Recent content by Helloooo

Oh okay! I manged to solve the integration with the change and with the substitution u=sin(t) Thank you!

∫zds=∫acos(t)*( (acos(2t))^2+(2asin(t))^2+(-asin(t))^2 )^1/2 dt , (0≤t≤pi/2) Simplified : ∫a^2cos(t)*(cos^2(2t)+5sin^2(t) )^1/2 dt , (0≤t≤pi/2) However here i get stuck and i can´t find a way to rewrite it better or to integrate as it is. Can i please get some help in this?

I see, thank you so much for your help!

Does that mean that the tangent plane and straight tangent line is the same? If so, why are they called differently or is it just in this particulary problem?

Firstly i noticed an error in my writing. The tangent plane i got was 4x+2y-3=0. I´m sorry for that So i redid the tangent plane to 4x+2y-6=0 However i still don´t know how to fint the straight tangent line. Thank you

I have solved the gradient: gradf(2,-1)=(4,2) and have the tangent plane: 4x+2y+3=0 Somehow the answer is: 3=2x+y And i really don´t understand why.