Weeks 3 & 4 Tutor Review Quiz

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March 4, 2016

1 Statistics

1.1 How to Make a Histogram

Describe the steps needed to create a histogram.Be thorough.

1.1.1 Solution

1. Find the range of the data. Recall thatRange = Maximum�Minimum.

2. Decide upon the number of classes you wouldlike. This is typically a number between 5 and20. You want to pick a number large enough sothat you can see the variation in your data, butsmall enough that most classes have members.

3. Divide the Range by the number of classes se-lected. Round this number up. The result iscalled your class width.

4. Construct your classes by starting at the min-imum data value. Each subsequent class willthen begin at a number one class width higher.For example, if your minimum was 3 and yourclass with was 2, then your classes would be:[3; 5); [5; 7); [7; 9); etc.

5. Make a frequency chart for your data using theconstructed classes. I recommend using a tallysystem rather than a hunt and peck system.

6. Draw a histogram from your data. It is okay toput tick marks either at the class boundaries(numbers that form the edge of two classes) orat the class midpoints (the average of a class'sboundaries).

Note also, while these steps are fairly general andin fairly common use they are not the only way tocreate a histogram. There are very few rules that

are immutable. The most important thing is thatyour histogram clearly conveys whatever informa-tion you wish to display about your distributionwithout being misleading in other ways.

1.2 Misleading Graphs

Go on the web and �nd an example of a graph thatis being misused in some way. Draw a represen-tation of that graph and describe how it is beingmisused.

1.2.1 Solution

I found the following image on a Google imagesearch for keywords: Misleading Graph.

The �rst set of bar graphs at a very quick glance ap-pears to show pizza being over 4 times as preferredas hamburgers and hot dogs being over 7 times aspreferred as hamburgers. This quick glance is mis-leading. The vertical axis does not show the origin.In the scale of the �rst graph, the axis would benearly 20 times the height of the entire graph be-low the bottom edge! Without being able to usethe bottom edge of the graph as a point of refer-ence we lose all sense of scale. The second graphincludes the data bars all the way to the origin andgives an appropriate sense of scale. This improvedgraph shows preference across the three groups tobe nearly equal.

1

1.3 What E�ect Does Changing One

Data Point Have?

Suppose there are 100 people employed in a com-pany. How will each of the following be a�ected ifthe person with the highest salary gets a $10k raise(be speci�c)? Why?

1. Median salary at the company

2. Mean salary at the company

3. Range of salary at the company

4. Interquartile range of salary at the company

1.3.1 Solution

1. The median salary is the middle data value ona ranked listing. As only the maximum datumis changed, no rankings change nor does thelocation of the middle value. The median valueis una�ected.

2. The mean salary depends on every data value.As data values either stayed the same or wentup, we expect the mean to go up. Let xi be thesalary of the ith person in an ordered ranking.Then we have

�x =x1 + x2 + : : :+ x99 + x100

100

�xnew =x1 + x2 + : : :+ x99 + x100 + 10000

100

�xnew =x1 + x2 + : : :+ x99 + x100

100+

10000

100�xnew = �x+ 100

Thus, the mean will go up by $100.

3. The range depends on the extreme data val-ues. Since the maximum increases by $10000we expect the range to increase by the sameamount. That is to say

range = max�min

rangenew = max+ 10000�min

rangenew = max�min+ 10000

rangenew = range+ 10000

So, the range increased by 10000.

4. The IQR depends on the 1st quartile and the3rd quartile. Neither of these are a�ected bychanging the maximum datum. Thus the IQRis una�ected.

1.4 Common Histogram Errors

Describe some of the common mistakes in makinghistograms and how you might help people correctthem.

1.4.1 Solution

� Often, people do not clearly label their classes.Any histogram should clearly indicate whethertheir horizontal axis ticks are for class mid-points or for class boundaries. In either case, itshould also be indicated how data exactly on aclass boundary is assigned to each class (givencompletely to the lesser class, completely tothe upper class, or split between them).

� Histograms should not have spaces betweenthe class bars (unless one of the classes isempty). That is, the horizontal axis is just asimportant as the vertical axis in a histogram.

� Histograms should always include the originsof each axis when an absolute comparison ofvalues is relevant.

1.5 Pros and Cons of Various Mea-

sures of Location and Spread

Discuss the advantages and disadvantages of usingmedian vs mean vs mode for measures of the lo-cation of a distribution. Do the same for IQR vsstandard deviation vs range for measures of spreadof a distribution.

1.5.1 Solution

Location. For unimodal symmetric distributionsall three of mean, median, and mode will roughlycoincide. For skewed distributions the mean willtypically be a�ected by more than the median.Which is appropriate depends upon whether onewishes to give equal weight to outliers or to focuson the central portion of the data. Means havethe advantages of being easier to calculate for con-tinuous distributions. Medians and modes tend tobe easier for discrete distributions (though the ad-vent of computers makes computation of statisticsof discrete data sets somewhat of a non-issue mostof the time.)Dispersion. The standard deviations tends to

2

have the same advantages that the mean does.While the IQR and range tend to share advantageswith the median. There is no obvious analog tomodes for dispersion.

2 Trigonometry

2.1 Graphing Trigonometric Func-

tions

Sketch graphs of the following trigonometric func-tions.

1. y = 3 sin(4x� �) + 1

2. y = � sec(2�x� �=3)

3. y = tan(2x� �=2)

2.1.1 Solution

1. The underlying graph for this problem is a sinegraph. We will use the concepts of amplitude,period, phase shift, and vertical shift to mod-ify the standard sine graph into this particulargraph.Amplitude. The amplitude of a sine graph isgiven by the absolute value of the coe�cient ofthe sine function. In this case j3j = 3 and sothe amplitude is 3.Period. The period of a sinusoidal graphtimes the angular frequency will always equalthe period of the standard version of thatgraph. The angular frequency, often called ! ,is the coe�cient of x, in this case 4. Thus wesolve 4T = 2� and �nd that T, the period, is�=2.Phase Shift. The phase shift is nothing moreor less than where the argument of the sinu-soidal graph is 0. We set 4x� � = 0 and solvefor x to �nd x = �=4 and so the phase shiftis �=4. This is where our graph will "begin"(that is, the behavior at 0 on a standard graphis shifted to this value on our graph).Vertical Shift. Any vertical shift will locatethe midline of our sinusoidal graph up or downfrom the x-axis. Our vertical shift in this caseis 1, thus the graph will be moved up 1 to becentered on the line y = 1.

We can begin by creating a graph with the cor-rect midline. Then we can plot points every1=4 of a period starting at our phase shift.

-2 p -3 p

2-p -

p

2

p

2p

3 p

22 p

-3

-2

-1

1

2

3

4

5

Now we can connect our plot with a sinusoidalcurve and repeat it with the correct period.

-2 p -3 p

2-p -

p

2

p

2p

3 p

22 p

-3

-2

-1

1

2

3

4

5

2. This next one is similar to the �rst except thatwe cannot call our vertical stretch an ampli-tude; however it is still found as j � 1j = 1.Our Period is 1. The phase shift is 1=6. Thereis 0 vertical shift. So, starting with a basicy = � secx graph and modifying similarly weget the following sequence as we build a graph.

-

4

3-

13

12-

5

6-

7

12-

1

3

1

6

5

12

2

3

11

12

7

6

17

12

-3

-2

-1

1

2

3

3. This one again follows the same pattern. Thereis no vertical stretch factor. The period of astandard tangent function is only � instead of2�. Thus we instead solve !T = � and �ndthat our period, T is �=2. Our phase shift is

3

�=4.

-p -3 p

4-

p

2-

p

4

p

4

p

2

3 p

4p

-3

-2

-1

1

2

3

2.2 Building a Trigonometric Func-

tion from Limited Data

Find a sinusoidal function with largest possible pe-riod that has a local maximum at (�2; 3) and alocal minimum at (4; 1).

2.2.1 Solution

A maximum and minimum of a sinusoidal trigono-metric function will be at least 1=2 period apart.Therefore, the maximum period will be twice thisgap. Our points have an x-separation of 6 so ourperiod will be 12. So, ! � 12 = 2� gives ! = �=6.The average of the y-values of a maximum andminimum will be the location of the midline andhence the amount of our vertical shift, in this case3�12 = 1. The amplitude is the distance between

the midline and an extrema, in this case 3� 1 = 2.Our phase shift will depend on what kind of si-nusoidal function we wish to chose; any will work.I will select a cosine function. At the beginningof a period, cosine graphs start at their maxi-mum. We know a maximum occurs at x = �2and so we can select a phase shift of -2. Thusx = �2 is a solution to �=6x � ' = 0 and so' = ��=3. We then have the formula for our func-tion: y = 2 cos(�=6x+ �=3) + 1.

2.3 Inverse Issues

For what values of x are the following formulaetrue?

1. sin(arcsinx) = x

2. arcsin(sinx) = x

2.3.1 Solution

Arcsine and sine are inverses only upon restrictingthe domain of our angle to be the principle domain,which for sine is [��

2 ;�2 ]. The corresponding sine

values range from -1 to 1.

1. Here, x represents a sine value, thus we musthave x 2 [�1; 1].

2. Here, x represents an angle value, thus we musthave x 2 [��

2 ;�2 ].

2.4 Evaluating Trigonometric Com-

positions

Evaluate the following.

1. cos(arcsin(7�=6))

2. arctan(sin(3�=2))

3. cos(arcsin(3=5))

2.4.1 Solution

Parts 1 & 3 can be done with diagrams on the unitcircle. Part 2 is most easily done with direct eval-uation using memorized values.

1. We draw our standard position representationof the angle 7�

6 . This angle has a referenceangle of �

6 . We (should) have the side lengthsfor this reference triangle memorized. From

the picture we see cos 7�6 = �

p3

2 .

7 p

6

-1

2

-3

2

-1 1

-1

1

4

2. sin 3�2 = �1 and so arctan(sin 3�

2 ) =arctan(�1) = ��

4 .

3. arcsin(3=5) is a �rst quadrant angle and so wecan draw our diagram in the �rst quadrant andthen use the Pythagorean Theorem to �ll inthe missing sides of our triangle. We can thensee cos(arcsin(3=5)) = 4=5.

sin-13

5

3

5

4

5

-1 1

-1

1

2.5 Establishing Identities

Establish the following identity.sec �

1� sin �=

1 + sin �

cos3 �

2.5.1 Solution

This is an exercise in algebraic manipulation.There are many possible routes that work. Oneis shown below.

sec �

1� sin �=

1 + sin �

cos3 �1 + sin �

1 + sin �� sec �

1� sin �=

sec �(1 + sin �)

1� sin2 �=

sec � � 1 + sin �

1� sin2 �=

1

cos �� 1 + sin �

cos2 �=

1 + sin �

cos3 �=

1 + sin �

cos3 �X

3 Pre-calculus

3.1 Asymptotes of a Rational Func-

tion

Find all straight line asymptotes of y =x3 + x2 � 12

x2 � 4.

3.1.1 Solution

Straight line asymptotes can come from twosources: the end behavior (slant or horizontalasymptote) or a vertical asymptote. We will in-vestigate each separately.Vertical Asymptotes. Vertical asymptotes of ra-tional functions can be found by fully reducing theratio and examining where the remaining factors ofthe denominator have roots.

y =x3 + x2 � 12

x2 � 4

y =(x� 2)(x2 + 3x+ 6)

(x� 2)(x+ 2)

y =x2 + 3x+ 6

x+ 2; x 6= 2

Thus we have a vertical asymptote when x+2 = 0,that is, x = �2.End Behavior. To �nd the long term behavior weactually carry out the division in the de�nition ofy.

(x3 + x2 � 12)� (x2 � 4) = x+ 1 +4

x+ 2

Since 4x+2 !1 as x! 0 we see that y � x+ 1 as

x!1. Thus y = x+ 1 is a slant asymptote of y.

3.2 Solving an Inequality

Solvex3 � x

x� 1� x3

3.2.1 Solution

This inequality is of the form f(x) � g(x). Wecan investigate such an inequality by rewriting it

5

as 0 � g(x)� f(x) and then making a sign chart.

x3 � x

x� 1� x3

0 � x3 � x3 � x

x� 1

0 � x3(x� 1)� (x3 � x)

x� 1

0 � x3(x� 1)� x(x2 � 1)

x� 1

0 � (x� 1)(x3 � x(x+ 1))

x� 1

0 � (x� 1)x(x2 � x� 1)

x� 1

0 � (x� 1)x(x� 1+p5

2 )(x� 1�p5

2 )

x� 1

x (�1;1�p5

2) ( 1�

p5

2; 0) (0; 1) (1; 1+

p5

2) ( 1+

p5

2;1)

rhs � + � � +

Thus we see the solution is ( 1�p5

2 ; 0] [ [ 1+p5

2 ;1).

3.3 Factoring a Polynomial

Factor 4x4 � 3x3 � 4x+ 3.

3.3.1 Solution

We begin by using the rational root theorem to �ndpossible rational roots and hoping we have some.The rational root theorem says if the polynomial inquestion has only integer coe�cients (such as ours)the only possible rational roots will be of the form

� < a factor of the constant term >

< a factor of the leading coe�cient >

In this problem, the factors of the constant aref1; 3g and the factors of the leading coe�cientare f1; 2; 4g thus our possible rational roots aref�1;� 1

2 ;� 13 ;�3;� 3

2 ;� 34g.

It is easiest to check for a root with synthetic divi-sion (see the next problem). We start trying num-bers.

4 -3 0 -4 31 4 1 1 -3

4 1 1 -3 0

Because the last digit was a 0 we see that 1 is aroot (we got lucky on the �rst try). Let's look formore.

4 1 1 -31 1 2 3

4 2 4 0

1 is a root a second time. This illustrates the im-portance of checking for multiple roots. At thispoint we can factor our expression using two linearfactors and one quadratic.

4x4 � 3x3 � 4x+ 3 = (x� 1)2(4x2 + 2x+ 3)

We could check 1 again, and then the remainderof the possible roots, but none of them will endup working (try one). However, we have othertechniques of �nding roots of a quadratic. Thequadratic formula is employed below to �nd them.

x =�2�

p22 � 4(4)(3)

2(4)=�1� i

p11

4

And so the complete factorization of our polyno-mial is

4x4�3x3�4x+3 = 4(x�1)2(x��1 + ip11

4)(x��1� i

p11

4)

Don't forget to have the leading 4.

3.4 Synthetic Division

Divide x4 � 6x3 � 12x+ 2 by x� 2 using syntheticdivision.

3.4.1 Solution

This is simply a matter of recalling the operationof synthetic division. Write the coe�cients ofthe dividend in descending order making sure toinclude 0's for any omitted orders. Write the rootof the divisor (which must be monic and linear)outside. The pattern is to add the top entry tothe middle entry to get the bottom entry. Thenmultiply that bottom entry by the root outside toget the middle entry of the next column. The �rstblank middle value is a 0.

1 -6 0 -12 22 2 -8 -16 -56

1 -4 -8 -28 -54

Now, the bottom row is the coe�cients of the quo-tient in descending order, and the last is the coef-�cient of the remainder. 1 -4 -8 -28 -54corresponds to x3 � 4x2 � 8x � 28 � 54

x�2 . This isthe result of the division.

6

3.5 Finding & Graphing an Inverse

Find the inverse of f(x) = e3x+2 � 1. Sketch bothf and f�1 on the same set of axes.

3.5.1 Solution

To �nd an inverse we substitute f(x)! x and x!f�1(x) then solve the result for f�1(x).

f(x) = e3x+2 � 1

x = e3f�1(x)+2 � 1

x+ 1 = e3f�1(x)+2

ln(x+ 1) = 3f�1(x) + 2

ln(x+ 1)� 2

3= f�1(x)

Below is a simultaneous plot of both functions. Theoriginal function is on top, the inverse is on bottom,and the dotted line is just to emphasize the re ec-tion nature that inverses will always have.

-4 -2 2 4

-4

-2

2

4

4 Calculus I

4.1 Possible Properties of Functions

True or false:

1. If f(x) > M has a non-empty solution set nomatter how big M is then limx!1 f(x) =1

2. A function cannot have two horizontal asymp-totes.

3. Functions cannot cross their asymptotes.

4.1.1 Solution

1. False. The stated condition says that a graphof f would go above the line y = M at leastonce no matter how big M is. However, goingabove a line is not equivalent to staying abovethat line. For example, y = x sinx will eventu-ally be as large as we wish, however, it also re-visits 0 twice within any interval of length 2�.The function wiggles back and forth with everincreasing amplitude. Such a function cannothave any limit, even an in�nite one.

2. False. A horizontal asymptote is a horizontalline that the graph approaches in either direc-tion. Since we have two directions to travel,left and right, we can have two di�erent hori-zontal asymptotes. y = tan�1x is such a func-tion, its two asymptotes being y = ��

2 .

3. False. There is no requirement whatsoever forfunctions to not cross their asymptotes. y =sin xx

is a function that crosses its asymptote,y = 0, in�nitely many times.

4.2 Mutual Tangent Lines

There are exactly 2 lines tangent to both y = x2

& y = 1� (x� 2)2. Find equations for those lines.[Bonus Challenge: Can you solve this again withoutusing calculus?]

4.2.1 Solution (Calculus)

A picture of the situation is immensely helpful.

-2 -1 1 2 3 4

-2

-1

1

2

3

4

Hx2,y2L

Hx1,y1L

Hx2,y2L

Hx1,y1L

7

Here we can see the two lines. They are each de-termined by a pair of points at which the lines aretangent to the two curves respectively. We will �ndconditions for the four variables: x1; y1; x2; and y2.

� The point (x1; y1) must be on the curve y = x2.Thus,

y1 = x21

� Similarly, the point (x2; y2) must be on thecurve y = 1� (x� 2)2 and so

y2 = 1� (x2 � 2)2

� The slope of the line must be the slope of thecurve y = x2 at the point (x1; y1) thus

y2 � y1x2 � x1

= 2x1

� Also, the slope of the line must be the slope ofthe curve y = 1� (x� 2)2 at the point (x2; y2)so

y2 � y1x2 � x1

= �2(x2 � 2)

This is four equations with four unknowns. Wewill solve this as a system of equations.

y1=x21

These are ourinitial equations.

y2=1� (x2 � 2)2

y2�y1x2�x1=2x1

y2�y1x2�x1=�2(x2 � 2)

1�(x2�2)2�x21x2�x1 =2x1 Substitute the y's

out using the �rsttwo equations.1�(x2�2)2�x21

x2�x1 =�2(x2 � 2)

1�(x2�2)2�(x2�2)2x2�(x2�2) =�2(x2 � 2) Subtract one equa-

tion from the otherto obtain x1 = 2 �x2 then substituteinto the top equa-tion.

x2=2�p2

2 Solve for x2

1st Solution 2nd Solution

x1=2+p2

2 x1=2�p2

2

x2=2�p2

2 x2=2+p2

2

y1=3+2

p2

2 y1=3�2

p2

2

y2=�1�2

p2

2 y2=�1+2

p2

2

Back substituteeach solution(separately) toobtain the twosolutions.

Now we can �nd the equation of the line that goesthrough each pair of points. It is convenient tonote that the slope of the line is equal to 2x1 byour original equations.

y = (2 +p2)(x� 2 +

p2

2) +

3 + 2p2

2

y = (2�p2)(x� 2�p

2

2) +

3� 2p2

2

With a little e�ort we can rearrange these as

y = (2 +p2)(x� 1) +

1

2

y = (2�p2)(x� 1) +

1

2

These are the two lines.

8

4.2.2 Solution (Algebra)

In this very di�erent approach to the problem, wemake use of two observations:

� The two lines intersect at one point. If we can�nd that point directly we will be half waydone.

� The only lines that intersect a parabola exactlyonce are lines tangent to the parabola or par-allel to the parabola's line of symmetry (whichin this case is vertical and not a possibility).

We begin by �nding the point of intersection ofthe two lines. The point of intersection is a pointof rotational symmetry of the combined graph ofthe two parabolas. That is, if we rotate the twoparabolas 180� around the point of intersection weget the same image. The below diagram will helpillustrate this.

-2-1 1 2 3 4

-2

-1

1

2

3

4

-2-1 1 2 3 4

-2

-1

1

2

3

4

-2-1 1 2 3 4

-2

-1

1

2

3

4

-2-1 1 2 3 4

-2

-1

1

2

3

4

-2-1 1 2 3 4

-2

-1

1

2

3

4

This means that any line connecting correspondingpoints will rotate into itself. The line connectingthe verticies is such a line and thus the midpointof the line segment containing the two verticiesmust be the point of intersection.The two verticies are (0; 0) and (2; 1) the midpointof the segment connecting these two points is (1; 12 )and this must be our common point of intersectionof the sought lines.To �nd the appropriate slopes we try a verydi�erent approach. It is generally true that anycurve which is either always concave up or alwaysconcave down will intersect its own tangent linesexactly once (at the point of tangency). We will�nd all such lines that go through our earlierde�ned common point. Let y = m(x � 1) + 1

2 beour sought line. We wish for this to intersect withour curves (we will concentrate upon y = x2 forsimplicity) exactly one time.

y = m(x� 1) +1

2

y = x2

x2 = m(x� 1) +1

2

0 = x2 �mx+m� 1

2

This quadratic equation has discriminatem2 � 4(m � 1

2 ). A quadratic equation has aunique solution exactly when its discriminate is 0.

m2 � 4m+ 2 = 0

(m� 2)2 � 2 = 0

jm� 2j =p2

m = 2�p2

Thus the only two possible slopes are 2 +p2 and

2 � p2. Since we know there are indeed two such

lines, they must have these slops. This yields thelines

y = (2 +p2)(x� 1) +

1

2

y = (2�p2)(x� 1) +

1

2

Which are, of course, the same as we got from thecalculus method.

4.3 Di�erentiating Using the De�ni-

tion

Use the de�nition of the derivative to di�erentiatethe following functions.

1. f(x) = x3 � x

2. g(x) =1

(x� 2)2

3. h(x) =px2 � 4

9

4.3.1 Solution

No fancy tricks on this one, just some direct evalu-ation of limits.

1.

f 0(x) = limh!0

f(x+ h)� f(x)

h

f 0(x) = limh!0

(x+ h)3 � (x+ h)� (x3 � x)

h

f 0(x) = limh!0

3x2h+ 3xh2 + h3 � h

h

f 0(x) = limh!0

(3x2 + 3xh+ h2 � 1)

f 0(x) = 3x2 � 1

2.

f 0(x) = limh!0

f(x+ h)� f(x)

h

f 0(x) = limh!0

1(x+h�2)2 � 1

(x�2)2

h

f 0(x) = limh!0

(x�2)2�(x+h�2)2(x+h�2)2(x�2)2

h

f 0(x) = limh!0

�2xh� 4h+ h2

h(x+ h� 2)2(x� 2)2

f 0(x) = limh!0

�2x� 4 + h

(x+ h� 2)2(x� 2)2

f 0(x) =�2x� 4

(x� 2)4

f 0(x) =�2

(x� 2)3

3.

f 0(x) = limh!0

f(x+ h)� f(x)

h

f 0(x) = limh!0

p(x+ h)2 � 4�p

x2 � 4

h

f 0(x) = limh!0

(x+ h)2 � 4� (x2 � 4)

h(p(x+ h)2 � 4 +

px2 � 4)

f 0(x) = limh!0

2xh+ h2

h(p(x+ h)2 � 4 +

px2 � 4)

f 0(x) = limh!0

2x+ hp(x+ h)2 � 4 +

px2 � 4

f 0(x) =xp

x2 � 4

4.4 Finding Example Functions

Find speci�c examples of the following.

1. A function that has a tangent line at everypoint on its graph but is not di�erentiable onall of R.

2. A function with f(x) > 0 & f 0(x) < 0 for everyx in R.

3. A bounded function with f 0(x) > 0 for everyx in R.

4.4.1 Solution

1. First, note that \not di�erentiable on all of R"means that there is at least one point whereit fails to be di�erentiable. It does not saythat the function is not di�erentiable at everypoint of R. There is only one way a functioncan fail to be di�erentiable, but still have atangent line at a particular point, the case of avertical tangent. Any function with a verticaltangent line at a particular point will su�ce.f(x) = x

13 is such a function with the interest-

ing location at the origin.

2. f(x) > 0 means f is above the x-axis andf 0(x) < 0 means f is decreasing. So we wanta function that is always decreasing but nevertouches or goes below the x-axis. There aremany such functions. f(x) = e�x is one suchexample. So is f(x) = � tan�1 x+ �

2 .

10

3. f 0(x) > 0 means f is increasing. We wanta function that is bounded (trapped betweentwo values) but is always increasing. Again,there are many examples. f(x) = tan�1 x isone such example.

4.5 Another Derivative Via the Def-

inition

Find ddx

sinx via the de�nition.

4.5.1 Solution

Again, we will use no special tricks and just breakapart the limit into pieces we know.

d

dxsinx = lim

h!0

sin(x+ h)� sinx

hd

dxsinx = lim

h!0

sinx cosh+ cosx sinh� sinx

hd

dxsinx = lim

h!0

�cosx sinh

h+

sinx(cosh� 1)

h

�

d

dxsinx = cosx � 1 + sinx � 0

d

dxsinx = cosx

5 Calculus II

5.1 Evaluating an Inde�nite Integral

EvaluateR

1px2�2xdx.

5.1.1 Solution

This is a challenging Calculus II integral. We willmake use of hyperbolic trigonometric functions toevaluate this integral. Here are a few relevant factsabout hyperbolic trigonometric functions.

� De�nitions

coshx =ex + e�x

2

sinhx =ex � e�x

2

� Some Formulas

ddx

coshx = sinhxddx

sinhx = coshx

cosh2 x� sinh2 x = 1

� Graphs

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

cosh x

sinh x

� Inverses If x � 0 then we can invert coshx as

y = coshx

y =ex + e�x

20 = ex � 2y + e�x

0 = (ex)2 � 2yex + 1

ex =2y �

p4y2 � 4

2

ex = y �py2 � 1

x = ln�y �

py2 � 1

�

We chose the + branch for the principle inversehyperbolic cosine.

y = sinhx

y =ex � e�x

20 = ex � 2y � e�x

0 = (ex)2 � 2yex � 1

ex =2y �

p4y2 + 4

2

ex = y +py2 + 1

x = ln�y +

py2 + 1

�

Here we are forced to chose the + branch if wewant a real inverse so that the exponential inthe second to last line is positive.

11

Okay, time for the integral.

Z1p

x2 � 2xdx =

Z1p

(x� 1)2 � 1dx

=

Z1p

(u)2 � 1du

Here we wish to make the substitution u = cosh tso that we can make use of the hyperbolic trigono-metric identity and simplify the radical in the de-nominator. Unfortunately, to do so we need payattention to the sign of u.

u < �1 1 < u

Let t � 0

u = � cosh t u = cosh tdt

du = � sinh t du = sinh tdt

R � sinh tdtp(� cosh t)2 � 1

R sinh tdtp(cosh t)2 � 1

Then, because cosh2 t� 1 = sinh2 t

R � sinh tdtpsinh2 t

R sinh tdtpsinh2 t

And since sinh t � 0 for t � 0

R � sinh tdt

sinh t

R sinh tdt

sinh tR �dt Rdt

�t+ c1 t+ c2

� cosh�1�u+ c1 cosh�1 u+ c2

� ln��u+p

u2 � 1�+ c1 ln

�u+

pu2 � 1

�+ c2

At this point, we back substitute u = x � 1 andobtain

Z1p

x2 � 2xdx

=

8<:� ln

��x+ 1 +

p(x� 1)2 � 1

�+ c1; x < �2

� ln�x� 1 +

p(x� 1)2 � 1

�+ c2; x > 2

5.2 What is a Di�erential Equation?

What is a di�erential equation? What does solvinga di�erential equation mean?

5.2.1 Solution

A di�erential equation is a relation between a func-tion and its own derivative(s).A solution to a di�erential equation is a functionde�ned on a set containing an open interval thatsatis�es the given di�erential equation at all pointsin its domain.

5.3 Using a Slope Field

Plot a slope �eld for dydx

= �x+1y�2 . Overlay several

typical solution curves. Without solving the equa-tion guess the general solution. Check this usingimplicit di�erentiation.

5.3.1 Solution

The most direct way to plot a slope �eld is to gener-ate a grid of (x; y) points and evaluate the slope ateach point, plotting it as a small segment with thatslope at the grid point used. For example, in thisproblem, chosing a grid point of (2; 1) would yielda slope of dy

dx= �2+1

1�2 = 1 and so at the point (2; 1)we would draw a small segment of slope 1. Doingthis for many grid points will generate a picture likebelow left.

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

We can now sketch in several typical solution curvesby selecting an initial condition and then drawing apath that is tangent to the slope �eld at all pointsalong the curve, as above right.We sketched circles. It would be di�cult to deter-mine with certainty from a slope �eld alone if thetrue solutions are exactly circles, or indeed evenclosed curves. However, we can check if this is the

12

case easily because we have the original di�erentialequation. Checking is simply a matter if seeing ifthis family of concentric circles satis�es the di�er-ential equation.Our circles appear to be centered at (1; 2) and sothe family can be described as

(x� 1)2 + (y � 2)2 = r2 r > 0

Di�erentiating implicitly and solving for dydx

we get

2(x� 1) + 2(y � 2)dy

dx= 0

dy

dx=�x+ 1

y � 2

Thus our family of curves do solve the di�erentialequation.

5.4 Mixing Problem

A vat holds 20 gal of pure water. At time t = 0 webegin pumping in a .3 kg/gal solution of brine ata rate of 6 gal/min. The solution is thoroughlymixed and continuously drained at a rate of 3gal/min. What volume of solution will be in thetank when the concentration of the solution is .2kg/gal (assume the tank is as large as necessary forno spilling)?

5.4.1 Solution

Mixing problems can be attacked by setting up amodel in the form of a di�erential equation. Wewill do so. [Note, I have chosen to keep unitsthroughout the entire computation. This make itsomewhat messier. It is not necessary to do so,but I wished to exhibit that you can do it if youwant/need to].

Let S(t) = the mass of salt in the tank attime t. Then the law of conservation of mass says

dS

dt= hsalt coming ini � hsalt going outi

We can then model the ow of salt in or out of thetank as the concentration of solution owing timesthe rate of ow. For us,

dS

dt=

:3kg

gal

6gal

min� hconcentration in tanki 3gal

min

Unfortunately, we do not know the concentrationin the tank, but we can write an expression for itinvoling S.

dS

dt=

:3kg

gal

6gal

min� S

hvolume in tanki3gal

min

Because we start with 20gal of solution and we arepumping in 3gal more per minute than we take out,our volume of solution as a function of time mustbe expressed by 20gal + 3gal

mint. Upon substitution

we have

dS

dt=

:3kg

gal

6gal

min� S

20gal + 3galmin

t

3gal

min

dS

dt=

1:8kg

min� 3S

20min+ 3t

This is a linear di�erential equation. We solve it assuch (assume 20 + 3 t

min> 0).

dS

dt+

3

20min+ 3tS =

1:8kg

min

�(t) = eR

320min+3tdt = e

Rduu = u = 20min+ 3t

(20min+ 3t)dS

dt+ 3S =

1:8kg

min(20min+ 3t)

d

dt[(20min+ 3t)S] = 36kg + 5:4kg

t

minZd

dt[(20min+ 3t)S] dt =

Z36kg + 5:4kg

t

mindt

(20min+ 3t)S = 36kg t+ 2:7kgt2

min+ C

S =36kg t+ 2:7kg t2

min+ C

20min+ 3t

Because S(0) = 0 (the tank holds only pure waterat the begining) we must have C = 0.

S =36kg t+ 2:7kg t2

min

20min+ 3t

S =36 t

min+ 2:7( t

min)2

20 + 3 tmin

kg

13

The concentration in the tank is thenS

<volume in tank> .

:2kg

gal=

�36 t

min+ 2:7( t

min)2

20 + 3 tmin

kg

�

��20gal +

3gal

mint

�

:2 =36 t

min+ 2:7( t

min)2

(20 + 3 tmin

)2

:2(20 + 3t

min)2 = 36

t

min+ 2:7(

t

min)2

0 = 4:5(t

min)2 + 60

t

min� 400

And so, t = �18:2137min or t = 4:88034min.Clearly we chose the positive time since a negativetime makes no sense in this problem.

5.5 Euler's Method

Use Euler's Method with a step size of 14 to approx-

imate the value of y(1) if y0 = x2 + y2 & y(0) = 0.

5.5.1 Solution

Euler's method gives us a way to approximate y(x)given an initial condition (x0; y0) and a di�erentialequation dy

dx= f(x; y). It works by following the

slope �eld over a �xed horizontal distance h untilwe have stepped far enough to reach the x-valuedesired. Be sure to select a step size that will stepyou through the desired x-value.

xn+1 = xn + h

yn+1 = yn + h � f(xn; yn)

Given our initial point (0; 0) it is simply a matterof stepping through Euler's method until we are atan x-value of 1.

x0 = 0

y0 = 0

x1 = 0 +1

4=

1

4

y1 = 0 +1

4(02 + 02) = 0

x2 =1

4+

1

4=

1

2

y2 = 0 +1

4(1

4

2

+ 02) =1

64

x3 =1

2+

1

4=

3

4

y3 =1

64+

1

4((1

2)2 + (

1

64)2) =

1281

16384

x4 =3

4+

1

4= 1

y4 =1281

16384+

1

4((3

4)2 + (

1281

16384)2)

=236587521

1073741824� 0:220339

Thus, we have y(1) � 0:220339 according to Euler'sMethod.

14