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Answer:
The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.
Approximation at points (1.1,1.9) is 0.7
Step-by-step explanation:
Given:
Tangent plane to a surface z=5x+2y-10 as the function at point (1,2)
To find :
f(x,y) at (1,2)
partial derivatives of function w.r.t. (x and y) and value of that function at given points.
Solution:(refer the attachment also)
Now we know that
the equation of tangent plane at given points to the surface is given by,
f(x1,y1,z1) and z=f(x,y)
z-z1=Fx(x1,y1)*(x-x1)+Fy(x1,y1)*(y-y1)
here Fx(x1,y1) and Fy(x1,y1) are the partial derivatives of x and y.
now
taking partial derivative w.r.t. x we get
Fx(x1`,y1)=
=5.
Then w.r.t y we get
Fy(x1,y1)=
=2.
The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.
Using the Linearization or linear approximation we get
L(x,y)=f(x1,y1)+Fx(x,y)*(x-x1)+Fy(x,y)(y-y1)
=-1+5(x-1)+2(y-2)
=5x+2y-10
Approximation at F(1.1,1.9)
=5(1.1)+2(1.9)-10
=5.5+3.8-10
=0.7
Approximation at points (1.1,1.9) is 0.7
Let and
. Then
and
The expression under the square root can be rewritten as
Recall that
so that
and assuming and
, we end up with
so that
as required.