We're asked to find the linear approximation
to f of x comma y equals 4xy minus three x squared y
at the point one comma two comma two.
Then use the linear approximation
to approximate the function value f of 1.1 comma 2.1.
When making a linear approximation
for a function of two variables,
we are using a tangent plane to make the approximation.
We're using the change in z along the tangent plane
to make the approximation for the true function value.
If we take a look at this graphically,
the function f of x comma y is the purple surface.
The point one comma two comma two is this red point,
and this blue plane is tangent to the surface
at the red point.
To make a linear approximation, we're using the change in z
along the blue tangent plane to make the approximation
for the true function value given by the purple surface.
Going back to our work if we take a look below,
the linear approximation for f of x comma y
at the point a comma b is L of x comma y
which equals f of a comma b
which is the function value at the point of tangency.
Plus all of this is just the change in z
along the tangent plane which is really differential z.
Which is equal to the partial of f evaluated
at a comma b times the quantity x minus a
plus the partial of f with respect to y
at a comma b times the quantity y minus b.
If we go back and look at some earlier notes,
remember differential z is equal
to the partial of f times dx plus the partial of y times dy
which we see here in our linear approximation.
So again this is really just the function value
at the point of tangency plus differential z,
the change of z along the tangent plane.
To get started, we're determining the linear approximation
at one comma two.
Which means for our formula a is equal to one,
and b is equal to two.
So f of x comma y is approximately equal to L of x comma y
which is equal to f of one comma two
plus the partial of f with respect to x at one comma two
times the quantity x minus a.
Which is x minus one which is really differential x.
Plus the partial of f with respect to y evaluated
at one comma two times the quantity y minus b
which is y minus two which is really differential y.
For the next step, let's go ahead and find
these partial derivatives, and then evaluate them
at one comma two.
To find the partial of f with respect to x,
you differentiate f with respect to x
treating y as a constant.
The derivative of 4xy with respect to x is 4y.
So we have 4y minus the derivative of three x squared y
with respect to x is 6xy.
Now we need to evaluate this at one comma two
which is four times two minus six
times one times two which is equal
to eight minus 12 or negative four.
Now I need to find the partial of f with respect to y.
To do this, we differentiate f with respect to y.
Treating x as a constant, the derivative of 4xy
with respect to y is 4x.
4x minus the derivative of three x squared y
with respect to y is three x squared.
I need to evaluate this at one comma two
which is four times one minus three times one squared
which is four minus three which equals one.
Now we have f of x comma y is approximately equal to
L of x comma y.
Which equals f of one comma two is equal
to the z-coordinate here which is two.
So we have two plus the partial of f with respect to x
at one comma two is negative four.
So we'll have minus four times the quantity x minus one,
and then plus the partial of f with respect to y
at one comma two is one times the quantity y minus two.
So here is our linear approximation for f of x comma y
at the point one comma two,
and now we can approximate the function value.
F of 1.1 comma 2.1 is approximately equal
to L of 1.1 comma 2.1.
What is equal to two minus four
times the quantity 1.1 minus one
plus one times the quantity 2.1 minus two.
Simplifying, we have two minus four times 0.1
plus one times 0.1
which gives us two minus 0.4 plus 0.1
which equals 1.7.
Again, here is the linear approximation
for f of x comma y at the point one comma two.
And here is the approximation for f of 1.1 comma 2.1.
I hope you found this helpful.