Question #279177

Use taylors formula to find quadratic and cubic approximations of f(x, y) =exsin y at origin. Estimate the error of the approximations if modx less than equal to 0.1 and mod y less than equal to 0.1

1
Expert's answer
2021-12-14T08:09:07-0500

Solution:


  • The given function is f(x,y)=exsin(y)f(x,y)=e^xsin(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 x0.1| x | ≤ 0.1 and y0.1.| y | ≤ 0.1 .


Calculate First partial derivatives at (0,0)(0,0) .


fx=exsin(y)f x =e^xsin(y)

fx(0,0)=0f x ( 0 , 0 ) = 0

fy=excos(y)f y = e^xcos(y)

fy(0,0)=1f y ( 0 , 0 ) = 1


Calculate second derivatives at (0,0)(0,0)


fxx=exsin(y)f x x = e^xsin(y)

fx(0,0)=0f x ( 0 , 0 ) = 0

fy=excos(y)f y = e^xcos(y)

fy(0,0)=1f y ( 0 , 0 ) = 1

fxy=exsin(y)f x y = -e^xsin(y)

fxy(0,0)=0f x y ( 0 , 0 ) = 0


Therefore quadratic approximation of f(x,y)f(x,y) at origin using Taylor formula is:


T(x,y)9+12[9x29y2]992[x2+y2]T ( x , y ) ≈ 9 + \frac{1}{2} [ − 9 x^2 − 9 y^2 ] ≈ 9 − \frac{9}{2} [ x^2 + y^2 ]


Next Calculating error at the origin using x0.1| x | ≤ 0.1 and y0.1| y | ≤ 0.1 as:


E(x,y)12×9[(0.1)2+(0.1)2]0.08685E ( x , y ) ≤ \frac{1}{2} × 9 [ ( 0.1 )^2 + ( 0.1 ) ^2 ] ≤ 0.08685


Thus, the Quadratic polynomial using the Taylor formula is 992[x2+y2]9 − \frac{9}{2} [ x ^2 + y ^2 ] and error in the approximation is less than 0.08685.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS