Recent content by letalea

Yes, that makes perfect sense. Thank you!

Could you explain this mistake to me? As I am not seeing how I come to get the coefficients. But yes you are right, when you factor in the coefficients it does give me one of the possible answers the prof gave us!

Homework Statement The volume underneath the surface z= y/ (1+xy) and above the square {(x,y)| 2≤x≤3 , 3≤ y≤ 4} is:Homework Equations Please see attachment. The Attempt at a Solution Please see attachment for solution. My professor had provided us with 8 possible solutions (where only one...

Homework Statement The field generated in Fig. 3 is due to a battery connected between the center conductor and the sheath. What is the voltage rating of the battery? (i.e. what is the electric potential of the positive terminal with respect to the negative terminal of the battery?) I...

Thank you I will give that a try!

Okay I hope this is what you were wanting me to answer... \partialz/\partialu = (\partialz/\partialx) (\partialx/\partialu) + (\partialz/\partialy)(\partialy/\partialu) =(2xysin(u)tan(v))(\partialx/\partialu) +(x^2sin(u)tan(v))(\partialy/\partialu) \partialz/\partialv=...

Yes sorry, it was a posting mistake, the question was z=x^2ysin(u)tan(v). Can you see any errors in how I solved the partial derivatives, as I have tried multiple times and can not seem to find the correct solution.

Homework Statement Let x=x^2ysin(u)tan(v), where x(u,v) and y(u,v) are smooth functions that, when evaluated at u=1 and v=-3 satisfy x=2.112, y=4.797, \partialx/\partialu = -3.491, \partialx/\partialv = -2.230 , \partialy/\partialu = 1.787 , \partialy/\partialv = 1.554. Then the...