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Multivariable Linear Approximation – Proving an expression with a square root in 2 variables – Exercise 3404

Exercise

Prove that the following holds for x and y that are close to 1

\sqrt{x^3+3y^2}\approx \frac{3}{4}x+\frac{3}{2}y-\frac{1}{4}

Proof

Since the equation we need to prove has the estimator sign, we probably need to use the linear approximation formula. Since there are two variables, 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 we set

x_0=1, y_0=1

Because we are asked to prove the equation for points close to 1.

x_0=1, y_0=1

The function will be the right side in the equation that we need to prove

f(x,y)=\sqrt{x^3+3y^2}

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

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

f'_y(x,y)=\frac{1}{2\sqrt{x^3+3y^2}}\cdot 6y

We put all the data in the formula and get

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

=\sqrt{1^3+3\cdot 1^2}+\frac{1}{2\sqrt{1^3+3\cdot 1^2}}\cdot 3\cdot 1^2\cdot (x-1)+\frac{1}{2\sqrt{1^3+3\cdot 1^2}}\cdot 6\cdot 1\cdot (y-1)=

=\sqrt{4}+\frac{1}{2\sqrt{4}}\cdot 3\cdot (x-1)+\frac{1}{2\sqrt{4}}\cdot 6\cdot (y-1)=

=2+\frac{3}{4}\cdot (x-1)+\frac{6}{4}\cdot (y-1)=

=2+\frac{3}{4}x-\frac{3}{4}+\frac{6}{4}y-\frac{6}{4}=

=\frac{3}{4}x+\frac{3}{2}y-\frac{1}{4}

Hence, we get

\sqrt{x^3+3y^2}\approx \frac{3}{4}x+\frac{3}{2}y-\frac{1}{4}

As required.

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