# Multivariable Linear Approximation – A multiplication with integer powers in 3 variables – Exercise 4223

Exercise

Find an approximate value of

$$1.002\cdot 2.003^2\cdot 3.004^3$$

Final Answer

$$108.972$$

Solution

We will use the 3-variable linear approximation formula

$$f(x,y,z)\approx f(x_0,y_0,z_0)+f'_x(x_0,y_0,z_0)\cdot(x-x_0)+$$

$$+f'_y(x_0,y_0,z_0)\cdot(y-y_0)+f'_z(x_0,y_0,z_0)\cdot(z-z_0)$$

Therefore, we need to define the following

$$x, y, z, x_0, y_0, z_0, f(x,y,z)$$

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

$$x_0,y_0,z_0$$

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

In our exercise, we will define

$$x=1.002, y=2.003,z=3.004$$

because these values appear in the question. And we set

$$x_0=1, y_0=2,z_0=3$$

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

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

$$f(x,y,z)=xy^2z^3$$

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

$$f'_x(x,y,z)=y^2z^3$$

$$f'_y(x,y,z)=2xyz^3$$

$$f'_z(x,y,z)=3xy^2z^2$$

We put all the data in the formula and get

$$f(1.002,2.003,3.004)\approx f(1,2,3)+f'_x(1,2,3)\cdot(1.002-1)+$$

$$+f'_y(1,2,3)\cdot(2.003-2)+f'_z(1,2,3)\cdot(3.004-3)=$$

$$= 1\cdot 2^2\cdot 3^3+2^2\cdot 3^3\cdot 0.002+2\cdot 1\cdot 2\cdot 3^3\cdot 0.003+3\cdot 1\cdot 2^2\cdot 3^2\cdot 0.004=$$

$$= 108+108\cdot 0.002+108\cdot 0.003+108\cdot 0.004=$$

$$= 108(1+0.002+0.003+0.004)=$$

$$= 108\cdot 1.009=$$

$$= 108.972$$

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