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.
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.
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?
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:
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??
Please PLEASE read the following and solve this problem:
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!!!!!!
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????
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.
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.*
Mathsyperson: Your right this isnt the formula for the Surface area.........but according to my question i had to substitute the height H of Volume to the surface area.
as the formula is S.A= Pi*R*S ...........and s= square of [R2+H2]......so i substituted the height in this formula. Which came to the above term. can it be simplified more???