Solution:
- The given function is f(x,y)=exsin(y)
- The objective is to find the quadratic approximation of a given function using the Taylor formula and also estimate the error in the approximation if ∣x∣≤0.1 and ∣y∣≤0.1.
Calculate First partial derivatives at (0,0) .
fx=exsin(y)
fx(0,0)=0
fy=excos(y)
fy(0,0)=1
Calculate second derivatives at (0,0)
fxx=exsin(y)
fx(0,0)=0
fy=excos(y)
fy(0,0)=1
fxy=−exsin(y)
fxy(0,0)=0
Therefore quadratic approximation of f(x,y) at origin using Taylor formula is:
T(x,y)≈9+21[−9x2−9y2]≈9−29[x2+y2]
Next Calculating error at the origin using ∣x∣≤0.1 and ∣y∣≤0.1 as:
E(x,y)≤21×9[(0.1)2+(0.1)2]≤0.08685
Thus, the Quadratic polynomial using the Taylor formula is 9−29[x2+y2] and error in the approximation is less than 0.08685.
Comments