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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...

Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Legend of Zelda: Spirit Tracks From The Cutting Room Floor Jump to: navigation, search The Legend of Zelda: Spirit Tracks Also known as : Zelda no Densetsu: Daichi no Kiteki (JP) Developer : Nintendo Publisher : Nintendo Platform : Nintendo DS Released in JP : December 23, 2009 Released in US : December 7, 2009 Released in EU : December 11, 2009 Released in AU : December 10, 2009 This game has unused areas. This game has unused graphics. This game has unused models. This game has unused music. This game has debugging material. This game has regional differences. This game has anti-piracy features. Spirit Tracks is the sequel to Phantom Hourglass and uses a modified version of its predecessor's engine. Yes, Zelda really is a spirit, and she's even...

0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...