Answer
(a) ${f_x}\left( {5,3} \right) = 2$, ${f_y}\left( {5,3} \right) = - 1$ and $f\left( {5,3} \right) = 6$
(b) $f\left( {5.2,2.9} \right) \simeq 6.5$
Work Step by Step
(a) We have the tangent plane to the graph of $z = f\left( {x,y} \right)$ at $P = \left( {5,3} \right)$ given by
$z = 2x - y - 1$
By Theorem 1 of Section 15.4, the equation of the tangent plane to $f\left( {x,y} \right)$ at $P = \left( {5,3} \right)$ is
$z = f\left( {5,3} \right) + {f_x}\left( {5,3} \right)\left( {x - 5} \right) + {f_y}\left( {5,3} \right)\left( {y - 3} \right)$
So,
$2x - y - 1 = f\left( {5,3} \right) + {f_x}\left( {5,3} \right)\left( {x - 5} \right) + {f_y}\left( {5,3} \right)\left( {y - 3} \right)$
$2x - y - 1 = f\left( {5,3} \right) + x{f_x}\left( {5,3} \right) - 5{f_x}\left( {5,3} \right) + y{f_y}\left( {5,3} \right) - 3{f_y}\left( {5,3} \right)$
$2x - y - 1 = x{f_x}\left( {5,3} \right) + y{f_y}\left( {5,3} \right) + f\left( {5,3} \right) - 5{f_x}\left( {5,3} \right) - 3{f_y}\left( {5,3} \right)$
Since $x$ and $y$ are independent variables, equating the terms in the left- and the right-hand sides gives
${f_x}\left( {5,3} \right) = 2$, ${\ \ \ }$ ${f_y}\left( {5,3} \right) = - 1$
$f\left( {5,3} \right) - 5{f_x}\left( {5,3} \right) - 3{f_y}\left( {5,3} \right) = - 1$
Substituting ${f_x}\left( {5,3} \right) = 2$ and ${f_y}\left( {5,3} \right) = - 1$ in the last equation gives
$f\left( {5,3} \right) - 5\cdot2 - 3\cdot\left( { - 1} \right) = - 1$
So, $f\left( {5,3} \right) = 6$.
Thus, ${f_x}\left( {5,3} \right) = 2$, ${f_y}\left( {5,3} \right) = - 1$ and $f\left( {5,3} \right) = 6$.
(b) Let $x=a+h$ and $y=b+k$. The linear approximation to $f\left( {x,y} \right)$ (Eq.(3) of Section 15.4) is given by
(1) ${\ \ \ }$ $f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
Write $\left( {a,b} \right) = \left( {5,3} \right)$ and $\left( {h,k} \right) = \left( {0.2, - 0.1} \right)$. Using equation (1), we get
$f\left( {5.2,2.9} \right) \simeq f\left( {5,3} \right) + {f_x}\left( {5,3} \right)\cdot0.2 + {f_y}\left( {5,3} \right)\cdot\left( { - 0.1} \right)$
Substituting ${f_x}\left( {5,3} \right) = 2$, ${f_y}\left( {5,3} \right) = - 1$ and $f\left( {5,3} \right) = 6$ in this equation gives
$f\left( {5.2,2.9} \right) \simeq 6 + 2\cdot0.2 + \left( { - 1} \right)\cdot\left( { - 0.1} \right) = 6.5$
