The thing to realize about dummy indices is that they have no intrinsic meaning except that
they are to be summed over. Thus, in the product
z_{abc}x^a x^b x^c
you can replace a with d and the meaning is the same. You might want to systematically replace each of your given letters with...
Let me first fix your expression by putting in the "tex" commands:
That doesn't look right.
Now,
y = (9 - x^{\frac{2}{3}})^{\frac{3}{2}}
has a derivative that looks like
\frac{dy}{dx} = -(9 - x^{\frac{2}{3}})^{\frac{1}{2}} x^{-\frac{1}{3}}
Now when you square that mess so...
You forgot that when you do partial integration, say with respect to y, you have to add an arbitrary function of x. Similarly when you integrate partially with respect to x you have to add an arbitrary function of y. This will give you the necessary freedom to determine your unknown function.
Your equation above is wrong. It should be
\int_a^b \sqrt{1 + (\frac{dy}{dx}})^2 dx
When you use this form of the equation for the arc length, you will find a couple of things simplify.
First, Thank you for taking the time to come up with an example and the suggested exercise. Thanks also to D.H. for his post. One more question: How is this generalized cross product related to the wedge product of geometric algebra?
I am curious. What is the relationship between the cross product defined above and the cross product of two vectors, a and b defined by:
(1) a X b is perpendicular to the plane defined by a and b,
(2) a X b has magnitude ab sin(theta),
(3) the vectors a, b, and a X b form a right-handed...
Before you make use of the previous post, multiply numerator and denominator by the complex conjugate of the denominator. Then make use of Yungman's post. From there it's a standard manipulation.
First, this will give you info on the relationship between Lommel functions and Bessel functions:
http://en.wikipedia.org/wiki/Lommel_function
As you probably noticed Lommel's D.E. is a nonhomogeneous Bessel equation.
Second, Maple is a symbolic manipulation program, similar to...
Well, here's what Maple produces when asked to integrate J_2(x)
x*BesselJ(2,x)*LommelS1(-1,1,x)-x*BesselJ(1,x)*LommelS1(0,2,x)
According to Maple,
LommelS1( \mu, \nu, z) and LommelS2( \mu, \nu, z)
solve the following differential equation:
z^2 y'' + z y' + (z^2 - \nu^2) y = z^(\mu...
I just calculated the integral of J_0(x) for fun and it's not bad. I haven't looked at the Bessel identities in years (many years), but I'll take a look at them for a change of pace. Will let you know if I come up with something.
I'm curious that you are trying to calculate an indefinite integral for J_2(x) . Since Bessel functions are so widely used in applications, could you recast the problem as a definite integral?
It seems that you've achieved some simplification in the last step. Couldn't you write J_0(x)...
This seems to be a pretty straight forward problem. Rewrite your metric in spherical coordinates. Identify your g_{\mu \nu} . Go to a book on General Relativity and find the definition of the Riemann tensor in terms of the Christoffel symbols and calculate it out. It will take a little time...
If you remember that
1 + tan^2 \theta = sec^2 \theta
then you'll find that the integrand simplifies quite nicely. Don't forget to change your integration limits appropriately.
Let's start over. You're going about this poorly. We'll put the radar station at our origin, let the vertical direction be y and the horizontal direction be x. Let the direct distance of the plane from the radar station along (the hypotenuse) be s. Then
s^2 = x^2 + y^2
When you...
Notice that the triangle is a right triangle and two of those sides are changing in length. What equation relates these lengths to each other? What process evaluates the rate of change? Use it on the equation and substitute in the instantaneous values given in your problem.
Yes there is a shorter way to do this. You can write it as
\frac{(x+2)^3}{(\sqrt[3]{x})^3}
and notice that that leaves you with
\frac{(x+2)^3}{x}
which is considerably easier to differentiate.
The electrostatic force is a conservative force. This means that the work done by the electrostatic force while moving a charge from one point to another is independent of the path and only depends on the end points. The magnitude of the work done by the electrostatic force will be the...
Well, I get T = 139N. Summing the forces in the y direction, I get
T = \frac{Mg}{cos(45)} = \frac{98.1}{\frac{1}{\sqrt{2}}} = 138.7
which I'd round off to 139N. I don't see a reason for the discrepancy right now.
Is the picture you gave the whole picture?
Draw a diagram of the forces acting on the mass. Since there is no motion, the forces must be in equilibrium. That means the vector sum of the forces in the x direction mus be equal to zero and similarly for the vector sum the forces in the y direction. You'll have to write the tension T in...
Integration by parts is based on the formula for the derivative of a product. Here's how it goes. Suppose you take the derivative of the product of f(x) times g(x):
\frac{d}{dx} (f(x)g(x)) = f(x) \frac{d}{dx}g(x) + g(x) \frac{d}{dx} f(x)
or in the notation of differentials and...
Well, you could remember that (1 - sin^2 \theta) = cos^2 \theta so that your integrand becomes cos^6 \theta and then look that integral up in a table of integrals, or look up a reduction formula for cos^6 \theta . You'll be able to integrate the expression that results from the reduction...
The fact that you shaded in an area indicates that you are thinking in terms of area, not length along the curve that was suggested to you. You should have an equation for arc length to work with. Try this one:
S = \int_0 ^ \pi \sqrt{ dx^2 + dy^2} dt
Now it's up to you to figure out dx...