Course:MATH110/Archive/2010-2011/003/Groups/Group 11

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Math 110/003 - Group 11
Members: Abdul-Sherif Abdul-Rahaman
Shannon Lee
Amir Mohamad Nazri
Matt Vetter

Course:MATH110/003/Groups/Group_11/Basic_Skills:_Trigonometric_Functions

Basic Skills Project-Group Contribution

http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_11/Basic_Skills

Computation of Trigonometric functions

Let us use the function

cos x = sin(pi/2-x)

y = cos x = sin(pi/2-x),

then, using the chain rule, we get

dy/dx= cos(pi/2-x) d/dx(pi/2-x)

=(-1)cos(pi/2-x) (since pi /2 is constant, and d/dx(-x) = -1)

(using the formula cos(pi/2- x) = sin x)

       =-sin x, is what we get




Solve :F(x)= cosx/1+sinx

We will apply the quotient rule

F’(x)= (1+sinx) (cosx)-(cosx) (1+sinx)/(1+sinx)^2

        =(1+sinx)(-sinx)-cosx(cosx)/(1+sinx)^2
        = -sinx-sin^2x-cos^2x/(1+sinx)^2
        = -sinx-(sin^2x + cos^2x)/(1+sinx)^2

And recall the trigonometry identity sin^2x +cos^2x= 1, hence

        = -sinx -1/ (1+sinx)^2
        
        = -(1+sinx)/ (1+sinx)^2


        = -1/ 1+sinx

Homeworks

HOMEWORK 4[[1]]


Information on Basic Skills

Homework 3

Question 1

Well in both instance the travel time were exactly the same. One was given in hour and minute and the second was given in minutes. We know that an hour is compose of 60 minutes so at 1hour and 20 minute we have 80 total minutes. The Second trip back was 80 minutes so both trips took equal time.

Question 2

The the reason the police did not stop the woman was because she was walking on the street, she was not driving. If the officer was to not stop her as she drove on to oncoming traffic he would not be doing his job. If the officer did not try stopping her it is obvious she was not breaking any laws.

Question 3

Well in the first two boxes you could pick a fruit up and figured it is either apples or oranges, it is a homogenous mixture in the first two boxes, but the third is a heterogenous mixture. I would personally grab into each box three times so i figure it is all the same or a mixture of both

Question 4

He is a brother to the fiddler because he too is blind.


Question 5

The moving quarter has made one full revolution around the fixed coin.



Question 6

If we let the 3 kinds of apples be X, Y and Z respectively, we know from the question that all these types of apples will sum up to a value; let that value be a. Our goal is to have at least two of the same type of apple, which leads to our question-how many apples do we need to draw in order to get a matching pair of apples?

In our attempt to find a solution we found the probability of drawing any type of apple out of the basket, concluding that it is necessary to draw 3 times minimally to have any chance of a matching pair being taken out.

On the 1st draw: 1/a chances of getting any of the types of apples.

1 apple drawn from a basket filled with a apples.

On the 2nd draw: 1/(a-1) chances of getting any of the types of apples, however since you have drawn one out already, one of the 3 types is short an apple, reducing the total by 1.

Another apple drawn from the basket filled with (a-1) apples.

On the 3rd draw: 1/(a-2) chances of getting any type of the apples, with one group short of an apple, possible the same group that you drew an apple from in the 1st and/or 2nd group, resulting in a matching pair of apples.

A third apple drawn from the basket filled with (a-2) apples.

The necessary information that are lacking in this question are the total number of apples in the basket and the total number of each type of apple. Polya's method was only useful in sorting out our thinking. In order to apply probability and combinatoric concepts and further Step 3 of Polya's method, more information is required.--BernadetteHii 17:55, 16 October 2010 (UTC)

Question 7

We know that there are 80 socks in a drawer, 40 blue socks and 40 brown socks.

i.) Our goal in the first half of the question is to draw out a matching pair, how many socks need to be taken from the drawer to ensure a matching pair is found?

Similarly to the preceding question, we use probability of drawing any of the socks out of the drawer, concluding that it is necessary to draw 3 socks, minimally, to have a pair of matching socks.

On the 1st draw: The probability drawing a blue or brown sock is 1/80.

1/40 chance of drawing a blue sock 1/40 chance of drawing a brown sock

On the 2nd draw: 1/79 chance of drawing a blue or brown sock, however the amount of one colour of socks is not reduced by 1.

If we take the two scenarios: a.) 1st draw=blue sock, you now have a 1/39 chance of drawing a blue sock and a 1/40 chance of drawing a brown one on the 2nd draw. b.) 1st draw=brown sock, you now have a 1/39 chance of drawing a brown sock and a 1/40 chance of drawing a blue one on the 2nd draw.

On the 3rd draw: 1/78 chance of drawing a blue or brown sock, however the amount of one of the colours has been reduced again by 1 sock.

a.) 2nd draw=blue sock, you now have a 1/38 chance of drawing a blue sock on the 3rd draw, assuming the 1st draw also was a blue sock b.)2nd draw=brown sock, you now have a 1/38 chance of drawing a bluce sock on the 3rd draw assuming the 1st draw was also a brown sock.


As more socks are taken from the drawer, one by one, the amount of one colour of sock reduces. Since the picking is at random one colour will not necessarily decrease faster than the other, assuming that they are mixed in the drawer not nicely laid out. Therefore, to ensure a matching pair is picked, 3 socks minimally need to be taken from the drawer.

ii.) To draw an unmatched pair of socks from this drawer, simply draw two socks. Picking is at random and assuming the socks are mixed up and not laid out within the drawer, drawing two socks is sufficient enough to two different socks are taken out.

1st draw: 1/40 chance of getting a blue sock 1/40 chance of getting a brown sock

On the first draw there is an equal chance of drawing either a blue or brown sock.

2nd draw: 1/40 chance of getting a brown sock 1/40 chance of getting a blue sock

On the second draw there is also an equal chance of drawing either a blue or brown sock, assuming that the sock draw the first time, is the opposite colour of the sock drawn the second time.--BernadetteHii 17:55, 16 October 2010 (UTC)

Question 8

If we assume Reuben's birthday is December 31, then on December 31 he turned 21 years old. If it is January 1, 2 days ago the would have been December 30. On that day he was still 20 years old. By the end of this year he will turn 22 years old. One year from the day January 1 he will be 23.

Polya's method seems to help organize the wording in this question making it easier to interpret the information the question gives us.--BernadetteHii 17:55, 16 October 2010 (UTC)

Question 9

From the question we know that the distance between each rung is 1 foot. If 10 rungs are showing we can assume that these 10 rungs are hanging off the side of the boat. Also, we assume that these 10 rungs are not submerged in water yet. If the tide rises 5 feet, this is equivalent to the distance between the bottom rung and the 5th rung of the ladder, therefore 5 rungs are now submerged in water, leaving the 5 upper rungs out of the water.

Similar to question 8, Polya's method is helpful to organize the wording in this question and to deduce from the question an appropriate scenario, in this case, 10 rungs of the ladder on the side of a boat, not covered by water.--BernadetteHii 17:55, 16 October 2010 (UTC)

Question 10

Polya's method helps start the question off by helping us interpret what the question is telling us. i.) one-half the population of women can be written as (1/2) (1/2)=(1/4)

If we know half the population are women, then it follows the other half of the population are men. One-half the population also equals (1/2)(1/2)=(1/4)

Knowing that half the population are chocolate eaters, we an take (1/4) of male chocolate eaters and (1/4) of female chocolate eaters, adding these together we get

(1/4)+(1/4)=(1/2), which is consistent with the information given in the question.

ii.) One-half of all men is (1/2)(1/2)=(1/4)

therefore it follows that 1/4 the population of men are chocolate eaters.--BernadetteHii 17:55, 16 October 2010 (UTC)

Question 11

A woman, her older brother, her son, and her daughter are chess players. The worst player's twin, who is one of the four players, and the best player are of opposite sex. The worst player and best have the same age. If this is possible, who is the worst player?

The first step in solving this problem is understanding what the question is asking us and determining what information we are given. The question is asking us to solve out of four possible players, which on is the worst.

The information we are given is: -The twin of the worst player is the opposite sex of the best player -the worst and best player have the same age

Through this information we can determine that it is not possible to solve for the worst player, if one of the twins is the worst player and the best player is the same age and opposite gender of the other twin, there is no way of determining which twin is the worst player; also, no other player fits the description of the best player.

Question 12

A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at he train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

This question is asking us to conclude why it was that the man rarely visited his Bronx girlfriend in order to determine why he was broken up with.

The man's daily routine probably meant he arrived at the train station at a similar point every time, the train's also run on a consistent schedule. Therefore, it is likely that the Brooklyn train schedule matched the Manhattan fellow's typical schedule more often; consequently, he rarely visited his Bronx girlfriend.

Question 13

If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

x= #chimes/second

x= 5 seconds/ 5chimes = 1 chime/ second

from this we can conclude that the clock chimes at a rate of 1 chime per second; therefore, if we want to solve for the time it would take to strike 10:00 (with 10 equally spaced chimes), we divide number of chimes by the rate of chimes:

10chimes/ 1chime/s = 10 seconds

Question 14

One day in the maternity ward, the name tags for four girl babies become mixed up. i)How many different ways could two of the babies be tagged correctly and two incorrectly?

The question is asking us to find the number of ways two correct and two incorrect names can be arrangged. There are four names for the four babies. Let the correct order be ABCD. If there are two correct name orders and two incorrect these are our possible arrangements: ABDC, ADCB, ACBD, BACD, DBCA, CBAD. There are 6 possible arrangements given that 2 of the names are correct and the other two are not.

ii) In how many ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

This question wants to know how many, if at all, ways can three of the baby names be correct and one incorrect. If the correct order of the names is ABCD, and ABC are correct the only possible name assignment for the fourth baby is D; therefore there are no possible arrangements for three correctly ordered names and one incorrect.

Question 15

Alex says to you, "I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck." Should you accept his bet?

This question is asking if Alex's assumption that the number of red cards in the first half of a split deck as there are black cards in the second half of a split deck is correct.

To solve, let R= red cards B= black cards

R(in first half deck) = B(in second half deck)

52 cards = R+B, half a deck = 26 cards

if, 1st half of deck has 16R + 10B, that leaves 16B +10R for the 2nd half of the deck

or, 1st half of deck has 2R+24B, that leaves 2B +24R for the 2nd half of the deck

No, you should not accept his bet because you would owe him a lot of money. Alex's statement that the number of red cards in the first half of the deck is equal to the number of black cards in the second half of the deck is true. Half a deck = 26 cards, so if you have 14 red cards in your first half of the deck that means the remaining 12 cards are black; therefore, in the second half of the deck, there are 14 black cards and 12 red cards remaining. So the number of red in the first half of the deck is equal to the number of black in the second half of the deck.

Qustion 16

Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?

Assume that total kids: 7 Boys: 3 Girls: 4 From this each son would have twice as many sisters as he has brothers.

Question 17

The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.

Theoretically, if both Dan and Sarah stand on the scale, it should have read “110” kg. Since it reads 105kg instead of 110 kg, it shows that the scale reads too high since it should reduce about 2.5 kg for each individual for it to be 105kg. Therefore, the actual reading for Dan should be 57.5 kg and as for Sarah’s it should be 47.5 kg. Both of these weights make up to 105kg.


Question 18

Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?

In order to calculate the total pennies in the jar, one has to work backwards. Starting from the time Carla takes 1/3 of the pennies from the jar. In order to find the previous amount of pennies in the jar before Carla took 1/3 of it, the amount left must be divided by a factor of 1/3. This can be written below as “x” represents the amount of pennies before Carla took 1/3 of it; 1/3 of pennies = 40 Pennies = 120 This number would then be divided by a factor of 1/3 to calculate the amount before Bret took 1/3 of it; 1/3 of pennies = 120 Pennies = 360 Then 360 would be divided again by the same factor of 1/3 to find the original amount of pennies in the jar; 1/3 pennies = 360 Pennies = 1080 In conclusion, there were 1080 pennies originally in the jar.

Question 19

One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?

If Angela consumes a quarter of the total amount of milk, that means there could only be a minimum of 4 number of people in the family. Since (1/4) x 4 makes 1 carton of milk, it is reasonable to conclude that the least number of people in Angela’s family would be 4 members.

Question 20

Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?

At every hour, the first clock runs about 5 minutes faster, and the second clock runs 5 minutes per hour slow. Therefore, for each consecutive hour, the first clock would gain 5 minutes while the second clock would lose 5 minutes. As a result, the total of minutes gained by the first clock for every hour is 10 minutes compared to the second clock. Because there is 60 minutes in an hour, and the first clock gains a lead of 10 minutes every hour, the two clocks would be an hour apart once the first clock strikes 18:00.

Question 21

Sven was in ninth place among seventeen runners in the marathon. It states that Sven was in the exact middle of placement in the marathon, which automatically narrowed the number of racers down to an odd number. This is because any single odd number is surrounded by the same amount of even numbers, so in this case of 9th, there are 8 racers before him, and 8 racers after him, leading the number to seventeen racers. The insert about Dan finishing in 10th place ruled out 19 as another potential number of racers going with the same logic as before. And Lars finishing in 16th place made it obvious there was more than 16 racers. Now there could not have been more racers as with Sven finishing ahead of Dan in 10th place, and being in the exact middle, there could not possibly have been more than 17 racers.

Question 22

Given that the vacation can be no shorter than 13 days, as it states that the rain cannot endure for the entire day. In the morning if it rains, it is followed by a sunny afternoon, and if it rains in the afternoon it is sunny in the morning. There are two distinct time frames, morning and afternoon, per day. Since there are only 2 slots per day, it makes sense to add the numbers of each type of day; 13 periods of rain, 12 sunny afternoons, and 11 bright mornings. After the addition you come out to the number of 36, and since there are 2 time slots per day, 36/2 is equal to 18 days of vacation.

Question 23

Yes you would be able to find the ages of the children. It would be necessary however to know the date at least. The number set for dates alone is much to factor in, from the 1 - 31 at most. However definitely not as high as 31 since the product of the three children's ages is 36. The date narrows it down by quite a bit. It would be simple and very possible to come up with a list of viable ages through simple calculations, but to narrow that list down the date would help. For example; (9,2,2), (2,3,6), (18,2,1), (4,3,3), (6,6,1), (12,3,1) etc ...

Question 24

Considering the second candle took 3 hours to burn out, half the time it took the first candle to burn out (6 hours). I made each candle into a standard and equal fraction of 6/6. Then I began by reducing the first candle to 5/6. Since the first candle lasted twice as long as the second candle, it is safe to assume that the second candle would go down 2 fractional points compared to the first candle, making it 4/6. You do this for one more interval and you find that when the first candle is equal to 4/6 the second candle is down to 2/6, half the value of the first. So doing some simple adjustments, the first candle lasted for 6 hours, and it is double the height of the second candle at fractions 4/6, therefore it took 2 hours before the first candle was twice as big as the second candle.

Question 25

For this question, I started out by figuring out how long each candle had lasted. The L + 1 candle lasted for 6 hours, whereas the L candle lasted for 4 hours. It could then be concluded that the L + 1 candle lasted 3/2 longer than the L candle. And since (L + 1)/L = 6/4, by cross-multiplying this equation, you see that 6L = 4(L + 1), which in turn becomes 6L = 4L + 4. By subtracting 4L from both sides to get rid of it on the right side, the equation then becomes 2L = 4. With some simple division, you discover that L = 2