Find the exact length of the polar curve described by: r= 4e^(-6Θ) on interval 7pi/5 ≤ theta ≤ 9pi?
s = int[sqrt{[r(Θ)]² + (dr/dΘ)²}] dΘ s = int[sqrt{16e^(-12Θ) + (-24e^(-6Θ))²}] dΘ s = int[sqrt{16e^(-12Θ) + 576e^(-12Θ)}] dΘ s = int[sqrt{592e^(-12Θ)}] dΘ s = 4sqrt(37)int[e^(-6Θ)] dΘ s = -(2/3)sqrt(37)e^(-6Θ) from 7π/5 to 9π s = -(2/3)sqrt(37)[e^(-54π) - e^(-42π/5)] s = (2/3)sqrt(37)[e^(-42π/5) - e^(-54π)]
Express the complex number represented by P in the form re^io, giving the exact value of o and the value of r correct to 3 sig fig.?
I'm sure there are more efficient ways of working this out but the one that springs to mind is the following: The set of points A = {z: Arg(z - 2i) = π/6} can be expressed by the line L1: y = tan(π/6)x + 2 or L1: y = (1/√3)x + 2 The set of points B = {z: |z - 3| = |z - 3i|} is given by the line that bisects y = -x + 3 which is L2: y = x Since the point P lies at the intersection, we require (1/√3)x + 2 = x From which x = 3 + √3 y = 3 + √3 The point P is therefore represented by z = (3 + √3) + i (3 + √3) Since r = |z| = √(x² + y²) = √2(3 + √3)....(after a little arithmetic) θ = Arg(z) = arctan(y/z) = arctan(1) = π/4 radians in polar coordinates z = √2(3 + √3)e^(i π/4)
Complex numbers. Converting from rectangular to complex?
Represent the complex number -2-3i with polar coordinates. I got the following answer: 3.60(cos56-sin56i) However this is the wrong answer. I do not understand how I have gone wrong as: a=-2 and b=-3 and i used the formulas: r=sqrt(a)^2+(b)^2 and theta = tan^-1 (b/a) these did not give me the right answer. Could someone please do the question and explain (as if you were explaining to someone who had no idea what they were doing!) what you are doing as you go along. I cold really use the help. Thank you so much! Ps. Could you please explain whether to use a ''+'' or ''-'' in the final answer. eg. 3.60(cos56-sin56i) OR is it 3.60(cos56+sin56i). As i do not understand that either. Thank you :)
What is the equation that gives you a heart on the graph?
I actually like this equation:
Solid Mechanics: What are the differences between plane stress and plane strain conditions?
Plane strain: Consider a example where you have one of the dimension very long, say gravity dam as shown in figure.Now extract length and draw it seperatley, divide the length into small cubes 1 2 and 3 and their faces is a b and c respectively.Now , say if loading is vertically upward such that it tries to stretch the dam in 1–1 direction, then there will be contraction in 3–3.If you see cube 1 tries to contractat the same time cube 2 and cube 3 independently wants to contractthat means the face b goes right due to contraction of 1 and also at the same time goes left due to contraction of 2That means, two people with equal strength trying to pull face b, that means it will not move at alland as strain is change in length to original length, the strain will not develop and this is your “ Plane Strain CAse”Note that strain is prevented to happen thus in this case stress occurs2: Plane Stress: Consider example where the thickness tends to 0That means , to understand say, only 1 cube exist in thickness direction which can contract or expandThus Strain exist. BUT, untill and unless the strain is prevented to happen stress does not existSo it is a plane stress case where strain exist but not stress.
Is this number a root of x^3 - 3x + 1?
-------- --------- --------- ---------- Final comments: Thanks to everyone who responded. I found the solutions of ksoileau, NBL, and (Ω) Dr D especially nice. Here is my solution: Notice that (-4+4i√3) = 8(-1+i√3)/2 = 8(primitive cube root of unity) = 8(cos120+isin120), so its cube root is 2(cos40+isin40), and the real part is 2cos40. So the problem is to show that x=2cos40 is a root of x^3 - 3x + 1, and that follows easily from the identity cos(3t) = 4cos(t)^3 - 3cos(t), with t=40°. That geometry on the complex plane is connected with the roots of cubics even when they have three real roots is the point of this problem. For this problem a picture is worth 1000 words: http://i44.tinypic.com/2s7ay4o.png