Recent content by chrsr34

Thank you guys. It appears to me that this quad is not pos or neg definite then. This is my 3rd option. Thanks guys Chris

hmm, well what I am not sure of most is the statement "q(x,y) > 0 for all x not equal to 0" This was given for 2 variables, i have nothing for 3 variables. So in I am not even sure if we are only taking into consideration what x is or if there are 2 variables to take consideration of in 3D. This...

Hey guys. Im having a problem with deciphering positive definite and negative definite for this quadratic form i determined from a matrix. I don't quite understand how to evaluate if its pos def or neg def. From what I see in my notes, q(x,y) > 0 for all x not equal to 0. This doesn't help...

yea i just realized i did that. So, r = < 2cos(t), 2sin(t), 4-4cos^2(t) > This look good?

All yes your right. So my position vector should be: r = < 4cos^2(t), 4sin^2(t), 4-4cos^2(t) > Is this correct?

I see. Yes i could find z with those coordinates. But is there a reason you chose those? Are those cylindrical coordinates? (im a little rusty). Is there anything wrong with setting t=x^2?

Homework Statement Consider the curve of intersection of the cylinders [x^2+y^2=4] and [z+x^2=4]. Find parametric equations for this curve and use them to write a position vector. Homework Equations Thats what I am looking for. What to set t equal to. The Attempt at a Solution I set...

You need to add the 3 vectors vectorally. So add up the i components for your resultant x-value, and add up your j's for the resultant y-value. Then to get the direction, use the arctan. This should get you going. To get unit vectors you need to divide each coordinate by the magnitude of the...

Sorry, I think I was thinking a little unclear. Its been a while. I've done a problem like this in the past but just don't remember how. There should be someone here who knows, but ill look into it tomorrow. Chris

Im not sure how you got that answer. 36000 is correct. Although I should correct myself by saying that you need to subtract the 2 forces because they are both positive, therefore oppose each other. So, ((9x10^9)(4x10^-6))/1m^2 = 36000 ((9x10^9)(8x10^-6))/1m^2 = 72000 The force of q2 is...

I think the problem with starting your car in winter has more to do with the characteristics of your battery and the internal energy of it. Not to mention the CCA rating

You have 2 variables in this problem. Solve for one variable and then plug it back into the orginal equation leaving you only 1 variable to solve for. i.e. Vp = 769000Vax/ 249000. Now plug that in for Vp and solve for Vax. Chris

oh yes, your right, i was squaring the 100/20 by accident. I still don't fully understand the entire process but ill try and figure it out. Thanks again Rainbow Child.

Hmm, now I am even more confused. Plugging s and vi into that equation gives me 6.4 m/s = vf. I don't understand, if the initial vi is 4m/s, and then accelerates for 10m to a new vf, wouldn't you be adding vi + delta v to get final velocity? u_f^2-4^2=\frac{1}{20}\,s_f^2 solving for uf yields...

wow,i did not even think about integrating the v from 4-vf. Thank you so much. But just to be clear, you say that i can't add the initial velocity. The final answer of that integration is .58m/s so that would conclude that I do have to add the initial 4 m/s to get 4.58m/s. Isnt this correct?