Answer
Using the linear approximation:
$f\left( {0.01, - 0.02} \right) \simeq 0.98$
Using a calculator: $f\left( {0.01, - 0.02} \right) \simeq 0.980297$
Work Step by Step
We are given $f\left( {x,y} \right) = {{\rm{e}}^{{x^2} + y}}$. The partial derivatives are
${f_x} = 2x{{\rm{e}}^{{x^2} + y}}$, ${\ \ \ }$ ${f_y} = {{\rm{e}}^{{x^2} + y}}$
According to Eq. (3), the linear approximation to $f\left( {x,y} \right)$ is given by
$f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
For $\left( {a,b} \right) = \left( {0,0} \right)$ and $\left( {h,k} \right) = \left( {0.01, - 0.02} \right)$ we have
$x = a + h = 0.01$ ${\ \ }$ and ${\ \ }$ $y = b + k = - 0.02$
Thus,
$f\left( {0.01, - 0.02} \right) \approx f\left( {0,0} \right) + {f_x}\left( {0,0} \right)\cdot0.01 + {f_y}\left( {0,0} \right)\cdot\left( { - 0.02} \right)$
$f\left( {0.01, - 0.02} \right) \simeq 1 + 0 - 0.02 = 0.98$
Using a calculator, we obtain $f\left( {0.01, - 0.02} \right) \simeq 0.980297$
