I need to know how to find an unknown variable in dart to solve a math equation. B is the unknown variable and the known is the known variable.
double B = 0.0;
double known = 1423.0;
known = B -
((B * 0.105 + B * 0.005 + B * 0.055) +
((B - (B * 0.105 + B * 0.005 + B * 0.055)) * 0.16));
I am not aware of any such package to solve equations, but you can simplify the equation itself just like you do in mathematics. Below is such a solution-
known = B -
((B * 0.105 + B * 0.005 + B * 0.055) +
((B - (B * 0.105 + B * 0.005 + B * 0.055)) * 0.16));
known = B -
((B * 0.165) +
((B - (B * 0.165)) * 0.16));
known = B -
((B * 0.165) +
((B * 0.835) * 0.16));
known = B -
((B * 0.165) +
(B * 0.1336));
known = B -
(B * 0.2986);
known = B * 0.7014;
B= known / 0.7014;
Related
How would I find the maximum possible angle (a) which a rectangle of width (W) can be at within a slot of width (w) and depth (h) - see my crude drawing below
Considering w = hh + WW at the picture:
we can write equation
h * tan(a) + W / cos(a) = w
Then, using formulas for half-angles and t = tan(a/2) substitution
h * 2 * t / (1 - t^2) + W * (1 + t^2) / (1 - t^2) = w
h * 2 * t + W * (1 + t^2) = (1 - t^2) * w
t^2 * (W + w) + t * (2*h) + (W - w) = 0
We have quadratic equation, solve it for unknown t, then get critical angle as
a = 2 * atan(t)
Quick check: Python example for picture above gives correct angle value 18.3 degrees
import math
h = 2
W = 4.12
w = 5
t = (math.sqrt(h*h-W*W+w*w) - h) / (W + w)
a = math.degrees(2 * math.atan(t))
print(a)
Just to elaborate on the above answer as it is not necessarly obvious, this is why why you can write equation:
h * tan(a) + W / cos(a) = w
PS: I suppose that the justification for "why a is the maximum angle" is obvious
In this triange:
Given the areas of triangles UPZ ZPW and WPY, how do you calculate the total area?
I've already found the solution from the available submissions at the website. But I want to know how to derive that solution.
cin >> a >> b >> c;
// a is UPZ, b is ZPW, c is WPY
double n = b*(a+b)*(a+b+c);
double d = b*(a+b)-(a*c);
cout << (n / d) ;
Indeed, this question is kind of off topic, it is a geometry problem. The way to find the area of the big triangle UVW is to apply the link between areas and ratioes of lengths of the segments of the triangle UYW and then to apply Menelaus' theorem to derive the ratio
WY/WV which reveals the ratio between the areas of the triangle UYW and UVW.
Let h_p be the length of the height from point P to the edge UW. Then
a = UZ * h_p / 2 and b = ZW * h_p / 2
Thus:
a / b = (UZ * h_p / 2) / (ZW * h_p / 2) = UZ / ZW
Let h_W be the length of the height from point W to the line UY
a + b = Area(WPU) = PU * h_W / 2 and c = YP * h_w / 2
Thus:
c / (a + b) = (YP * h_W / 2) / (PU * h_W / 2) = YP / PU
By Menelaus' theorem for the triangle UWY and the line VZ, with P on VZ, we get:
1 = ( VW / VY ) * ( YP / PU ) * ( UZ / ZW ) = ( VW / WY ) * (c / (a + b)) * (a / b)
so
VY / VW = (c * a) / ( b * (a + b))
and therefore:
WY / VW = 1 - (VY / VW) = 1 - (c*a) / ( b*(a + b)) = (a*b + b^2 - a*c ) / (a*b + b^2)
Let h_U be the length of the height from the point U to the edge VW. Then
Area(UVW) = VW * h_U / 2
and
Area(UYW) = a + b + c = WY * h_U / 2
Hence
Area(UVW) / Area(UYW) = Area(UVW) / (a + b + c) = (VW * h_U / 2) / (WY * h_U / 2) = VW / WY
so
Area(UVW) / Area(UYW) = VW / WY = (a*b + b^2) / (a*b + b^2 - a*c)
Area(UVW) / Area(UYW) = Area(UVW) / (a + b + c) = (a*b + b^2) / (a*b + b^2 - a*c)
Finally, we obtain the formula:
Area(UVW) = (a + b + c) * (a*b + b^2) / (a*b + b^2 - a*c)
Area(UVW) = b * (a + b) * (a + b + c) / (b*(a + b) - a*c)
The nonlinear equation is following,A,B,C,D，E are know. I want to rearrange the formation of the equation. Let the X at the left of equation,and let other parameters all at the right of equation. such as X= A*B/D+E^2/C
Is there are some software to do this?such as R.
Try this -- ignore the warnings from that XML package that have started recently.
library(Ryacas)
A <- Sym("A")
B <- Sym("B")
C <- Sym("C")
D <- Sym("D")
E <- Sym("E")
X <- Sym("X")
Solve(E == A * B * (X + C) / (A + B * (X + C)) - A * B * (X + D + C) / (A + B * (X + D + C)), X)
giving:
expression(list(X == (root((2 * (E * A * B) + (2 * (E * B^2 *
C) + E * B^2 * D))^2 - 4 * (E * B^2 * (E * A^2 + (2 * (E *
A * B * C) + E * A * B * D) + (E * B^2 * C^2 + E * B^2 *
C * D) + A^2 * B * D)), 2) - (2 * (E * A * B) + (2 * (E *
B^2 * C) + E * B^2 * D)))/(2 * (E * B^2)), X == -(2 * (E *
A * B) + (2 * (E * B^2 * C) + E * B^2 * D) + root((2 * (E *
A * B) + (2 * (E * B^2 * C) + E * B^2 * D))^2 - 4 * (E *
B^2 * (E * A^2 + (2 * (E * A * B * C) + E * A * B * D) +
(E * B^2 * C^2 + E * B^2 * C * D) + A^2 * B * D)), 2))/(2 *
(E * B^2))))
An alternative to the above if you have specific values for A, B, C, D, E would be to numerically solve it using, for example, uniroot.
I have an algorithm written as follows but I need to write that code into R. I have included the algorithm and the R code. I am not sure if that is represented well enough. To write the R code in the sequential order is not straight forward. I am sorry for not providing all the values of the variables here. I am not sure of the output yet which is the reason I am unable to show the required. It is more of a theoretical question.
Algorithm
VBDMAX = (va - VG) * 0.79 * (dep / D) ^ -1.21
VBOWMAX = -0.7 * VBDMAX
VBOWX = 0
' SKIP BOW IF -10D<X<15D OR OUTSIDE EDGE OF BARGES
If Y > B / 2 Then GoTo 200
If X < -10 * D Then GoTo 200
If X >= 15 * D Then GoTo 200
VBOWX = X * VBOWMAX / (10 * D) + VBOWMAX
If X <= 0 Then GoTo 200
VBOWX = X * (VBDMAX - VBOWMAX) / (5 * D) + VBOWMAX
If X <= 5 * D Then GoTo 200
VBOWX = -X * VBDMAX / (10 * D) + 15 * VBDMAX / 10
200 ' end bow
This is the R code that I have written
VBDMAX = (va - VG) * 0.79 * (dep / D) ^ -1.21
VBOWMAX = -0.7 * VBDMAX
VBOWX = 0
# SKIP BOW IF -10D<X<15D OR OUTSIDE EDGE OF BARGES
VBOWX <- ifelse ((Y>B/2 | X < -10*D | X>=15*D), 0,X*VBOWMAX/(10*D)+VBOWMAX)
VBOWX <- ifelse (X<=0 , X * (VBDMAX - VBOWMAX) / (5 * D) + VBOWMAX,
ifelse(x <=5*D, -X * VBDMAX / (10 * D) + 15 * VBDMAX / 10))
You can use ifelse constructs but you will need to nest those:
VBDMAX = (va - VG) * 0.79 * (dep / D) ^ -1.21
VBOWMAX = -0.7 * VBDMAX
VBOWX =
ifelse(Y > B / 2 || X < -10 * D || X >= 15 * D,
0,
ifelse(X <= 0,
X * VBOWX / (10 * D) + VBOWMAX,
ifelse(X <= 5 * D,
X * (VBDMAX - VBOWMAX) / (5 * D) + VBOWMAX,
-X * VBDMAX / (10 * D) + 15 * VBDMAX / 10
)
)
)
Understanding your question as how to translate "goto" statements to R, there are the following posibilities (if really needed) besides or in adjunction to the (often more appropriate) if/ifelse constructions as you already did:
a) entire code (for severe errors or if problem is solved): if (condition) stop("explain why...") or stopifnot(condition)
b) from within loops: see next and break
c) from within function: if (condition) return(), stopping the function here
Imagine a Binary Erasure Channel as depicted on Wikipedia.
One equation describing the mutual information is following:
I(x;y)
= H(x) - H(x|y)
= H(x) - p(y=0) • 0 - p(y=?) • H(x) -p(y=1) • 0
Why is it "p(y=?) • H(x)" and not "p(y=?) • H(x|y=?)"?
It can be proved using Bayes' theorem.
The channel:
x y
1-f
0 --------> 0
\
\ f
+------> ?
/
/ 1-f
1---------> 1
Let input distribution be P(x) = {p(x=0)=g; p(x=1)=1-g}
Then:
p(x=0/y=?) = p(y=?/x=0) * p(x=0) / p(y=?)
p(x=0/y=?) = (f * g) / (f * g - f * (1 - g)) = g;
p(x=1/y=?) = p(y=?/x=1) * p(x=1) / p(y=?)
p(x=1/y=?) = (f * (1 - g)) / (f * g - f * (1 - g)) = 1 - g;
As result:
p(x=0/y=?) = p(x=0)
p(x=1/y=?) = p(x=1)
From the defenitions of entropy and conditional entropy:
H(X) = p(x=0) * log(1 / p(x=0)) + p(x=1) * log(1 / p(x=1))
H(X/y=?) = p(x=0/y=?) * log(1 / p(x=0/y=?)) + p(x=1/y=?) * log(1 / p(x=1/y=?))
So:
H(X) = H(X/y=?)