Recent content by MichaelT

Working on some optics problems and I am confused about the front and back of the lens. The sign conventions are key to finding the correct image location. Now, is the front always where the object is located? The problem I am working on involves a divergent lens and a concave mirror. The...

Now this is something that just came to mind. As an undergrad Physics major, I have already taken a few introductory Physics courses. We have studied gravity, electromagnetic forces, etc. All of the work we have done involved measuring these forces and understanding how a body will react while...

Sure does! Now think about the bounds of this triangle. Have you studied x-simple and y-simple regions? You can think of the triangle as both y-simple and x-simple because it ends in either a straight line or a point in both the x and y directions. So it is up to you to choose. In one direction...

Hmmm well in 3 dimensions the plane x + y = 2 is a plane. But if you want to look at it in 2 dimensions, just plot some points in the first quadrant of the xy coordinate system. i.e when y=0, x=2. If you take the points and connect them, a common geometric shape will be bounded between the x...

Hey negat1ve! I haven't done this problem with a triple integral, but I did do it with a double integral. If you would like to use that approach (a bit easier, as long as you have an integration table handy :wink:)...Think of the plane x + y = 2 in two dimensions. What shape does this form in...

I am definitely getting there with this hint. so I take the norm of c'(t) X (c"(t) X c'(t)) and get ||c'(t) X (c"(t) X c'(t))|| = ||c'(t)||* ||c"(t) X c'(t)||sin(theta) where theta = pi/2 so ||c'(t) X (c"(t) X c'(t))|| = ||c'(t)||* ||c"(t) X c'(t)|| Now I need to relate this to k =...

Homework Statement If c is given in terms of some other parameter t and c'(t) is never zero, show that k = ||c'(t) x c"(t)||/||c'(t)||3 The first two parts of this problem involved a path parametrized by arc length, but this part says nothing about that, so I assume that this path is not...

I'm pretty sure I got it right, since a classmate of mine solved for the volume using a triple integral and we both got the same answer. So I don't think I need it checked, but if anyone wants to see the whole thing I would be happy to post it. As to polar and cylindrical coordinates...We...

Ok this is how I went about this. I used a double integral of the cylinder over the plane. I solved the equation of the cylinder for z, and that was the integrand. This was integrated over [0,2] X [0, 2-y] , dx dy. After some lengthy calculations (and a very helpful table of integrals...

That is most definitely what I got when I just re-did the problem! Yay! Thank you very much, I will go and check it now :biggrin:

yeah it looks like I did miss something! Does it look like I am approaching this the right way? Just re-did the problem, and this is what I got T'(t) = (c"(t)/||c't||) - [c'(t)/||c'(t)||3] (c'(t) dot c"(t))

Oh wait, I was wrong about something. T(t) will only equal c'(t) if it is parametrized by the arc length. This is what I got, if anyone cares to check (please do!) T'(t) = [||c'(t)||(c"(t)) - c'(t)(c'(t) dot c"(t))]/||c'(t)||3

Oh wait, I was wrong about something. T(t) will only equal c'(t) if it is parametrized by the arc length. This is what I got, if anyone cares to check (please do!) T'(t) = [||c'(t)||(c"(t)) - c'(t)(c'(t) dot c"(t))]/||c'(t)||3

We need to find the volume of the solid bounded by the cylinder with the equation z^2 + y^2 = 4 and the plane x + y = 2, in the first octant (x,y,z all positive). Firstly, I am trying to visualize the graphs. From what I can tell, the cylinder is centered around the x-axis and has a radius...

So we are given T(t) = c'(t)/||c'(t)|| as well as ||T|| = 1 We also know T(t)dotT(t) = 1 and T'(t)dotT(t) = 0 The problem asks us to find T'(t) I tried differentiating c'(t)/||c'(t)|| treating ||c'(t)|| as the square root of the dot product of c'(t) with itself. I used the product...