Answer
a) $1+x$
b) $-y+\dfrac{\pi}{2}$
Work Step by Step
Since, we have $L(x,y)=z_0+f_x(x-a)+f_y(y-b)$ ...(1)
a ) Now,
$f_x=1$; and $f_y=0$
Now eq. (1) becomes $L(0,0)=1+1(x-0)+0=1+x$
b) Since, we have $L(x,y)=z_0+f_x(x-a)+f_y(y-b)$ ...(1)
Now,
$f_x=0$; and $f_y=-1$
Now eq. (1) becomes $L(0,\pi/2)=0+0(x-0)+(-1)0(y-\pi/2)=-y+\dfrac{\pi}{2}$
Hence, a) $1+x$ b) $-y+\dfrac{\pi}{2}$
