jump to navigation

SCIE 4101 Calculus Review December 3, 2009

Posted by Ms. Miller in UTA.
comments closed

I will try to put a new review packet up today, but in the mean time, check out the following powerpoints:

Tour of Differential Calculus
Tour of Integral Calculus

These were given at a recent workshop I went to for math teachers as a refresher (sound familiar?).

SCIE 4101 Math Review Packet #6 November 19, 2009

Posted by Ms. Miller in UTA.
comments closed

Here is the next packet.

Answers

  1. There are 36 possible characters, so (a) 36^8=2.82\times 10^{12} possibilites with repeating characters, and (b) P(n,r)=\frac{36!}{(36-8)!}=\frac{36!}{28!}=1.22\times 10^{12} possibilities. Note: \frac{36!}{28!}=\frac{1\cdot 2\cdot 3\cdot ...\cdot 28\cdot 29\cdot ...\cdot 36}{1\cdot 2\cdot 3\cdot ...\cdot 28}=29\cdot 30\cdot ...\cdot 36
  2. C(n,r)=\frac{12!}{3!(12-3)!}=\frac{12!}{3!(9!)}=\frac{10\cdot 11\cdot 12}{1\cdot 2\cdot 3}=220
  3. There are four aces and 16 possible 10-point cards in a deck. Therefore, the probability of drawing an ace followed by a 10-point card is \frac{4}{52}\cdot \frac{16}{51}=\frac{16}{663}. The probability of drawing the 10-point card first and then the ace is \frac{16}{52}\cdot \frac{16}{51}=\frac{16}{663}, so the probability is the same.
  4. Since each penny flip is an independent event, the probabilities are also independent. No matter how many heads come up in a row, the probability that the next flip will come up heads is still \frac{1}{2}.

    5. To find an inverse of a function (assuming that an inverse can be found), the way that is usually taught is to swap the x and y variables and solve for y. A better way, especially if the equation is already in function notation, is to remember that f(f^{-1}(x))=x, and solve for f^{-1}(x).

  5. f(f^{-1}(x))=x
    2f^{-1}(x)-3=x
    2f^{-1}(x)=x+3
    f^{-1}(x)=\frac{x+3}{2}
  6. g(g^{-1}(x))=x
    (g^{-1}(x))^2 -16=x
    |g^{-1}(x)| = \sqrt{x+16}
    g^{-1}(x)=\pm\sqrt{x+16}
  7. f(x): Domain: all real numbers \geq\frac{4}{3}; Range: all real numbers. f^{-1}: Domain: all real numbers; Range: all real numbers \geq\frac{4}{3}
  8. a. P(4) represents the bird population in 2007 (2003+4).
    b. P^{-1}(4) represents the number of years since 2003 when the bird population hits 4000 birds.
  9. T=\frac{1}{4}R+40
    \frac{1}{4}R=T-40
    R=4T-160=f^{-1}(T)
  10. f\circ g(x)=f(g(x))=f(x^2)=4x^2+1
  11. g\circ f(x)=g(f(x))=g(4x+1)=(4x+1)^2
    =16x^2+8x+1
  12. A tricky part here is realizing that the number of months between January, 2005 and December, 2009 is 59, not 60 (the interest is first compounded in February, 2005). After that, just plug the numbers into the formula: Q=1000(1+0.035)^{59}=1000(1.035)^{59}=7,611.68.
  13. Q=200(1-0.114)^t=200(0.886)^t
  14. We know that the function will have the form h(x)=ab^x.
    h(0)=ab^0=3 h(1)=ab^1=15
    a=3 3b=15\Rightarrow b=5

    Therefore, h(x)=3(5)^x

  15. f(3)=ab^3=-\frac{3}{8} f(-2)=ab^{-2}=-12
    \frac{ab^3}{ab^{-2}}=\frac{-\frac{3}{8}}{-12}
    b^5=\frac{1}{32}
    b=\sqrt[5]{\frac{1}{32}}=\frac{1}{2}
    a(\frac{1}{2})^3=-\frac{3}{8}\Rightarrow a=-3

    Therefore, f(x)=-3(\frac{1}{2})^x

  16. Q=1000e^{(0.035)(59)}=7,885.30
  17. 8% compounded quarterly \Rightarrow 2% each quarter
    8%: (1.02)^4 \approx 1.0824 or 8.24%
    7.95%: e^{0.0795}\approx 1.0827 or 8.27%

    Therefore, 7.95% compounded continuously is better.

  18. a. 10^N=1
    N=0    		 b. 10^N=1
    N=1    		 c. 10^N=\sqrt{10}=10^\frac{1}{2}
    N=\frac{1}{2}
    d. 8.5 e. x+3 f. \frac{1}{2}
  19. 5e^{2x}=50
    e^{2x}=10
    \ln e^{2x}=\ln 10
    2x=\ln 10
    x=\frac{\ln 10}{2}\approx 1.151

    20. One of the properties of logarithms is \log x^a=a\log x (or \ln x^a=a\ln x.

  20. 500=25(1.1)^{3x}
    (1.1)^{3x}=20
    \log (1.1)^{3x}=\log 20
    3x\log 1.1=\log 20
    x=\frac{\log 20}{3\log 1.1}\approx 10.477
  21. Pe^{-0.00121t}=\frac{1}{2}P
    e^{-0.00121t}=\frac{1}{2}
    \ln e^{-0.00121t}=\ln \frac{1}{2}
    -0.00121t=\ln \frac{1}{2}
    t\approx 5728.49
  22. To convert between the two forms, remember e^k=b. Therefore, e^{0.3}=b\approx 1.35, which means the formula is Q=7(1.35)^t.
  23. e^k=0.886
    \ln e^k=\ln 0.886
    k\approx -0.121

    Since it’s a negative number, it’s a continuous decay rate of 12.1%.

    24. To convert between radians and degrees, use the relation that 180\ \mathrm{degrees}=\pi\ \mathrm{radians}.

  24. 30\ \mathrm{radians}\cdot\frac{180\ \mathrm{degrees}}{\pi\ \mathrm{radians}}=\frac{30(180}{\pi}=1718.87\textdegree
  25. \frac{\pi}{6}\ \mathrm{degrees}\cdot\frac{\pi\ \mathrm{radians}}{180\ \mathrm{degrees}}=\frac{\pi^2}{6(180)}=0.00914\ \mathrm{rad.}
  26. 3t=2\pi
    t=\frac{2\pi}{3}\Rightarrow \mathrm{Period}=\frac{2\pi}{3}
  27. y=\sin\theta:\ (-\infty,+\infty)\rightarrow[-1,1]
    y=\sin^{-1}x:\ [-1,1]\rightarrow[-\frac{\pi}{2},\frac{\pi}{2}]
    The arcsine of x is the angle whose sine is x.

    28. Use Law of Sines if you have a side and two angles; use Law of Cosines if you have two sides and one angle.

  28. In thirty minutes (0.5 hours), plane #1 has traveled 254.5 miles and plane #2 has traveled 251.5 miles.
    c^2=254.5^2+251.5^2-2(254.5)(251.5)\cos 103\textdegree
    c^2\approx 156819.27\Rightarrow c\approx 396.0\ \mathrm{miles}
  29. If the vertex angle is 39°, then the other two angles are each \frac{180-39}{2}=70.5\textdegree.
    \frac{425}{\sin 39\textdegree}=\frac{x}{\sin 70.5\textdegree}
    x\sin 39\textdegree=425\sin 70.5\textdegree
    x\approx 636.6\ \mathrm{feet}
  30. When you diagram this problem, you will notice that you have two triangles: one with the 60° angle of elevation and a 210′ side (the other angles are 110° and 10°), adjacent to a right triangle with the 70° angle of elevation. To solve this problem, you will need to first find the length of the shared side (y) using Law of Sines. Then, use sine to find the height.
    \frac{210}{\sin 10\textdegree}=\frac{y}{\sin 60\textdegree}
    y\sin 10\textdegree=210\sin 60\textdegree
    y\approx 1047.3
    \sin 70\textdegree=\frac{x}{1047.3}
    x\approx 984\ \mathrm{feet}
  31. x^2=97^2+82^2-2(97)(82)\cos 125\textdegree
    x^2\approx 25257.45\Rightarrow x\approx 158.9\ \mathrm{feet}

SCIE 4101 Review Packet #5 November 14, 2009

Posted by Ms. Miller in UTA.
comments closed

Math Review Packet 5

I don’t expect y’all to construct a complicated proof from scratch, but I do think it’s important that you be (re)acquainted with basic geometric properties.

To give you some help, here are the basic algebraic and geometric properties.

Also, this packet reviews trigonometric ratios. There are a lot of different mnemonic devices for remembering the order of things (SOH-CAH-TOA is one), but one I like because my students seem to be able to remember it better is “Some Old Hippie Caught Another Hippie Tripping On Acid.” Either way, we have

\sin = \frac{opp}{hyp} \cos = \frac{adj}{hyp} \tan = \frac{opp}{adj}

For a right triangle, if you know either two sides or one side and one acute angle, you can find out everything else about that triangle. If you would like some additional practice, here are some worksheets I use with my students:

Trig Ratio Practice
Solving Triangles Practice

Answers to Packet

1.
m\angle 1+m\angle 2=180 def. linear pair
m\angle 2 +m\angle 3=180 def. linear pair
m\angle 1+m\angle 3=m\angle 2+m\angle 3 trans. prop. =
m\angle 1=m\angle 2 subtr. prop. =
\angle 1 \cong \angle 2 def. of \cong
2.
\angle 1 and \angle 2 are supplementary given
m\angle 1+m\angle 2=180 def. of supplementary
m\angle 1+m\angle 3=180 def. linear pair
m\angle 1+m\angle 2=m\angle 1+m\angle 3 trans. prop. =
m\angle 2=m\angle 3 subtr. prop. =
\angle 2\cong\angle 3 def. of \cong
a \parallel b converse of corresponding angles
3.
C midpoint of \overline{BD} and \overline{AE} given
AC=ED, DC=BC def. of midpoint
\overline{AC}\cong\overline{EC}, \overline{DC}\cong\overline{BC} def. of \cong
\angle 1\cong\angle 2 vertical angles
\triangle ACB\cong\triangle ECD SAS
\angle A\cong\angle E CPCTC
4.
\overline{IR}\cong\overline{RE} def. of rhombus
\angle IFS\cong\angle IRS isosceles triangle
\overline {IE} bisects \overline {FR} parallelogram
\overline {FS}\cong\overline {RS} def. of bisector
\triangle FIS\cong\triangle RIS SAS
\angle FSI\cong\angle RSI CPCTC
m\angle FSI+m\angle RSI=180\textdegree def. linear pair
m\angle FSI+m\angle FSI=180\textdegree substitution prop. =
m\angle FSI=90\textdegree div. prop. =
m\angle RSI=90\textdegree trans. prop. =
\overline {FR}\perp\overline {IE} linear pair congruent

This is by no means the only way to construct this proof. You could also use vertical angles to prove angles congruent; another (maybe easier) method would be to use the fact that the diagonals bisect each other to use SSS.
5. v=20; w+5=\frac{1}{2}(w-7+3w-11)\Rightarrow w=14
6. (f+2)(f-2)=(15)(4)\Rightarrow f^2-4=60\Rightarrow f=8
7. 0.4452 8. 0.2756 9. 0.1564
10. 68\textdegree 11. 22\textdegree 12. 34\textdegree
13. (missing leg=48)
a. \frac{48}{50}=\frac{24}{25}		 b. \frac{14}{50}=\frac{7}{25} c. \frac{48}{14}=\frac{24}{7}
d. \sin^{-1}\frac{48}{50}\approx 74\textdegree e. 90\textdegree-74\textdegree=16\textdegree f. \frac{48}{50}=\frac{24}{25}
14. \sin 21\textdegree=\frac{VS}{3\times 10^8}\Rightarrow VS\approx 107,510,385\ (1.08\times 10^8)\ \mathrm{km}

SCIE 4101 Review Packet #4 Answers November 2, 2009

Posted by Ms. Miller in UTA.
comments closed

Here are the answers for the other review packet I handed out on Thursday, 10/29. The quiz over this packet will be on 11/12.

14. B (1 sq. yard = 9 sq. feet) 15. B (\frac{180-176}{180})
16. B 17. D
12. G 13. B
14. H 1. B
54. H 30. H (\tan 30\textdegree=\frac{2400}{x})
24. G 4?. G (S=\frac{120}{360}\pi 36^2)
39. D (\frac{5280}{15840}=\frac{6600}{6600+x}) 27. D (\overline{AB} \parallel \overline{CD}; y-2=-\frac{4}{3}(x-5))
33. A (\frac{3\pi + 7\pi}{16\pi}) 28. B (angle is bisected)

SCIE 4101 Review Packet #3 Answers October 28, 2009

Posted by Ms. Miller in UTA.
comments closed

Assume perimeter is in \mathrm{units}, area is in \mathrm{units^2}, and volume is in \mathrm{units^3} unless stated otherwise.

1a. To find the length, use the Pythagorean Theorem (a^2+b^2=c^2): 5^2+b^2=13^2, which means that the length is 12.

P=2l+2w=2(12)+2(5)=34
A=bh=(12)(5)=60

1b. Using the Pyth. Theorem as before, we have 2a^2=(60\sqrt{2})^2, which works out to a=60.

P=60+60+60\sqrt{2}=120+60\sqrt{2}
A=\frac{1}{2}bh=\frac{1}{2}(60)(60)=1800

2a.

C=2 \pi r=2(16)\pi=32\pi \approx 100.53
A=\pi r^2=256\pi

2b.

C=\pi d=21\pi \approx 65.97
A=\pi(10.5)^2=110.25\pi \approx 346.36

3a. Since the perimeter is 64 and we know two sides are 25, we know that the third side is 14. Half of that is the short leg of a right triangle, so we can use the Pyth. Theorem to solve for x: 7^2+x^2=25^2, which means that x=24.

3b.

(x+4)(x-10)=240
x^2-6x-40=240
x^2-6x-280=0
Factoring
(x-20)(x+14)=0		 Quadratic Formula
x=\frac{6 \pm \sqrt{(-6)^2-4(1)(-280)}}{2(1)}
x=\frac{6 \pm \sqrt{1156}}{2}
x=3 \pm 17
x=20,-14 x=20,-14

(It pays to remember how to factor!)
3c.

\pi r^2=121\pi
r=11\ \Rightarrow \ y=2r=22

3d. There are actually two different ways to work this.

Method #1 Method #2
Divide the trapezoid into a rectangle flanked by two right triangles. This means that the longer base can be broken up into three lengths: 3, 9, and 3, and the right triangles have one leg that is 3 and a hypotenuse of 5, and the height is the other leg.
3^2+h^2=5^2
h=4		 A=\frac{1}{2}h(b_1+b_2)
48=\frac{1}{2}h(9+15)
48=\frac{1}{2}(24)h
48=12h
h=4

4a. S=Ph+2B (P is the perimeter of the base, h is the height of the prism, and B is the area of the base.)

S=[2(20)+2(8)](12)+2(20)(8)=992

Notice that if we selected the other rectangle as our base, we’d still get the same answer:
S=[2(8)+2(12)](20)+2(8)(12)=992

V=Bh

V=[(8)(20)](12)=1920

Again, it doesn’t matter which rectangle we select as our base.
4b. The formula for the surface area of a cylinder is S=Ch+2\pi r^2.

S=(8\pi)(9)+2\pi(4^2)=72\pi+32\pi=104\pi.

The volume of a cylinder is the same as a prism: V=Bh, where B is the area of the base, in this case a circle.

V=\pi(4^2)(9)=144\pi

4c.

S=4\pi r^2
S=4\pi (8^2)=256\pi
V=\frac{4}{3}\pi r^3
V=\frac{4}{3}\pi(8^3)=\frac{2048\pi}{3}

4d. To find the surface area of a pyramid, you must have the “slant height” (l). To find the slant height of this pyramid, look at the right triangle created by the height, half the base, and the lateral face. 8^2+6^2=l^2\ \Rightarrow \ l=10

S=\frac{1}{2}(4)(12)(10)+12^2=384
V=\frac{1}{3}(12^2)(8)=384

Remember, even thought the two numbers are the same, one represents surface area (square units) and the other volume (cubic units).

5. An interior angle of a regular polygon is found using the formula \frac{(n-2)(180)}{n}, where n is the number of sides.
a. For a regular hexagon, each interior angle is \frac{4(180)}{6}=120\textdegree.

7x+15=120
7x=105
x=15

b. For a regular pentagon, each interior angle is \frac{3(180)}{5}=108\textdegree.

x^2-3x=54 (angle is half of 108°)
x^2-3x-54=0
(x-9)(x+6)=0
x=9,-6

6. If all the dimensions are doubled, the surface area will increase by a factor of 22, or 4. The volume will increase by a factor of 23, or 8.

S=(150)(4)=600\ \mathrm{cm^2}
V=(225)(8)=1800\ \mathrm{cm^3}

7.

4\pi r^2=180\pi
r^2=\frac{180\pi}{4\pi}=45
r=\sqrt{45}=3\sqrt{5}

V=\frac{4}{3}\pi(\sqrt{45})^3
V=\frac{4}{3}\pi(45\sqrt{45})
V=\frac{4}{3}\pi(135\sqrt{5})
V=180\sqrt{5}\pi\ \mathrm{m^3}

SCIE 4101 Review Packet #3 October 28, 2009

Posted by Ms. Miller in UTA.
comments closed

Math Review Packet 3

I will try to post the answers for this on Wednesday.

SCIE 4101 Review Packet #2 Answers October 22, 2009

Posted by Ms. Miller in UTA.
comments closed

Transformations
A transformation is a change in the position, size, or shape of a figure. The original shape is called the preimage, and the resulting shape is called the image. There are four types of transformations:

  • Translation (slide) – all of the coordinates move the same distance in the same direction
  • Reflection (flip) – each point and its image are the same distance from the line of reflection
    • Across the x-axis: the x-coordinate stays the same, the y-coordinate changes sign. (x,y) \longrightarrow (x,-y).
    • Across the y-axis: the y-coordinate stays the same, the x-coordinate changes sign. (x,y) \longrightarrow (-x,y)
    • Across the line y=x: the coordinates trade places. (x,y) \longrightarrow (y,x)
  • Rotation (turn) – each point and its image are the same distance from the point of rotation
    • 90° about the origin: (x,y) \longrightarrow (-y,x)
    • 180° about the origin: (x,y) \longrightarrow (-x,-y)
  • Dilation (size) – if centered about the origin, multiply each coordinate by the scale factor

Linear Functions
A function is a rule that is applied to an independent variable. The most common form of this is called the slope-intercept form of a line: y=mx+b, where m is the slope and b is the y-intercept. Another form is the point-slope form: y-y_{1}=m(x-x_{1}).

Slope (also called rate of change) is the ratio of the difference in the y-values to the difference in the x-values.

Direct and Inverse Variation
A related type of function is variation. Direct variation is a linear equation of the form y=kx: x and y both increase or both decrease. Inverse variation is a nonlinear relationship of the form y=\frac{k}{x}: x and y move in opposite directions. In both cases, k is the constant of variation.

Vertical Asymptotes
Vertical asymptotes are boundaries on rational functions at points where the function is undefined. Generally, to find a vertical asymptote, evaluate the denominator of the function to find out where/if it can be equal to zero. Whenever a function evaluates to \frac{0}{0}, you have a “hole”. As the function gets closer and closer to the vertical asymptote, its value becomes more and more extreme. It does not behave this way around a hole.

Discriminants
Discriminants are an indicator of whether a quadratic function will have imaginary roots or zeros. If you’ll recall, for a quadratic equation of the form ax^x+bx+c=0, the quadratic formula is x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}. The discriminant is the middle part: b^2-4ac.

  • If this is less than zero, then the equation will have imaginary roots, and the graph does not cross the x-axis.
  • If it is equal to zero, the equation will have one solution, which is where the graph touches the x-axis.
  • If it is greater than zero, the equation will have two real roots, where the graph crosses the x-axis.

Answers

1. U'(-2,1), T'(2,-4),A'(3,1); U'(2,-1), T'(-2,4), A'(-3,-1)
2. H'(5,2), A'(8,-1), N'(6,-11), D'(2,-3)
3. M'(-3,2\frac{1}{2}), A'(4,-2\frac{1}{2}), V'(-3,-2\frac{1}{2})
4. M'(-5,-6), A'(5,8), V'(5,-6)
5. 5n-2 6. y=4x-3
7. {0,1,4,5} 8. 6350
9. \frac{x^2-ax}{2} 10. y\geqslant 2x-3
13. x=4; solutions are 0 and 8
14a. direct; y=3x 14b. inverse; y=\frac{2.4}{x}
15. k=96 16a. vertical asymptote at x=-2 and a hole at x=-3
16b. vertical asymptote at x=-4 and a hole at x=-4 16c. vertical asymptote at x=3 and a hole at x=-2
17a. D=61; two real roots 17b. D=-31; two imaginary roots
17c. D=0; one real root 17d. D=64; two real roots

SCIE 4101 Math Review Packet #2 October 19, 2009

Posted by Ms. Miller in UTA.
comments closed

Here is Math Review Packet 2(#16 is updated!). I do not plan to give a quiz over it until 10/29, but you may want to look over it before class.

SCIE 4101 FYI October 19, 2009

Posted by Ms. Miller in UTA.
comments closed

The problems from the Preparation Manual that correspond to the first review packet are Competencies 001-003 (or problems 1-6).

This Thursday, we will be looking at Domain II, which is Competencies 004-007 (or problems 7-13). If you have any questions about any of these problems, be sure to ask me about them in class.

SCIE 4101 Math Review Packet #1 October 16, 2009

Posted by Ms. Miller in UTA.
comments closed

It was good to finally meet with y’all last night. I hope my introduction made some sort of sense to you.

Math Review Packet 1

Answers

1. N, Z, Q, R 2. Q, R 3. R 4. N, Z, Q, R 5. Q, R
6. R 7. Z, Q, R 8. N, Z, Q, R 9. C 10. R
11. Example: \frac{3}{7} or 2.5 12. Example: -15 13. 12
14. Example: \pi or \sqrt{5} 15. Example: \sqrt{-1} 16. Example: 20
17. Yes 18. No; 5-6 19. Yes
20. No; \frac{-3}{4} 21. Yes 22. No; \frac{20}{4}
23. R is always closed except for even fractional exponents (square roots, fourth roots, etc.).
Group Properties Addition in the Reals Multiplication in the Reals
Commutative p+q=q+p ab=ba
Associative p+(q+r)=(p+q)+r a(bc)=(ab)c
Identity There exists a number 0 such that for all p, p+0=0+p=p There exists a number 1 such that for all a, a \cdot 1=1 \cdot a=a
Inverse For all p, there exists a number -p such that p+-p=0. For a\neq 0, there exists a number \frac{1}{a} such that a \cdot \frac{1}{a}=1.
25. \frac{3}{4} 26. 7 27. 2
28. \frac{5}{6} 29. \frac{5}{2} 30. -\frac{13}{36}
31. 8.7654 \times 10^7 32. 5.477 \times 10^{-4} 33. 9.7 \times 10^13
34. 54,487,000 35. 0.000000000684 36. 0.000165585
37. -7 38. 24
39. 2^{64}=18,446,744,073,709,551,616\ \mathrm{gold}\ \mathrm{pieces}
40. Since there are at most 366 days in a year, if there are 370 people, at least two of them must share a birthday.
41. 38=34+3+1