Multivariable Linear Approximation – An expression with arctan function in 2 variables – Exercise 4221

Exercise

Find an approximate value of

$\arctan(\frac{1.01}{0.98})$

Final Answer

Show final answer

Solution

We will use the 2-variable linear approximation formula

$f(x,y)\approx f(x_0,y_0)+f'_x(x_0,y_0)\cdot(x-x_0)+f'_y(x_0,y_0)\cdot(y-y_0)$

Therefore, we need to define the following

$x, y, x_0,  y_0, f(x)$

And place them in the formula. x and y will be the numbers that appear in the question, and the points

$x_0,y_0$

will be the closest values to x and y respectively, which are easy for calculations.

In our exercise, we will define

$x=1.01, y=0.98$

because these values appear in the question. And we set

$x_0=1,y_0=1$

because they are the closest values to x and y respectively that are easy for calculations.

After we have defined x and y, it is easy to find the function. Just put x instead of the number we set to be x and put y instead of the number we set to be y. In this way, we get the function

$f(x,y)=\arctan(\frac{x}{y})$

In the formula above we see the function partial derivatives. Hence, we calculate them.

$f'_x(x,y)=\frac{1}{{(\frac{x}{y})}^2+1}\cdot\frac{1}{y}=$

$=\frac{y^2}{x^2+y^2}\cdot\frac{1}{y}=$

$=\frac{y}{x^2+y^2}$

$f'_y(x,y)=\frac{1}{{(\frac{x}{y})}^2+1}\cdot(-\frac{x}{y^2})=$

$=\frac{y^2}{x^2+y^2}\cdot(-\frac{x}{y^2})=$

$=\frac{-x}{x^2+y^2}$

We put all the data in the formula and get

$f(1.01,0.98)\approx f(1,1)+f'_x(1,1)\cdot(1.01-1)+f'_y(1,1)\cdot(0.98-1)=$

$=\arctan(\frac{1}{1})+\frac{1}{1^2+1^2}\cdot 0.01+\frac{-1}{1^2+1^2}\cdot(-0.02)=$

$=\frac{\pi}{4}+\frac{1}{2}\cdot 0.01+\frac{-1}{2}\cdot(-0.02)=$

$=\frac{\pi}{4}+0.005+0.001=$

$=\frac{\pi}{4}+0.005+0.001=$

One can set

$\pi=3.1415$

And get

$=\frac{3.1415}{4}+0.005+0.001=0.791$

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