Chapter 4.3, Problem 68E

Applied Calculus for the Manageria...

10th Edition
Soo T. Tan
ISBN: 9781285464640

Chapter
Section

Applied Calculus for the Manageria...

10th Edition
Soo T. Tan
ISBN: 9781285464640
Textbook Problem

CONCENTRATION OF A DRUG IN THE BLOODSTREAM The concentration (in milligrams per cubic centimeter) of a certain drug in a patient’s bloodstream t In after injection is given by C ( t ) = 0.2 t t 2 + 1 Sketch the graph of the function C, and interpret your results.

To determine

To sketch: The graph of the function C and interpret the result.

Explanation

Given:

The concentration of certain drug in a patientâ€™s bloodstream t hr after injection is given as,

C(t)=0.2tt2+1 (1)

Calculation:

As the function is given in rational form, thus the vertical asymptote for a rational number is the point where the function is undefined, that is the denominator is zero.

Since the denominator of the given function is never zero, thus there is no vertical asymptote for the given function.

The formula to calculate the horizontal asymptote (ie, the line y=b ) is either, limxâ†’âˆžf(x)=b or limxâ†’âˆ’âˆžf(x)=b .

Take limits t tends to âˆž and use both sides of equation (1).

limtâ†’âˆžC(t)=limtâ†’âˆž(0.2tt2+1)=limtâ†’âˆž(0.2tt2t2(1+1t2)t2)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰[Divideâ€‰Numeratorâ€‰andâ€‰Denominatorâ€‰byâ€‰t2]=limtâ†’âˆž(0.2t(1+1t2))=0.2âˆž(1+1âˆž)

Further solve the above equation.

limtâ†’âˆžC(t)=01+0=0

Take limit t tends to âˆ’âˆž and use both sides of equation (1).

limtâ†’âˆ’âˆžC(t)=limtâ†’âˆ’âˆž(0.2tt2+1)=limtâ†’âˆ’âˆž(0.2tt2t2(1+1t2)t2)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰[Divideâ€‰Numeratorâ€‰andâ€‰Denominatorâ€‰byâ€‰t2]=limtâ†’âˆ’âˆž(0.2t(1+1t2))=0.2âˆ’âˆž(1+1âˆ’âˆž)

Further solve the above equation.

limtâ†’âˆ’âˆžC(t)=01+0=0

Since limtâ†’âˆžC(t)=0 and limtâ†’âˆ’âˆžC(t)=0 , thus the horizontal asymptote is y=0 .

Differentiate equation (1) with respect to t.

Câ€²(t)=(t2+1)ddt(0.2t)âˆ’(0.2t)ddt(t2+1)(t2+1)2=(t2+1)(0.2)âˆ’(0.2t)(2t)(t2+1)2=0.2t2+0.2âˆ’0.4t2(t2+1)2=0.2âˆ’0.2t2(t2+1)2 (2)

This shows that for Câ€²(t)=0 , the critical point comes out to be,

0.2âˆ’0.2t2=00.2t2=0.2t2=1t=1

Thus, the intervals are (0,1) and (1,âˆž) .

Take the test points for different intervals in the table below to determine the signs of Câ€²(t) .

 Intervals Test point c Câ€²(c) Sign of Câ€²(t) (0,1) 0.5 0.096 + (1,âˆž) 3 âˆ’0.016 âˆ’

Thus, the function decreases on the interval (1,âˆž) and increases on the interval. (0,1)

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