Q. 56 Suppose聽100聽mg聽of聽a聽drug聽i... [FREE SOLUTION]

Chapter 4: Q. 56 (page 326)

$Suppose 100 mg of a drug is administered to a patient each morning in pill form and it is known that after 24 hours the body processes 80 % of the drug from such a pill, leaving 20 % of the drug in the body . The amount of the drug in the body right after the first pill is taken is A (1) = 100 m g . 24 hours later, after the second pill has been taken, the amount in the body is A (2) = 100 (0.2) + 100 = 120 m g . 48 hours later, the amount in the body right after taking the third pill is A (3) = 100 (0.2) (0.2) + 100 (0.2) + 100 = 124 m g .$
$(a) Explain the calculations for A (1), A (2), and A (3) described in the exercise . Which term in A (3) corresponds to the drug left from the first pill ? (b) Interpret the given graph in the context of this problem . What do the marked points represent ? (c) Express A (n) in sigma notation . (d) Calculate the amount of drug in the body after the 4^{t h} through 10^{t h} pills . Do you notice anything special about A (n) as n gets larger ?$

Short Answer

Expert verified

$(a) . T he term is : 100 (0.2) (0.2) (b) . T he points shown below in the graph represents the amount of drug remains from the previous pills . (c) . {鈭�}_{k = 1}^{n} 100 ({(0.2)}^{n} + 1) (d) . A (4) = 124.8 m g, A (5) = 124.96 m g, A (6) = 124.992 m g, . ., A (10) = 124.9999872 m g$

Step by step solution

Step 1. Given Information

$100 mg of drug is administered to a patient each morning in pill form, after 24 hours the body processes 80 % of the drug leaving 20 % in the body . Amount of Drug in 1 st day A (1) = 100 m g After 24 hours, the amount in the second day is, A (2) = 100 (0.2) + 100 = 120 m g On third day, the amount is A (3) = 100 (0.2) (0.2) + 100 (0.2) + 100 = 124 m g .$

Step 2. Solution (a) : The objective is to explain the calculations for A(1), A(2), A(3) and then to determinethe term in A(3) corresponding to the drug left from the first pill.

$The amount of drug left after consuming the first pill is 20 % = 0.2 of the amount contained in the 100 mg of the drug . So, A (1) = 100 A (2) = 100 (0.2) + 100 = 120 m g The term in A (3) corresponding to the amount of drug left from the first pill is, 100 (0.2) (0.2) Therefore, the term is : 100 (0.2) (0.2)$

Step 3. Solution (b) : The objective is to interpret the graph in the content of the problem and to determine what the marked points represents.

$The graph represents the amount of dose remains in the body each day after consuming the pills . The marked points at 100 represents the amount of drug contained by the pill . And, the points shown below in the graph represents the amount of drug remains from the previous pills .$

Step 4. Solution (c) : The objective is to express A(n) in sigma notation.

$The given terms are, A (1) = 100 m g A (2) = 100 (0.2) + 100 = 120 m g A (3) = 100 (0.2) (0.2) + 100 (0.2) + 100 = 124 m g Therefore, the sigma notation is : {鈭�}_{k = 1}^{n} 100 ({(0.2)}^{n} + 1)$

Step 5. Solution (d) : Objective is find A(4), A(5),...,A(10) and determine the relation between A(n) and n.

$Using sigma notation : A (4) = 124.8 m g A (5) = 124.96 m g A (6) = 124.992 m g . . A (10) = 124.9999872 m g On observing we find that o n increasing n after 3, A (n) doesnt increase significantly .$

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Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution (a) : The objective is to explain the calculations for A(1), A(2), A(3) and then to determinethe term in A(3) corresponding to the drug left from the first pill.

Step 3. Solution (b) : The objective is to interpret the graph in the content of the problem and to determine what the marked points represents.

Step 4. Solution (c) : The objective is to express A(n) in sigma notation.

Step 5. Solution (d) : Objective is find A(4), A(5),...,A(10) and determine the relation between A(n) and n.

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