Post Your Problem Here (1 Viewer)

[victorling]

no problem!

 

[...]

Originally posted by Seraph
Um....

....

.....
Crap? !!!

thx

thats the problem being an arsenal supporter

 

[victorling]

arsenal is fine

 

[...]

arsenal is gay


nuff said!!!

 

[Numero Uno]

Oh New soccer thread

Arsenal is gay


and so is ... Henry

 

[victorling]

which team do u like then?

 

[KeypadSDM]

Originally posted by Seraph
yes , i need help with approaching these type of questions.. cause i really get lost with them

A wire piece is 6m long it is cut into two parts , one is used to form a square and the other piece a rectangle whose length is three times its width. Find the length of the two parts if the sum of the area is a maximum.....

Wait a second. This question's fundamentally incorrect.

Do you mean when the area's a minimum? Because the maximum area (assuming you have to cut it into 2 pieces) is just below 9 m^2

 

[victorling]

thx maths god

 

[Seraph]

oh god

Using Simpsons rule

Integrate (x = 2 x =1 ordinates) 1/(2 x Sqrtx)dx

argh so frustrating i get 0.37 as my answer book says its like 0.41 or something

anyone help me out here??? and can anyone suggest some easier ways to get through the calculations of these fast.....

 

[Seraph]

*tappa tap tap *

 

[Heinz]

Originally posted by Seraph
oh god

Using Simpsons rule

Integrate (x = 2 x =1 ordinates) 1/(2 x Sqrtx)dx

argh so frustrating i get 0.37 as my answer book says its like 0.41 or something

anyone help me out here??? and can anyone suggest some easier ways to get through the calculations of these fast.....

There isnt an easy method to do these questions, just draw a table of values up and try not to skip any steps because you might make an error. integral = h/3(ends + 4(odd) + 2(even). Im assuming the questions asking you to use 3 function values. It becomes a little too difficult otherwise.

 

[Affinity]

it's better to remember simpson's rule as
h/3 *(1424242424242424....1)

 

[Xayma]

Originally posted by Affinity
it's better to remember simpson's rule as
h/3 *(1424242424242424....1)

Hmm I just do multiple applications since (b-a) is a constant it ends up the same it just makes it easier to remember since they aren't going to ask for something with 19 function values or something insane like that since that just waists time.

 

[Seraph]

wait

isnt there 2 versions of the simpsons rule

one that is h/6 and h/3

the h/3 is for sub-intervals or something

btw heinz thanks for the working out , how did u get 1/5 as h though

h = (b- a )/ n
n being the number of function values

i thought it would be (2 -1) / 3

= 1/3
?????

and also when you group the odd and even numbers do we have to actually substitute the values into the equation first to see whether they are odd or even ?

 

[Heinz]

Originally posted by Seraph
wait

isnt there 2 versions of the simpsons rule

one that is h/6 and h/3

the h/3 is for sub-intervals or something

btw heinz thanks for the working out , how did u get 1/5 as h though

h = (b- a )/ n
n being the number of function values

i thought it would be (2 -1) / 3

= 1/3
?????

If you keep memorising formula, you just confuse yourself. Think of h as the difference between one function value and another. (its width). so if its 3 function values between 1 and 2 obviously your function values are 1, 1.5 and 2 so the distance between 1 and 1.5 is 0.5. if it were 5 fuction values, the distance would be 0.25 or if it were 3 functions between 0 and 5, h would be 2.5 and so on.

n is actually the number of intervals so h is (2-1)/2 = 1/2 .thus h/3 becomes 1/6 or (0.5/3). As for the different rules, i really cant remember. Ive always used h/3(sum of beginning and end function values + 4(sum of odd function values) +2(sum of even function values). the FOTE method (four odd two even)

 

[~*HSC 4 life*~]

i just use b-a/b (f(first) + 4xf(middle) + f(last))

so much easier 4 me

 

[CM_Tutor]

Going back to Seraph's question about the 6 m wire cut into two parts, and then bent into a square and a rectangle, KeypadSDM makes a good point. The question (as stated) asked for the MAXIMUM area. There are two possibilities:

Case 1: The question should have said MINIMUM, in which case victorling's answer is correct, the pieces are 18/7 m (forms the square) and 24/7 m in length, and the minimum total area is 27/28 sq m. (Note that, in this case, victorling's solution should have noted that A'' = 7/32 for all x, and thus the solution found was indeed a minimum - exam markers expect an answer to include a test for the nature of the MAX/MIN.)

Case 2: The question did indeed mean MAXIMUM, in which case some more work is needed. This is an important trick to look for - it comes up in exams.

Since we are talking about a piece of wire, there is an inherent constraint within the problem, that is that x lies in the closed interval [0, 6] (ie. 0 <= x <= 6). If the only stat. point in this domain is a MIN, then the MAX must be at one of the end points -try drawing a graph if you aren't convinced.

If x = 0, then the entire length of wire is used to form the rectangle (ie. area of sqaure is 0 sq m), and the total area is 1.6875 sq m.

If x = 6, then the entire length of wire is used to form the square (ie. area of rectangle is 0 sq m), and the total area is 2.25 sq m.

And thus the answer to the question is: the maximum area is 2.25 sq m, and occurs when the entire 6 m length of wire is used to form the square.

WATCH OUT FOR THIS - As I said above, it does pop up in exams occasionally, and it occurs in two forms (Form 1: asked for a MAX / Min, but calculus only shows a MIN / MAX. Form 2: calculus shows a MAX / MIN as desired, but it lies outside the constraint domain), and has also been asked in the HSC.

 

[KeypadSDM]

I would have to disagree.

The maximum area is nearly, but not equal to 2.25 m^2 as the question states that some wire must be used to form both the square and the rectangle.

I.e. 0 < x < 6, not 0 <= x <= 6

 

[m1nx]

sequences + series make me dizzy. any help would be much appreciated!

The sum of the first 10 terms of an arith. series is 60 and the sum of the next 10 terms is 460. Find the 6th terms of the series.

 

[Calculon]

sum of n terms = (n/2) * (a+ l) with a being the smallest term and l being the largest and d being the common difference

60 = (10/2)*(2a+9d) __460 = (10/2)*(2a+29d)

12 = 2a+9d________ 92 = 2a+29d

(12-9d)/2 = a_______ 92 = (12-9d) + 29d

__________________ 92 = 12 + 20d

__________________ 80 = 20d

__________________ 4 = d

12 = 2a + 9*4

-12 = a

6th term = a + 5d

= -12 + 5*4
= 8