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## #1 Re: Help Me ! » Same Problem!!!!! » 2006-02-28 02:26:17

Comeon people i seriously need help!!!!! Its the last time i will ever ask .......Ricky ur help will be appreciated, i know u know this pretty well, so whenever u find time post a reply. Thanks

## #2 Re: Help Me ! » Same Problem!!!!! » 2006-02-27 02:47:57

Heres the Picture:

## #3 Help Me ! » Same Problem!!!!! » 2006-02-27 02:43:47

johny
Replies: 5

Well this question has been posted previously, but the previous one was inaccurate and this one has changed values.

A Lorry has a downwards wiping windscreen Wiper on a 40cm long metal arm. The rubber wiper part is at the end of the arm and is an articulate double module that alters in length during motion. The rubber alters between 30cm and 20cm, during a wiping "sweep" of the glass.

The full angle of the sweep is 60 degrees.

The bottom end of the rubber follows a path according to the equation

y = 1/30x (x-10* square root of 3)

The upper end of the rubber follows a path according to the equation:

y = 1/15 (x-5*square root of 3)^2 + 25

And we need to calculate the area of glass cleared in one full sweep of the arm.

## #4 Re: Help Me ! » One problem!!!!! » 2006-02-20 02:04:19

Thanks alot Irspow!!!!! Sorry i couldnt reply before as my PC wasnt working........and i asked a maths teacher about the Windscreen washer problem which u said wasnt correct, well got to say u were right as the Question did contain inaccuracies!!!! Sorry for bothering u soo much.

## #5 Re: Help Me ! » One problem!!!!! » 2006-02-15 10:45:36

Ok, i get it now, and yes i am using radians!!!!! and just one last question......how would i differentiate this equation which would finally give me an equation for Velocity of the burner at any time t (as velocity is rate of change of displacement)???

## #6 Help Me ! » One problem!!!!! » 2006-02-15 09:46:48

johny
Replies: 5

Sorry guys recently i have been asking too many questions but i have been struggling with such topics!!!!! So whenever anyone finds time, could hopefully post a solution to the following question:

A small high-intensity gas burner runs on a narrow rail, forwards and backwards underneath plastic ski brush components.

Each components is heated for one cycle of the burner, and this lasts for 30 seconds (the period T)

The displacement of the burner is given by the equation (shown in the pic).

So ignoring the thickness of the burner block, what is the minimum length of narrow rail needed?

## #7 Re: Help Me ! » Integration problem!!!!! » 2006-02-13 20:36:43

Well i havent checked  anything yet, as soon as i find out i will post it in the same topic

## #8 Re: Help Me ! » Integration problem!!!!! » 2006-02-13 05:50:32

Thanks once again irspow!!!!

Pic:

## #10 Help Me ! » Integration problem!!!!! » 2006-02-11 22:56:04

johny
Replies: 5

Hello guys, well recently i have been really struggling with integration. Heres the question:

A piece of machinery has a cutter which is cooled by a water soluble oil. The cooling solution is held in a reservoir in the shape of frustrum of a cone.

We need to find the volume by integration of the reservoir.

Heres the pic:

## #11 Re: Help Me ! » A problem!!!! » 2006-02-06 08:12:14

Seriously guys thanks alot for giving soo much time to it!!!!! If i find something out then i wil let u know.........and once again thanks

## #12 Re: Help Me ! » A problem!!!! » 2006-02-04 23:07:34

I dont want to be pain, but just one last question becuase i too am getting a bit confused.....so i need to simplify the equation given on post no. 18???? And that would give me the area????

## #13 Re: Help Me ! » A problem!!!! » 2006-02-04 08:09:00

i doubt u guys are thick heads but i sure am, so if u have time then it would be awesome if u post the area and i would be really greatful

## #14 Re: Help Me ! » A problem!!!! » 2006-02-04 07:46:37

Ok so when i put the values in the equation u gave me before would give me the area of glass cleared???

## #15 Re: Help Me ! » A problem!!!! » 2006-02-04 06:51:58

I guess the wiper is fixed to the blade below 10cm from the top, meaning at the center.

## #16 Re: Help Me ! » A problem!!!! » 2006-02-04 06:31:45

Hey ricky.....i know it does seem confusing but i posted the exact question and diagram from the book!!!

## #17 Re: Help Me ! » A problem!!!! » 2006-02-04 05:22:10

Hey irspow, well it doesnt say any rate so it might aswell be constant. The only think it says is the rubber alters between 30cm and 20cm, thats about it.....so i would think its constant rate of change. If any more questions, not a prb !!!!

Heres the pic!

## #19 Re: Help Me ! » A problem!!!! » 2006-02-03 22:46:22

Yeah i am really sorry i forgot to post the pic!!!! i will upload it!!!!

## #20 Help Me ! » A problem!!!! » 2006-02-03 06:37:10

johny
Replies: 38

Heres the question:

A lorry has a downwards wiping windscreen wiper on a 40cm long metal arm. The rubber wiper part is at the end of the arm and is an articulate double module that alters in length during motion. The rubber alters between 30cm and 20cm, during a wiping sweep of the glass.

The full angle of sweep is 120 degrees and motion is symmetrical about the y-axis.

The bottom end of the rubber follows a path according to the equation y= 1/50x^2

The upper end of the rubber follows a path according to the equation y= 1/50x^2 + 30.

We need to calculate the area of the glass cleared in one full sweep of the arm.

I couldnt really do the question, but i believe it does involve integration. Can anyone please helo me out here??

## #21 Help Me ! » A mighty Confusion!!!! » 2005-12-04 06:35:03

johny
Replies: 1

I am basically doing a frustrum of a cone, what i need to do in the question is re-arrange the volume for cone to make H the subject and substitute this H in the Surface area formula for cone. If u get what i mean!!!!!!!!! Now the problem is this, i made a diagram of a truncated cone, and put some dotted lines to make it into a whole cone.........I said the dotted lines to be as 1/3 of the radius and height. SO when i did the volume expression it came something like this:

V = 1/3*pi*r^2*h    -   1/3*pi*r^2/9*h/3.  ........................(i)

When u simplify the equation it come up something like this:

h = 81V/26*pi*r^2

I had to make volume fixed to 600.

so:  h =  1869 / pi*r^2.

Now this is where the problem comes:

Now when i substitute this equation into surface area as i said i had to:

S.A. = Pi*R*S   where S= square root of H^2 + R^2

S.A. = Pi*r  Square root of 1869/pi*r^2  + r^2

Well here it is, if i have already subtracted 1/3 of height and radius in the volume formula in equation (i)
, then do i need to do, divide remaining r^2 with 3??? Or have i already subtracted the r and h in equation (i)
.
S.A. = Pi*r/3   square root of 1869/Pi*r^2  +  r^2/3          if u get what i mean???

Please try to quickly explain this, as i need to hand this in by tommarrow!!!!!!

## #22 Re: Help Me ! » Differentiation!!!!! » 2005-12-04 01:57:37

I didnt want to start a new topic on this but i am very confused about the following:

= Pi* R * S          where S= Square root of H^2 + R^2

U know i was doing a question for truncated cone.........so what i did was i put some dotted lines on that truncated cone to make it like a cone and i said that the dotted lines were 1/3 of the whole cone, as i wanted to find the surface area, meaning the radius is 1/3, height is 1/3 and slant height (s) is 1/3.  So can we write:

= Pi * R/3 * Square root of R^2 + H^2  /  3  ????    (note: 3 is not in square root, its seperate)

In simple numbers i have done :  Pi*r/3*s/3

I hope u understand the problem????

## #23 Re: Help Me ! » Differentiation!!!!! » 2005-12-03 09:01:35

Thanks ryos and irspow for spending time on my question ......Irspow, well the formula was originally for the volume and surface area of a cone, my shape was a truncated cone, but as i had too many variables in the formula i.e height, radius. I had to remove one of them and i did it by saying the short part of the cone is 1/3 of the whole cone......what i did next was re-arranged the formula for volume to take height, then i put the equation of height in surface area.

I had volume fixed to 600cm^3.

Then i substituted the equation of height into the surface area. It came to something like this:

S.A= Pi*R * square root of R^2 + 2025/Pi*R^2

Then i had to subtract this equation by 1/3 R and 1/3 of H, when i simplified it came to the equation which i gave u!!!!! .

Hope u understand. And thanks once again.

## #24 Help Me ! » Differentiation!!!!! » 2005-12-03 01:33:01

johny
Replies: 5

I am seriously seriously stuck onto this problem!!!! I really need some help, i need to hand the whole thing by monday, the following question is the only part i am badly stuck on, please help!!!! I need to differentiate this:

= Square of Pi^2*r^6 + 4100625 / r    -  Square of Pi^2*r^6 + 36905625 / 3r.

*Note that the r and 3r dividing the square of above expression are not in square, they are seperate and just dividing.*