(1 pt) Suppose that χ ( s , t ) = − 4 s 2 − 2 t 2 , y \displaystyle \,\chi(s,\, t)\, =\, -4s^2\, -\, 2t^2,\,y\, χ ( s , t ) = − 4 s 2 − 2 t 2 , y ( s , t ) \displaystyle \,(s,\, t)\, ( s , t ) y ( 1 , 1 ) = 1 \displaystyle \, y(1,\, 1)\, =\, 1\, y ( 1 , 1 ) = 1 ∂ y ∂ t ( 1 , 1 ) = − 2. \displaystyle \, \dfrac{\partial y}{\partial t}\, (1,\, 1)\, =\, -2. ∂ t ∂ y ( 1 , 1 ) = − 2 . u = x y , v \displaystyle \, u\, =\, xy,\, v\, u = x y , v x , y \displaystyle \, x,\, y\, x , y ∂ v ∂ y ( − 6 , 1 ) = 4. \displaystyle \, \dfrac{\partial v}{\partial y}\, (-6,\, 1)\, =\,4. ∂ y ∂ v ( − 6 , 1 ) = 4 . f ( s , t ) = u ( x ( s , t ) , y ( s , t ) ) \displaystyle \, f(s,\,t)\, =\, u(x(s,\, t),\, y(s,\, t))\, f ( s , t ) = u ( x ( s , t ) , y ( s , t ) ) g ( s , t ) = v ( x ( s , t ) , y ( s , t ) ) . \displaystyle \, g(s,\, t)\, =\, v(x(s,\, t),\, y(s,\, t)).\, g ( s , t ) = v ( x ( s , t ) , y ( s , t ) ) . . . . . . ∂ f ∂ s ( 1 , 1 ) = − 32 , \displaystyle \dfrac{\partial f}{\partial s}\, (1,\, 1)\, =\, -32,\, ∂ s ∂ f ( 1 , 1 ) = − 3 2 , . . . ∂ f ∂ t ( 1 , 1 ) = 8 , \displaystyle \dfrac{\partial f}{\partial t}\, (1,\, 1)\, =\, 8,\, ∂ t ∂ f ( 1 , 1 ) = 8 , . . . ∂ g ∂ s ( 1 , 1 ) = − 16. \displaystyle \dfrac{\partial g}{\partial s}\, (1,\, 1)\, =\, -16. ∂ s ∂ g ( 1 , 1 ) = − 1 6 . ∂ g ∂ t ( 1 , 1 ) \displaystyle \, \dfrac{\partial g}{\partial t}\, (1,\, 1)\, ∂ t ∂ g ( 1 , 1 ) dv/dt=dv/dx*dx/dt+dv/dy*dy/dt dx/dt=-4t -> evaluate at (1,1) =-4 dv/dt=-4dv/dx+4(-2) dv/dt=-4dv/dx-8 How can I find the missing dv/dx in order to get a value for dv/dt? Thanks!
Last edited by a moderator: Nov 4, 2015
