   Chapter 4.3, Problem 68E ### Applied Calculus for the Manageria...

10th Edition
Soo T. Tan
ISBN: 9781285464640

#### Solutions

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 patient’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, limxf(x)=b or limxf(x)=b .

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

limtC(t)=limt(0.2tt2+1)=limt(0.2tt2t2(1+1t2)t2)[DivideNumeratorandDenominatorbyt2]=limt(0.2t(1+1t2))=0.2(1+1)

Further solve the above equation.

limtC(t)=01+0=0

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

limtC(t)=limt(0.2tt2+1)=limt(0.2tt2t2(1+1t2)t2)[DivideNumeratorandDenominatorbyt2]=limt(0.2t(1+1t2))=0.2(1+1)

Further solve the above equation.

limtC(t)=01+0=0

Since limtC(t)=0 and limtC(t)=0 , thus the horizontal asymptote is y=0 .

Differentiate equation (1) with respect to t.

C(t)=(t2+1)ddt(0.2t)(0.2t)ddt(t2+1)(t2+1)2=(t2+1)(0.2)(0.2t)(2t)(t2+1)2=0.2t2+0.20.4t2(t2+1)2=0.20.2t2(t2+1)2 (2)

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

0.20.2t2=00.2t2=0.2t2=1t=1

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

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

 Intervals Test point c C′(c) Sign of C′(t) (0,1) 0.5 0.096 + (1,∞) 3 −0.016 −

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

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