02t), we have:
N(t) = ∫ 2e^(-0.02t) dt
Now, let's consider the integral:
∫ e^(-0.02t) dt
To integrate e^(-0.02t), we can use substitution:
Let u = -0.02t
Then du = -0.02 dt
dt = du / -0.02
The integral becomes:
∫ e^(u) * (-1/0.02) du
Simplifying:
-50 ∫ e^(u) du
Using the integral formula:
∫ e^(u) du = e^(u) + C
We can now substitute back:
-50 (e^(u) + C)
Replacing u with -0.02t:
-50 (e^(-0.02t) + C)
Now, we have the indefinite integral:
N(t) = -50 (e^(-0.02t) + C)
To find the particular solution that matches the initial condition, when t = 0, N(t) = 45 words/minute:
N(0) = -50 (e^(-0.02*0) + C)
45 = -50 (e^0 + C)
45 = -50 (1 + C)
Solving for C:
45 = -50 - 50C
50C = -95
C = -95/50
C = -1.9
Substitute C back into the expression for N(t):
N(t) = -50 (e^(-0.02t) - 1.9)
This expression gives the dictation speed of the student after t weeks in the course.
n=INT n'= 100e^.02t +C
I assume your e-.02t should have been e^.02t
Now find C. You know 45=100e^0+C, so solve for C