Suppose that x(s, t) = - 4s^2 -3t^2, y a function of (s, t) with y(1, 1) = 1 and

Question

Suppose that x(s, t) = - 4s^2 -3t^2, y a function of (s, t) with y(1, 1) = 1 and partial differential y/partial differential t(1, 1) = -3. Suppose that u = xy, upsilon a function of x, y with partial differential v/partial differential y(-7, 1) = 4 Now suppose that f(s, t) = u(x(s, t), y(s, t)) and g(s, t) = v(x(s, t), y(s, t)) You are given partial differential f/partial differential s(1, 1) = -36, partial differential f/partial differential t(1, 1) = 15, partial differential g/partial differential s(1, 1) = 48. The value of partial differential g/partial differential t(1, 1) must be (a) 8 (b) 9 (c) 12 (D) 14 (e) 15

Explanation / Answer

given x(s,t)=-4s2-3t2 ,y(1,1)=1,(y/t)=-3, (v/y)=4 ,f/s=-36,f/t=15,g/s=48 , u=xy

u/y=x,u/x=y

(x/s) =-8s, (x/t)=-6t

at (1,1) ,(x/s) =-8, (x/t)=-6 , u/y=-7,u/x=1

f/s =(u/x)(x/s) +(u/y)(y/s)

f/t =(u/x)(x/t) +(u/y)(y/t)

g/s =(v/x)(x/s) +(v/y)(y/s)

g/t =(v/x)(x/t) +(v/y)(y/t)

-36=(1)(-8) +(-7)(y/s) =>(y/s)=(36-8)/7=4

15 =(1)(-6) +(-7)(-3)

48 =(v/x)(-8) +(4)(y/s) =>48 =(v/x)(-8) +(4)(4) =>(v/x)=-4

g/t =(v/x)(-6) +(4)(-3)

=>g/t =(-4)(-6) +(4)(-3)

=>g/t =24-12

=>g/t =12

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