Question

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)?

Answer 1

`L(x,y)=0.92539816`. See below.

Explanation:

If we stay near the point of tangency (`x_0,y_0`), then the tangent plane serves as a linear approximation of `f(x,y)`.

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 `L(x,y)` is the linear approximation of `f(x,y)`. Essentially, when `(x,y)~~(x_0,y_0)`, then `f(x,y)~~L(x,y)`.

Given that `f(x,y)=arctan(x*y^2)`, we can use linear approximation to estimate `f(0.94,1.17)`.

First, we pick "nice" points for the approximation which are near the original `x` and `y` values. Integers are great choices. Given `x=0.94` and `y=1.17`, I would use `x_0=1` and `y=1`. These are much more pleasant to work with.

`f(x,y)=arctan(x*y^2)`

We take the partial derivative of the function with respect to `x`:

`(del)/(delx)=y^2/(x^2y^4+1)`

Then with respect to `y`:

`(del)/(dely)=(2xy)/(x^2y^4+1)`

Now, just as we do when finding the tangent plane, we calculate `f_x(x_0,y_0)`, `f_y(x_0,y_0)`, and `f(x_0,y_0)`. From above, we've chosen `(x_0,y_0)=(1,1)`.

`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 `z=L(x,y)=>L(x,y)=x/2+y+pi/4-3/2`.

This means that when `(x,y)` is close to `(1,1)`, `arctan(x*y^2)~~x/2+y+pi/4-3/2`.

Finally, our linear approximation is found by putting the actual `x` and `y` values into the equation we found for `L(x,y)`:

`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 `arctan(x*y^2)` for `(0.94,1.17)`.

`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 `1.7%`.