Recent content by PensNAS

Thanks for the help! I actually went and talked to my professor, since this is part of a statics problem and not calculus. He was okay with just plotting a few points around where the curves get closest. @Mark44, I'm from Pensacola.

Am I wrong in thinking the formula for the distance between the functions is the top curve minus the bottom curve?

It is a parabola. The book gives us the graph of both functions, but limits it to the first quadrant. I should have soda that too, my bad.

It is a decreasing, concave curve is what I should have said. How does one show that there is not a solution for both equations?

The first equation looks similar to z=1/x. The radius of the circle is a/2. I've already sketched out the equations.

Homework Statement Show that these curves do not intersect. z=(1/a)(a-y)^2 y^2+z^2=a^2/4 Where a is the radius of the circle and other shape. Homework Equations There aren't any. The Attempt at a Solution I tried setting them equal to each other but got the equation...

I get that when I spend the extra 15 seconds to calculate |BA| further than 15.9. Yep, home to beaches and Naval Aviation.:biggrin:

4h+4 is correct, I dropped a negative. The denominator was a typo. I have it on my paper as \frac{90000(h^{2}-h+52.25)}{16h^{2}+8h+1}. I redid the calculation for the tension and got 79.5 lb. It would seem I got sloppy, this was the 8th time working through the problem.Thanks for the help!

Vector OB should be the one I need, which is <0,h+1,-6> Vector BA should be <4,0.5-h,-6>. This gives a new cross product of T_{BA}<\frac{-12h-3}{|BA|},\frac{-24}{|BA|},-\frac{4h-4}{|BA|}> Am I right in thinking I can then set the i and k components of M_{O} equal to the i and k components of...

Homework Statement Utility Pole BC is guyed by cable AB as shown. The tension in cable AB (TAB) acting at point B creates a moment about point O (MO) equal to -900i+Myj-315k lb-ft, where My is unknown. a) What is the length of the pole (LBC)? b) What is the magnitude of the tension, TAB...

I set x=scos(θ)t and y=ssin(θ)t-0.5gt^2. Solving for t with the x equation I got t=x/scosθ. Plugging that into the y equation I got the same equation of y=dtanθ-d^2(g/(2(scosθ^2)). I tried simplifying that some more and got (2(s^2)sin(2θ)d-gd^2)/2(s^2)cos(θ)^2. It still is not right and I don't...

1. Homework Statement During a tennis match, a player serves the ball at a speed s, with the center of the ball leaving the racquet at angle θ below the horizontal at a height y0. The net is a distance d away and has a height h. When the ball is directly over the net, what is the distance...