Please Help, How to Sketch a circular cone, height h and base r?
Consider the Surface: r = ucos(v)i + usin(v)j + (h*u/r)k 0<u<r and 0<v<2*pi
(should be less then equal to for constraints)
Show that this represents the curved surface of a circular cone height h and base radius r. Sketch this cone?
I have tried multiple times to solve this and cant seem to grasp a solution. PLEASE HELP
- ?Lv 48 years agoFavourite answer
I have posted a full solution to your question at one of the math help forums I collaborate with, so that I may give you an easy to read explanation using LaTeX:
Hope this helps
- Anonymous5 years ago
Cut the cone and the cylinder by using a vertical aircraft , so you're going to see at XY airplane .- you're going to se a slant facet or line of the cone and a vertical line of the cylinder .- The slant line (side of the cone passes with the aid of points ( four,zero) and (0,12) .- The equation is m= (12-0)/ (zero-four) = -three the line is y--zero= -three(x-4) y= -3x+12 on this method , at x , you've got a top y , so the quantity of the cylinder is V= pi x^2 y V= pi x^2 (-3x+12) V= pi (12x^2-3x^three) dV/dx= pi ( 24x -9x^2) d2V/dx2= pi ( 24-18x) Do dV/dx=0 24x-9x^2=0 x( 24-9x)=zero x=0 ( No answer, it is a min ) x= 24/9 = 8/three cm If it's a max , d2V/dx2 <zero , pi (24-18 *24/9) = -216pi/9 <0 , so x=24/9 makes V max So the radius of the cylinder is x= 8/three cm the peak is y= -three* eight/three +12 = -eight+12 = 4 cm Vmax = pi (eight/three)^2 *four Vmax= (256 pi/9)