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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