Question

For a certain​ drug, the rate of reaction in appropriate units is given by Upper R prime (t )equalsStartFraction 2 Over t plus 1 EndFraction plus StartFraction 1 Over Start Root t plus 1 End Root End Fraction where t is time​ (in hours) after the drug is administered. Find the total reaction to the drug over the following time periods.

a. From t=1 to t=12.
b. From t=12 to t=24

Answer 1

To find the total reaction to the drug over the time periods t=1 to t=12 and t=12 to t=24, we need to evaluate the definite integral of the rate of reaction function within these intervals. The total reaction from t=1 to t=12 is 2ln(12) + 2arctan(12^(1/2)) - 2. The total reaction from t=12 to t=24 is 2ln(24) + 2arctan(24^(1/2)) - 2ln(12) - 2arctan(12^(1/2)).

Step-by-step explanation:

To find the total reaction to the drug over the time periods t=1 to t=12 and t=12 to t=24, we need to find the definite integral of the rate of reaction function within these intervals.

a. From t=1 to t=12:

We integrate R'(t) from t=1 to t=12:

∫[2/t + 1 + 1/(sqrt(t) + 1)] dt

= 2ln|t| + t^(1/2) + 2arctan(t^(1/2)) evaluated from t=1 to t=12.

Substituting the upper and lower limits into the integral, we get:

2ln(12) + 12^(1/2) + 2arctan(12^(1/2)) - 2ln(1) - 1^(1/2) - 2arctan(1^(1/2))

= 2ln(12) + 12^(1/2) + 2arctan(12^(1/2)) - 2 - 2arctan(1)

= 2ln(12) + 2arctan(12^(1/2)) - 2

So, the total reaction to the drug from t=1 to t=12 is 2ln(12) + 2arctan(12^(1/2)) - 2.

b. From t=12 to t=24:

We follow the same steps as above, but substitute the upper and lower limits t=24 and t=12:

2ln(24) + 2arctan(24^(1/2)) - 2ln(12) - 2arctan(12^(1/2))

= 2ln(24) + 2arctan(24^(1/2)) - 2ln(12) - 2arctan(12^(1/2))

Therefore, the total reaction to the drug from t=12 to t=24 is 2ln(24) + 2arctan(24^(1/2)) - 2ln(12) - 2arctan(12^(1/2)).

Answer 2

a) 8.13

b) 4.10

Step-by-step explanation:

Given the rate of reaction R'(t) = 2/t+1 + 1/√t+1

In order to get the total reaction R(t) to the drugs at this times, we need to first integrate the given function to get R(t)

On integrating R'(t)

∫ (2/t+1 + 1/√t+1)dt

In integration, k∫f'(x)/f(x) dx = 1/k ln(fx)+C where k is any constant.

∫ (2/t+1 + 1/√t+1)dt

= ∫ (2/t+1)dt+ ∫ (1/√t+1)dt

= 2∫ 1/t+1 dt +∫1/+(t+1)^1/2 dt

= 2ln(t+1) + 2(t+1)^1/2 + C

= 2ln(t+1) + 2√(t+1) + C

a) For total reactions from t = 1 to t = 12

When t = 1

R(1) = 2ln2 + 2√2

≈ 4.21

When t = 12

R(12) = 2ln13 + 2√13

≈ 12.34

R(12) - R(1) ≈ 12.34-4.21

≈ 8.13

Total reactions to the drugs over the period from t = 1 to t= 12 is approx 8.13.

b) For total reactions from t = 12 to t = 24

When t = 12

R(12) = 2ln13 + 2√13

≈ 12.34

When t = 24

R(24) = 2ln25 + 2√25

≈ 16.44

R(12) - R(1) ≈ 16.44-12.34

≈ 4.10

Total reactions to the drugs over the period from t = 12 to t= 24 is approx 4.10