On part (b), have you perhaps "plugged in" the initial angular velocity where the time interval 0.192 seconds belongs?
I think you have misled yourself because you have omitted a factor (and wrote another incorrectly) in the typing of your equation in part (A):
I believe that...
You get 1 = 0 only by rather slippery neglect of the arbitrary constant. The full integration-by-parts would have to read
\int \frac{cos x dx}{sin x} = ( csc x )( sin x ) + \int (sin x) ( csc x cot x ) dx \mathbf{+ C } .
Someone had a post a month or so back (I don't know if I can...
...because \lambda is a positive number, so the first non-trivial zero of the equation will be in the next quadrant where \tan \lambda is positive, and you're after \lambda^{2}.
[EDIT: On thinking about this a little more, we could also go in the negative direction, since \tan \lambda...
I just did this problem with a student this past week (Stewart, Section 4.7, no?). The trick if you do this algebraically is not to simplify your expression too soon. It will also be better if you eliminate h , rather than x . Your Pythagorean result becomes
L^{2} = (x + 4)^{2} +...
You're welcome! And I quite understand your frustration with WebAssign. I work with students here who also have to wrestle with physics on that system...
Remember that you want the sets of points (x,y), so not everything hinges on x alone.
part A: Where would ln(xy) be undefined? Where is it zero?
part B: 2x - 3y is always defined, but where is this difference equal to zero?
I don't think the air is going through the tube if the tube has 14.5 meters of water in it. I think he is saying that atmospheric pressure is being applied to the top of the water in the tube. So the total pressure applied at the mouth of the tube at the point where it meets the water in the...
Why restrict the sets so much? The functions will not be continuous anyplace where the functions are undefined (or where the limit is not defined, or where the limit does not equal the function value).
The numerators are always defined, so they do not present a problem. So the ratios are...
The only thing I can think of that the problem might be looking for is that the pressure from the tube should be the sum of the hydrostatic pressure from the water, \rho gh , plus the atmospheric pressure of 101,300 N/m2 (which is the "hydrostatic pressure" of the air), and that this times the...
Disregard my first remark: I was thinking of the force value. I don't see anything obviously wrong. Are your numbers radii, and not diameters? What do you mean by saying, "The answer turned out wrong." Is there a given answer?
I agree up to this point.
Your "velocity-squared" equation is all right, and \omega_{0} = 0 . But doesn't that then lead to
\omega^{2} = 0^{2} + 4 \alpha \pi \Rightarrow \omega^{2} = 4 \pi \cdot \frac{2 \pi - \omega (0.69)}{(0.5)(0.69)^2} ,
with \omega being the...
That I had seen before. I think I generally overlook it because most of the problems where I've ever used this method had broader regions of convergence than in this problem.
Since your text is suggesting that there are positive and negative solutions, you are not restricted to the "principal circle" , 0 ≤ t < 2{\pi} . So you can "go around the circle" as many times as you like:
2t = \frac{\pi}{4} + k \cdot 2{\pi} , \frac{5\pi}{4} + k \cdot 2{\pi}...