How do you use differentials and the function #f(x,y) = arctan(x*y^2)# to approximate the value of f(0.94, 1.17)?
1 Answer
Explanation:
If we stay near the point of tangency (
The tangent plane is given by:
#z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)#
And so we have:
#z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)=L(x,y)#
Where
Given that
First, we pick "nice" points for the approximation which are near the original
#f(x,y)=arctan(x*y^2)#
We take the partial derivative of the function with respect to
#(del)/(delx)=y^2/(x^2y^4+1)#
Then with respect to
#(del)/(dely)=(2xy)/(x^2y^4+1)#
Now, just as we do when finding the tangent plane, we calculate
#f_x(1,1)=(1^2)/(2)=0.5#
#f_y(1,1)=2/2=1#
#f(1,1)=arctan(1)=pi/4#
Revisiting our equation for the tangent plane:
#z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)#
Putting in our values:
#z=pi/4+0.5(x-1)+1(y-1)#
#=>z=pi/4+x/2-1/2+y-1#
#=>z=x/2+y+pi/4-3/2#
Where
This means that when
Finally, our linear approximation is found by putting the actual
#f(0.94,1.17)~~(0.94/2)+1.17+pi/4-3/2=0.92539816#
We can check the accuracy of our approximation by comparing it to the actual value of
#f(0.94,1.17)=arctan((0.94)*(1.17)^2)#
#f(0.94,1.17)=0.910149371#
This is quite close to our approximation, with a difference of about