Given z=f(x,y),x=x(u,v),y=y(u,v), with x(3,5)=4 and y(3,5)=3, calculate zv(3,5) in terms of some of the values given in the table below.

Question

fx(3,5)=2fx(3,5)=2	fy(3,5)=dfy(3,5)=d	xu(3,5)=qxu(3,5)=q	yu(3,5)=byu(3,5)=b
fx(4,3)=rfx(4,3)=r	fy(4,3)=afy(4,3)=a	xv(3,5)=sxv(3,5)=s	yv(3,5)=−3yv(3,5)=−3

Answer 1

To calculate the partial derivative zv(3,5), you can use the chain rule. Given that z is a function of x and y, and x and y are functions of u and v, you can write the chain rule for z in terms of u and v as follows:

dz/dv = (dz/dx) * (dx/dv) + (dz/dy) * (dy/dv)

Now, you are given some values for the partial derivatives and functions at the point (3, 5):

fx(3, 5) = 2

fy(3, 5) = d

xu(3, 5) = q

yu(3, 5) = b

xv(3, 5) = s

yv(3, 5) = -3

You also need the values of dx/dv and dy/dv, which are denoted as xv(3,5) and yv(3,5), respectively. Given that:

xv(3, 5) = s

yv(3, 5) = -3

Now, plug these values into the chain rule:

dz/dv = (dz/dx) * (dx/dv) + (dz/dy) * (dy/dv)

dz/dv = (fx(3, 5)) * (xv(3, 5)) + (fy(3, 5)) * (yv(3, 5))

dz/dv = (2) * (s) + (d) * (-3)

Now, express zv(3, 5) in terms of the given values:

zv(3, 5) = 2s - 3d

So, the value of zv(3, 5) is 2s - 3d, where s represents xv(3, 5) and d represents fy(3, 5).