Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 29

a) $1+x$; b) $-y+\dfrac{\pi}{2}$

a. As we know that $L(x,y)=f(x,y)+f_x(x-m)+f_y(y-n)$ Here, we have $f_x=1$; and $f_y=0$ Thus, we get $L(0,0)=1+1(x-0)+0 \\ \implies L(0,0)=1+x$ b. As we know that $L(x,y)=f(x,y)+f_x(x-m)+f_y(y-n)$ Here, we have $f_x=0$; and $f_y=-1$ Thus, we get $L(0,\dfrac{\pi}{2})=0+0(x-0)+(-1)(y-\dfrac{\pi}{2}) \\ \implies L(0,\dfrac{\pi}{2})=-y+\dfrac{\pi}{2}$