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

Exercise

Find an approximate value of

$\sqrt{1.02^3+1.97^3}$

Final Answer

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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

$x=1.02, y=1.97$

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

$f(x,y)=\sqrt{x^3+y^3}$

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

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

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

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

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

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

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

$=3+0.01-0.06=$

$=2.95$

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