Need Math Help?

[Corpsecreate]

I think the third line of your dual should read:
y1 + 2y2 - 2y3 + y4 >= 4

LOL your absolutely right. Such a stupid error on my part. Thanks for pointing that out.

 

[JustNoOne]

I have a question about calculus... well... not really about calculus itself but here:

There are 2 poles 10 units high that are also 10 units apart. Find the shortest distance of a rope connecting the top of a pole to the point on a ground and to the top of the other pole.

My intuition says that the rope must touch the middle inbetween them to obtain the shortest length of the rope. So to start the question.

Let 'x' represent the distance of one of the poles to the point where the rope touches the ground
Let 'D(x)' represent the function describing the length of the rope based on the variable x

D(x) = √(x^2 + 100) + √(200 - 20x + x^2)

Then to find the shortest length of the rope, we need the derivative of D(x)
D'(x) = ½(2x)(x^2 + 100)^-½ + ½(2x - 20)(x^2 - 20x + 200)^-½

to find the shortest length, we need to set D'(x) at 0...

The problem is... how would you go on solving it without any sort of graphing technology. I spend 20 mins on a test trying to solve this out mathematically and I already knew the answer =.= Is there a faster method of solving for x?

Help pl0x

 

[Mind Trick]

I'm not sure if I get your question right, do you mean how to solve D'(x)=0 without graphing or just a faster way than writing it out?

I guess you could argue with some rule that D(4,9)>D(5) and because of symmetry x=5 should be the solution.

Or if you want to solve D'(x)=0, just bring one part to the other side of the equation, use a/b=c/d => ad=bc, then ^2 everything and the answer should roll out pretty quick as well.

 

[JustNoOne]

My question is how would you solve for D'(x) = 0 without using any graphing technology or calculators since it screwed me on my test today because I took too long solving that question.. =.=

So now I have to do a remake test.

 

[T-block]

Basically what Mind Trick said:

Let D'(x) = 0
x/sqrt(x^2+100) + (x-10)/sqrt(x^2-20x+200) = 0
x/sqrt(x^2+100) = -(x-10)/sqrt(x^2-20x+200)

Square both sides
x^2/(x^2+100) = (x^2-20x+100)/(x^2-20x+200)

Cross-multiply
x^4-20x^3+200x^2 = x^4-20x^3+100x^2+100x^2-2000x+10000

Cancel terms
2000x = 10000
x=5

 

[Blissard]

Can someone help me find/explain the Width to Height ratio of a silver rectangle?

A silver rectangle is a rectangle from which you can remove 2 squares with side lengths equal to the rectangle's height, and the remaining rectangle will be similar to the original rectangle. It's basically a golden rectangle, but with 2 squares instead.

Thanks in advance ^.^

 

[T-block]

I might be misunderstanding the question...

Let the height be x, width be rx, where r is the width:height ratio.

After removing the two squares, the remaining rectangle will have sides of length x and (rx - 2x). We want the ratio of sides for the remaining rectangle to also be r, so set

x/(rx - 2x) = r
1/(r-2) = r
r² - 2r - 1 = 0

Quadratic formula yields r = 1 + sqrt(2)

 

[terr13]

Say W is width and H is height, and say H>W. Then we know that when we take out 2 squares of length W, we're left with a similar triangle.
So (H-2W)/W = W/H, since the ratio is preserved due to similarity.
We want to find the ratio or W/H, so we can multiply this out and set x = W/H.
1/x -2 = x.
1-2x = x^2.
x^2+2x-1 = 0.
x^2+2x+1 = 2.
(x+1)^2 = 2.
x = -1+- sqrt(2).
Since a negative answer makes no sense, we're left with W/H = sqrt(2)-1.

 

[Blissard]

T-block is Kami

You got it spot on :D I was guessing that the answer was 1+sqrt2, but I wasn't sure, and I didn't really know how to explain it haha. Thanks again!

triangle

Triangle? Lol but thanks for the input anyway.

 

[terr13]

Woops, typo, the rest of the math is right anyways.

 

[Octave]

nvm got it

 

[Corpsecreate]

Ok guys this question might challenge your brains a bit. It certainly is challenging mine and I am unable to find a solution. The question is as follows:

A brand of Toilet Paper says that for each roll of paper they sell, they have 280 rectangular sheets of dimensions 10cm x 11cm that are linked together to form a long roll. This roll is wrapped around a cardboard cylinder of diameter 4.25cm. After being wrapped around, the total diameter from the centre of the cardboard cylinder to the outer edge of the paper is 10.5cm thus giving a 6.25cm thickness for all the paper. The width of the toilet paper is 10cm.

Find the thickness of each sheet.

My attempt (which I know is wrong):

The width of the toilet paper is 10cm and so the 11cm length is the dimension that is being wrapped around. The circumference of the cardboard cylinder is 17Pi/4. So one sheet of paper will go around (44/17Pi)*100% = 82.386% of the cardboard cylinder. So that means after 280 sheets have been wrapped around, the thickness of the paper will amount t0 0.82386*280 = 12320/17Pi. Now we can set 12320x/17Pi = 25/4 and we get x = 425Pi/49280 = 85Pi/9856 ~ 0.0271cm.

This is certainly incorrect because it assumes that as the paper is rolled around, the thickness doesnt change. Clearly it would change and that would drastically change the answer. I dont know how to solve this question.

 

[Mind Trick]

Everytime you cross the same point you can set the radius to 2,125+nx, so I think you need to use recurrence? (/gonna look at it later)

 

[terr13]

This is my guess, I don't want to work it all out, but here's my approach.

Since the width is 10cm, that means the entire length is 280*11cm, or 3080cm. The roll has diameter 17/4, so the circumference is 17pi/4. Let the thickness of the paper be x. With each wrap around the roll, the diameter increases by 2x, so the circumference becomes (17/4 +2x)pi, and we can write a recurrence relation, with a_n = circumference. a_n = a_(n-1)+2pix. Let n = the number of times we wrap around the roll, then we have nx =6.25cm, and sum_{i=1}^{i=n} 17i/4+2pix(i-1) = 3080cm.
We can use substitution, since x = 6.25/n. sum 17i/4 + 12.5pi(i-1/i) = 3080. I think n is around 30? so the thickness would be 6.25/30 or around .2?

 

[Corpsecreate]

I just solved this question last night and terr13 is doing it very similar to what I did. One small slip up that I made (which is the same mistake you made) is nx = 6.25. It should be 2nx = 6.25 because 6.25 is the difference in diameter, not the radius. I'm not sure where you got n = 30 though.

Anyway my final answer was

Spoiler

t = 0.023686, n = 131.934841

 

[terr13]

i mistyped something into wolfram alpha

 

[Corpsecreate]

Dont worry about part (a) and (b).

For part (c) you get a DE with constant coefficients so you get lambda = 1 multiplicity 2. So the general solution for w(x) must be:

w(x) = ce^x + kxe^x where c and k are the two integration constants.
dw/dx = ce^x + kxe^x + ke^x

For part (d) I get y(x) = e^-x + 1/(ce^2x+xe^x).
Wolfram gives (ce^x - xe^x) for the denominator. So either I am wrong, or the website is wrong. I have more faith in the website than myself and I have looked through all my working numerous times. What answer do you get?

 

[Kole]

Say I have a geometric sequence/progression.

If the 1st and 4th numbers are given, how do I find the common ratio?

 

[Corpsecreate]

Say I have a geometric sequence/progression.

If the 1st and 4th numbers are given, how do I find the common ratio?

The form of a geometric progression is:

a, ar, ar^2, ar^3, ar^4....ar^(n-1)

You are given the first number and the fourth number, that means you now have a formula for r.

a*r^3 = T4
r = 3rd root of (T4/a)

In general:

r = (Tn-1) root of (Tn/a)

 

[Corpsecreate]

3 numbers:

x1 = 0.20472
x2 = 0.65767
x3 = 0.13761

Find the ratio of x1 : x2 : x3 with whole numbers. It doesnt have to be PERFECTLY accurate but it should be close. For example, if the numbers were simpler.

x1 = 0.1
x2 = 0.6
x3 = 0.3

The ratio would be 1 : 6 : 3

Need the answer to this quick, I have an answer but my friend over here just wont accept it as being correct.

 

[terr13]

I guess 5:10:2 is pretty close.

 

[Corpsecreate]

I get 3:10:2

My friend gets 3:6:1

 

[Deleted member]

if you want it exact, the obvious answer would be:
20472:65767:13761
(their greatest common denominator is 1)

less precise you could do 10:33:7

also your friend is way off, just tell him the 3rd number is over half the value of the first so 3:1 cannot match there. and the secodn one is over 3 times greater then the first, so again 3:6 is off.

 

[Corpsecreate]

ok good. I actually lied and my friend is really my lecturer who took away some of my marks because apparently my ratio of 3:10:2 was wrong and his 3:6:1 was correct. Of course there are infinite different ratios that apply but 3:6:1 is quite far off.

 

[terr13]

Okay, I don't know what's wrong with me when I post, but I always type the wrong thing. 3:10:2 is what I meant.

 

[Corpsecreate]

It's ok, that happens to me as well and I usually lose marks for it. Like in my recent assignment I integrated 1 as 0.5x^2 instead of just x...

 

[Lore]

I have to memorize twenty trig identities by Tuesday. Any advice on where to start?

 

[Corpsecreate]

I dont even know of 20 trig identities O.o

Do you mean identities [like Sin(2x) = 0.5Sin(x)Cos(x)] or exact values [Sin(30) = 0.5]?

 

[Browny]

I have to memorize twenty trig identities by Tuesday. Any advice on where to start?

Dead serious, this is how I remember things, and my memory is very good, maybe give it a try

http://en.wikipedia.org/wiki/List_of_trigonometric_identities
scroll to related identities
then start from sin and recite what sin =, starting from the top left. Repeat it over and over until you dont mess it up. Then move onto the second one, but start remembering it from the start again, so repeat to yourself what sin 0 w/r to cos then sin = w/r to tan/ Once you can remember those 2 in order, move to the third, reciting 1, 2 3 in order all the way to cot. Repeat this for all of the sins and then again for the rest.

Constantly attempt to recite an entire line 100% accurately and if you mess it up once, start the process again.

Its ROTE learning at its finest and you wont remember it after you're done, but I find this method to be extremely effective, made it up myself Ive used it to memorize acronyms, countries, formulas, speeches, pretty much everything.

 

[Sasha]

@Browny

I don't think that's necessary. Just remember:

tan = sin/cos
cot = 1/tan
sec = 1/cos
csc = 1/sin

sin^2 + cos^2 = 1
tan^2 + 1 = sec^2
cot^2 + 1 = csc^2

You won't have to remember any other formulas. You'll just have to solve an equation to get it.

For example, I don't remember the formula for sec, but I do know:

sin^2 + cos^2 = 1
Hence
cos^2 = 1 - sin^2
Hence
cos = + - sqrt (1 - sin^2)
Hence
sec = + - 1/sqrt (1 - sin^2)

EDIT: I guess I'm a little late for Tuesday. Oh well, I tried.

 

[Browny]

Well my method would still work for memorizing those 7, but yeah for another 5 what you said would work better lol

 

[Elyssa Xey Hexen]

Anyway, the best way to memorize is to think about it for a period of time such as 10-15 seconds such as reciting it aloud or running it through your head. Then, clear it from your mind for a few minutes. After that, try to recall what you learned.

You have to practice retreiving the information after some time from viewing it or else you will forget it.

A further way to practice retriving the information is working through a quick problem that involves it.

Spoiler

Uhh... What is the basic trig stuff to memorize? I know the first 2 calculus courses, you really only need:
-the unit circle
-know what sec, cos, cot, tan are.
-sin^2(x) + cos^2(x) = 1
(from this you can derive two other identities simply dividing the whole thing by either cos^2(x) or sin^2(x), then simplfying)
-double angle formulas
(which can be derived from eulers formula and some further usage of sin^2(x) + cos^2(x) = 1)

Apart from those, the rest are pretty unused. I guess the addition and subtraction ones are useful.

 

[Life]

(I think Browny will get a sense of deja vu from this, but) while we're on the subject of memorizing stuff, anyone have a good mnemonic for Law of Cosines/Law of Sines? Or do I just have to memorize the old-fashioned way(s)?

 

[Elyssa Xey Hexen]

anyone have a good mnemonic for Law of Cosines/Law of Sines? Or do I just have to memorize the old-fashioned way(s)?

Law of sines say that these ratios are equal to each other:
(Sin A / a) = (Sin B / b) = (Sine C / c)
[collapse=image of triangle]

[/collapse]
So, if you notice, 1) each ratio is the sine of an angle divided by its opposing side length. And law of sines says that 2) each of those ratios are equal to each other. Remember those two things and your good to go. Easy, right? "Two sides-two angles"

Not sure for law of cosines for an easy way to remember it. Never use it much.

 

[Deleted member]

aren't you supposed to put the length of the sides outside the parentheses?

 

[Yonder]

I thought i'd need this topic and all the help I can get. Turns out, Math applications is insultingly easy and im passing with flying colors so maybe see ya next year haha

 

[Elyssa Xey Hexen]

aren't you supposed to put the length of the sides outside the parentheses?

My mistake. (sin(A)/a) = (sin(B)/b) = (sin(C)/c)
^That.
Always get sloppy with extra parathesis.

 

[ballin4life]

@Browny

I don't think that's necessary. Just remember:

tan = sin/cos
cot = 1/tan
sec = 1/cos
csc = 1/sin

sin^2 + cos^2 = 1
tan^2 + 1 = sec^2
cot^2 + 1 = csc^2

You won't have to remember any other formulas. You'll just have to solve an equation to get it.

I don't remember the sec^2 = 1 + tan^2 one off the top of my head (I forgot which side the +1 goes on), but I know you can get it from sin^2 + cos^2 = 1 and just divide both sides by cos^2. You can get the one below it similarly by dividing through by sin. So those are just versions of sin^2 + cos^2 = 1.

I've been using double angle formulas a bit recently for my complex analysis class.

 

[JustNoOne]

need a little help for vectors ^^

L1 : (3,4)+t(5,6)
L2: (-2,3)+s(2,7)

At what point do these lines intersect at?

 

[Corpsecreate]

need a little help for vectors ^^

L1 : (3,4)+t(5,6)
L2: (-2,3)+s(2,7)

At what point do these lines intersect at?

You should be able to generate a simultaneous equations with two unknowns and then you can solve it for t and s.

for L1 you know that its x position is 3 + 5t and its y position is 4 + 6t
for L2 you know that its x position is -2 + 2s and its y position is 3 + 7s

for the vectors to intersect, both their x and y positions must be equal. That should point you in the right direction.