We want to find the linear approximation
to f of x comma y at the point two comma four comma 10,
then use linear approximation to approximate
a new function value.
Notice how this function value is near the given point.
So linear approximation to a function of two variables
is L of x comma y given here
where it's going to be approximately equal
to f of x comma y as long as x comma y is near
the point where we find the linear approximation
which is the point a comma b.
So let's take a look at this graphically.
Just like in two dimension,
we can use a tangent line to approximate a function value.
In three dimensions, we can use a tangent plane.
So here we have a surface graphed in purple.
The given point is this red point,
and the yellow plane is the tangent plane.
So when making a linear approximation for a function
of two variables, we're actually using this tangent plane
to approximate the new function value.
So going back to our notes,
in our case our linear approximation L of x comma y
at the point two comma four
would be equal to f of a comma b,
in our case notice a comma b would be two comma four
and x comma y would be the new coordinates 2.17 and 4.29.
So f of a comma b would be f of two comma four
plus the partial with respect to x
evaluated at two comma four times x minus a,
that'd be times the quantity x minus two
plus the partial with respect to y
evaluated at two comma four times the quantity y minus b
or y minus four.
Now if we look at this equation more carefully,
notice how this is just the function value
at the point of tangency
and then plus the sum of these products
is just differential z.
So all of this is differential z.
The changes z along the tangent plane,
which we're using to approximate the true delta z,
the change of z using the true function.
Because remember,
differential z is defined here.
And if we compare this to our linear approximation,
notice how differential x is just x minus a
and differential y is just y minus b.
So our next step will be to find our partial derivatives
and then evaluate them at the point two comma four.
Let's first rewrite the given function.
We have f of x comma y equals,
notice how we have a square root of two in the denominator.
Let's write this as five divided by two to the 1/2,
and then we have times xy to the 1/2.
In this form, it'll be easier to find
our partial derivatives.
So the partial with respect to x,
we'll treat y as a constant
and differentiate with respect to x.
So notice how we would multiply by 1/2.
So let's write that as five over two times two to the 1/2,
and then we'd subtract one from the exponent,
so xy to the negative 1/2 times the derivative of xy
with respect to x, that would be y.
So simplifying, let's write five y in the numerator.
And then for the denominator, we have this two,
we move this down to the denominator to make xy to the 1/2.
We already have a two to the 1/2 there,
which would be equivalent to the square root of two xy.
Now we want to evaluate this at the point two comma four.
So we'd have five times four divided by two,
and then we have the square root of two times two times four
that's the square root of 16,
so this simplifies to 20 over 8, 20/8,
common factor of four, this simplifies to 5/2.
And then we want to find the partial with respect to y,
treating x as a constant
and differentiating with respect to y.
So again, we'd multiply by 1/2.
So we have five over two times two to the 1/2
subtract one from the exponent, xy to the negative 1/2
times the derivative of xy with respect to y
which would now be x.
So now we have five x divided by two
times the square root of two xy.
Again, evaluating at the point two comma four,
now we have five times two divided by two
times the square root of 16, so now we have 10/8,
which simplifies to 5/4.
So now we have all the information we need
in order to find the equation of our linear approximation.
We have this value, and this value.
So now we'll perform those substitutions and simplify
to find our linear approximation at the given point.
We would have L of x comma y,
again at two comma four.
That would be f of two comma four.
Well the function value here would be the z coordinate of 10
plus the partial with respect to x at two comma four,
that's 5/2 times the quantity x minus two
and then plus 5/4 times the quantity y minus four.
We will go ahead and simplify this,
but notice how for the second part,
to find a linear approximation,
we would just simply substitute the new x and y values
into L of x comma y.
To simplify we'll distribute,
so we'll have 10 plus 5/2 x minus five
plus 5/4 y minus five.
Notice how we have 10 minus five minus five,
that would be zero.
So our linear approximation of f of x comma y
at the given point is L of x comma y equals
5/2 x plus 5/4 y.
And now to use this linear approximation
to approximate our new function value,
we just need to find L of 2.17 comma 4.29.
So we'd have 5/2 times 2.17
plus 5/4 times 4.29.
Let's go to the calculator.
We have 5/2
times 2.17
plus 5/4
times 4.29
which gives us 10.7875 for our approximation.
So again, the idea here was that f of 2.17 comma 4.29
is approximately equal to L of 2.17 comma 4.29,
which is 10.7875.
So we found the linear approximation to our function
f of x comma y at the given point here,
and then we use this to approximate a new function value
which we found to be 10.7875.
I hope you found this helpful.