Multivariable Linear Approximation – An expression with a square root in 2 variables – Exercise 4219


Find an approximate value of


Final Answer



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


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

In our exercise, we will define

x=1.02, y=1.97

because these values appear in the question. And we set

x_0=1, y_0=2

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


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

f'_x(x,y)=\frac{1}{2\sqrt{x^3+y^3}}\cdot 3x^2=\frac{3x^2}{2\sqrt{x^3+y^3}}

f'_y(x,y)=\frac{1}{2\sqrt{x^3+y^3}}\cdot 3y^2=\frac{3y^2}{2\sqrt{x^3+y^3}}

We put all the data in the formula and get

f(1.02,1.97)\approx f(1,2)+f'_x(1,2)\cdot(1.02-1)+f'_y(1,2)\cdot(1.97-2)=

=\sqrt{1^3+2^3}+\frac{3\cdot 1^2}{2\sqrt{1^3+2^3}}\cdot 0.02+\frac{3\cdot 2^2}{2\sqrt{1^3+2^3}}\cdot (-0.03)=

=\sqrt{9}+\frac{3}{2\sqrt{9}}\cdot 0.02+\frac{12}{2\sqrt{9}}\cdot (-0.03)=

=3+\frac{3}{2\cdot 3}\cdot 0.02+\frac{12}{2\cdot 3}\cdot (-0.03)=

=3+\frac{3}{6}\cdot 0.02+\frac{12}{6}\cdot (-0.03)=

=3+\frac{1}{2}\cdot 0.02+2\cdot (-0.03)=



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