A place to air possible concerns or difficulties in understanding ProjectEuler problems. This forum is not meant to publish solutions. This forum is NOT meant to discuss solution methods or giving hints how a problem can be solved.
Forum rules
As your posts will be visible to the general public you are requested to be thoughtful in not posting anything that might explicitly give away how to solve a particular problem.
This forum is NOT meant to discuss solution methods for a problem.
In particular don't post any code fragments or results.
Don't start begging others to give partial answers to problems
Don't ask for hints how to solve a problem
Don't start a new topic for a problem if there already exists one
What you are required to do is to give [sum]x accurate to 9 decimal places.
This doesn't directly mean that the V(x) need to be accurate to 9 decimal places.
That depends on the function V(x).
War ruins the life and health of untold numbers of innocent children.
RishadanPort wrote:...
I compute V(0) for alpha = 30, and R = 3
and get 32.64838334800159 instead of 32.648388556.
I am assuming that 32.648388556 <-- is what should be expected after you do V(0) yes? ...
Yes. For alpha = 30 deg. and R = 3, V(0) = 32.648388556 when rounded to 9 decimals.
I'm very confused on how the volume missing is 32.648388556
With A = 30 degrees, Radius = 3, x =0 I get ~28.2743
In fact I get this answer with twomethods leading me to believe I have misunderstood the question.
I used shell method to integrate the empty space integrate( 2 * pi * x * (sin(A)*x), x = 0 to R) = 28.2743
Alternately I calculate the volume of cone (1/3 * pi * R^2 * (sin(A) * R) = 9*pi/2) and subtract that from a cylinder (pi * R^2 * (sin(A) * R) = 27*pi/2) = 27*pi/2 - 9*pi/2 = 18*pi/2 = ~28.2743
kingvash wrote:I'm very confused on how the volume missing is 32.648388556
With A = 30 degrees, Radius = 3, x =0 I get ~28.2743
In fact I get this answer with twomethods leading me to believe I have misunderstood the question.
I used shell method to integrate the empty space integrate( 2 * pi * x * (sin(A)*x), x = 0 to R) = 28.2743
Alternately I calculate the volume of cone (1/3 * pi * R^2 * (sin(A) * R) = 9*pi/2) and subtract that from a cylinder (pi * R^2 * (sin(A) * R) = 27*pi/2) = 27*pi/2 - 9*pi/2 = 18*pi/2 = ~28.2743
We shall let the amount of space wasted in cubic metres be given by $V(x)$. If $x = 1.114785284$, which happens to have three squared decimal places, then the amount of space wasted, $V(1.114785284) \approx 36$. Given the range of possible solutions to this problem there is exactly one other option: $V(2.511167869) \approx 49$.
It feels to me as though this question has been edited and lost some key explanation. Should this say something like the following?
We shall let the amount of space wasted in cubic metres be given by $V(x)$. If $x = 1.114785284$, which happens to have three squared decimal places, then the amount of space wasted, $V(1.114785284) \approx 36$, which is an integer square. Given this angle of repose and diameter, there is exactly one other value of $x$ which gives an integer square wastage: $V(2.511167869) \approx 49$.
