Recent content by P.O.L.A.R

nevermind i got it forgot about the 2 on the bottom doh.

d/dx[t/(1+t^2)^(1/2)] The answer say 1/(1+t^2)^(3/2) I can tget that for the life of me. Used product rule and I can't seem to simplify it I come up with 2/(1+t^2)^(3/2) Where does the 2 go? This is for a principal unit vector problem so a quick solution is all i need. thanks

Ok that makes total sense thanks

Homework Statement A ball is kicked with initial speed 20 m/s and initial angle 40 degrees up an incline of angle 15 degrees. Assume that the ball leaves the ground a the base of the incline at x_{0}=0 and y_{0}=0. How far up the incline does the ball initially land (not how far horizontally...

I agree if the functions were being integrated with respect to y it would make more sense for a to c. Then if you look the two graphs go beyond e which would leave to believe that there should be an f at the final intersection and if placed like the a then it to could be integrate with...

How did you get u=tan^2(x)?? because x is really [sqrt(x)]^2 having trouble seeng the tan^2(x).

I startd in intermediate (which I found to be the best thing ever) and I am now in Calc III and Engineering Physics I and I amdoing great. It doesn't matter where you start or anything you should do hat you love and if physics is it do it. A counselor shouldn't try to steer you away from...

On that note at my school and most you can go CE then it requires only an extr 30 hours to get a second bachelors. After you have your CE degree you only need 10 hours of CS based classes and then just take 20 hours of other interesting classes or push-overs if you wish and boom you have both...

Ok I think I have the trick is to draw it out an use the law of sine a/sin(A)=b/sin(B)=c/sin(C) for part a i got 6.83km much diff than the rest. You have to place the angle across from the side like the picture...

I already proved it myself people were asking proof questions and I found this one to be a good one. Starting that way didnt work for me. I couldn't figure out where to go after taking the tan.

Ok I got you it just seemed that way. I got the same thing by the way.

Yes but when you enter arctan in a calculator it gives you the answer in either the first or 4th quadrent.

What mistake are you pointing out because you said to ignore the sign and you don't get the get the answer. If it is -75 you have to add 180 to get the angle in the right position which is 105 definitely different than 75. IS the answer negative or positive?? one more thing is that r=a=b or...

first and foremost you can't ignore the sign because negative arctan is in the 4th quadrent and the positive of that is in the 2nd quadrent. Ignoring the negative is something you cannot do.

There is a theorm for this because what you have is an \infty / \infty correct? (L'Hopital) Note: If this is a calc one qustion ignore this.