They're getting harder!
1. Z=-13+5i. A= the digit in the third decimal place of arg Z in radians (positive solution).
2. The matrix M has 4 in the top left position, -4 in the top right position, -1 in the bottom left position and 2 in the bottom right position. B= the number in the bottom right position of the inverse of M.
3. dy/dx +2y*Tan x = Sin(2x) has a solution y=2 at x=pi/3. At x=pi/6, y=6-(3/4)LnC. Find C.
4. r=1+2CosG, (where G is usually represented as ‘theta’), and takes values from zero to pi/2. At the point P, the tangent to the curve is parallel to the initial line. Given that O is the pole, the length OP=(U + sqrt(W))/V, where sqrt(W) means the square root of W. D=W/U-V.
5. Cosh (2x) – 5 Sinh (x) = 4 has two solutions. The larger of the two is Ln (3 + sqrt(F)). Find F.
2: A quick Google and I have B=4.
5: Courtesy of Wolfram, F=10.
The rest: Not even sure I understand the questions!
By the way, I particularly liked your permutation explanation. I could handle the "all different" case but didn't know what to do with the repeated letters.
#1: The function "arg" of a complex number is the angle you get when you connect the number zero to the number a+bi in the complex plane
. It is given by arctan(b/a), in this case arctan(-5/13), which is 2.7744188.. (note: if the calculator says -0.36716... you need to add pi to this to get a positive value). Note that Wolfram can do this directly
#2: There is a fairly straightforward way to compute the inverse of a matrix, which I assume you found somewhere like this
#3: There is a standard method
to solve this differential equation. At this point, there is nothing clever going on -- just a basic calculus problem from second-year engineering math. If you are very careful
about how you type in the problem, Wolfram is able to solve it
. But if you aren't careful about the syntax, it won't work correctly.
#4: This is just a standard calculus problem using polar coordinates
. It's easy to calculate distance in polar coordinates, the only question is where is the point P? You can see the graph here
Now what you want is to refer to the formula here
, which tells you how to compute derivatives in polar coordinates, which means first you compute:
r'(t) = -2sin(θ)
dy/dx = [ r' sin(θ) - r cos(θ) ] / [ r' cos(θ) - r sin(θ) ]
By "parallel to the initial line" I am going to assume they mean "horizontal" in which case you want the numerator of dy/dx to be zero, i.e.
-2sin(θ)sin(θ)+(1+2cos(θ))cos(θ) = 0
You can simplify this to:
cos(θ) + 2 cos(θ)^2 - 2 sin(θ)^2 = 0
but by replacing sin(θ)^2 with 1 - cos(θ)^2, we get:
4 cos(θ)^2 + cos(θ) - 2 = 0
Now, we can let u = cos(θ) and just solve for that:
4 u^2 + u - 2 = 0
u = (-1+√(33))/8 (you can ignore the negative solution with -√(33))
Now, that's really saying:
cos(θ) = (-1+√(33))/8
But what is the quantity we want? It's not θ (thank goodness), it's:
r(θ) = 1 + 2cos(θ) = 1 + 2 ((-1+√(33))/8) = (3+√(33))/4
This gives you U=3, W=33, V=4, and so D = 33/3-4 = 7.
#5: What you have to do here is apply a standard identity:
cosh(2x) = 2 cosh(x)^2 - 1 = 2 sinh(x)^2 + 1
You want to use the second version, so that it's all in terms of sinh(x):
2 sinh(x)^2 + 1 - 5 sinh(x) = 4
You just let u=sinh(x) and you have a quadratic:
2u^2 - 5u - 3 = 0
(2u+1)(u-3) = 0
u = -1/2 or u = 3. Then you can just compute:
x = arcsinh(3/2) = http://www.wolframalpha.com/input/?i=arcsinh%28-1%2F2%29
x = arcsinh(1) = http://www.wolframalpha.com/input/?i=arcsinh%283%29
You know which one of these is correct already.