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

Question

Suppose that x(s, t) = s^2 - 2t^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, v a function of x, y with partial differential v/partial differential y(-1, 1) = -5. 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) = -2, partial differential f/partial differential t(1, 1) = -1, partial differential g/partial differential s(1, 1) = -24. The value of partial differential g/partial differential t(1, 1) must be (a) 19 (b) 23 (c) 24 (d) 25 (e) 26

Explanation / Answer

given x(s,t)=s2-2t2 ,y(1,1)=1,(y/t)=-3, (v/y)=-5 ,f/s=-2,f/t=-1,g/s=-24 , u=xy

u/y=x,u/x=y

(x/s) =2s, (x/t)=-4t

at (1,1) ,(x/s) =2, (x/t)=-4 , u/y=-1,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)

-2=(1)(2) +(-1)(y/s) =>(y/s)=4

-1 =(1)(-4) +(-1)(-3)

-24 =(v/x)(2) +(-5)(y/s) =>-24 =(v/x)(2) +(-5)(4) =>(v/x)=-2

g/t =(v/x)(-4) +(-5)(-3)

=>g/t =(-2)(-4) +(-5)(-3)

=>g/t =8+15

=>g/t =23

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