Exercise
Prove that the following holds for x and y that are close to 1
x 3 + 3 y 2 ≈ 3 4 x + 3 2 y − 1 4 \sqrt{x^3+3y^2}\approx \frac{3}{4}x+\frac{3}{2}y-\frac{1}{4} x 3 + 3 y 2 ≈ 4 3 x + 2 3 y − 4 1
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 ) ≈ f ( x 0 , y 0 ) + f x ′ ( x 0 , y 0 ) ⋅ ( x − x 0 ) + f y ′ ( x 0 , y 0 ) ⋅ ( y − y 0 ) 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) f ( x , y ) ≈ f ( x 0 , y 0 ) + f x ′ ( x 0 , y 0 ) ⋅ ( x − x 0 ) + f y ′ ( x 0 , y 0 ) ⋅ ( y − y 0 )
Therefore, we need to define the following
x , y , x 0 , y 0 , f ( x ) x, y, x_0, y_0, f(x) 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 x_0=1, y_0=1 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 x_0=1, y_0=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 ) = x 3 + 3 y 2 f(x,y)=\sqrt{x^3+3y^2} f ( x , y ) = x 3 + 3 y 2
In the formula above we see the function partial derivatives. Hence, we calculate them.
f x ′ ( x , y ) = 1 2 x 3 + 3 y 2 ⋅ 3 x 2 f'_x(x,y)=\frac{1}{2\sqrt{x^3+3y^2}}\cdot 3x^2 f x ′ ( x , y ) = 2 x 3 + 3 y 2 1 ⋅ 3 x 2
f y ′ ( x , y ) = 1 2 x 3 + 3 y 2 ⋅ 6 y f'_y(x,y)=\frac{1}{2\sqrt{x^3+3y^2}}\cdot 6y f y ′ ( x , y ) = 2 x 3 + 3 y 2 1 ⋅ 6 y
We put all the data in the formula and get
x 3 + 3 y 2 ≈ f ( 1 , 1 ) + f x ′ ( 1 , 1 ) ⋅ ( x − 1 ) + f y ′ ( 1 , 1 ) ⋅ ( y − 1 ) = \sqrt{x^3+3y^2}\approx f(1,1)+f'_x(1,1)\cdot(x-1)+f'_y(1,1)\cdot(y-1)= x 3 + 3 y 2 ≈ f ( 1 , 1 ) + f x ′ ( 1 , 1 ) ⋅ ( x − 1 ) + f y ′ ( 1 , 1 ) ⋅ ( y − 1 ) =
= 1 3 + 3 ⋅ 1 2 + 1 2 1 3 + 3 ⋅ 1 2 ⋅ 3 ⋅ 1 2 ⋅ ( x − 1 ) + 1 2 1 3 + 3 ⋅ 1 2 ⋅ 6 ⋅ 1 ⋅ ( 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)= = 1 3 + 3 ⋅ 1 2 + 2 1 3 + 3 ⋅ 1 2 1 ⋅ 3 ⋅ 1 2 ⋅ ( x − 1 ) + 2 1 3 + 3 ⋅ 1 2 1 ⋅ 6 ⋅ 1 ⋅ ( y − 1 ) =
= 4 + 1 2 4 ⋅ 3 ⋅ ( x − 1 ) + 1 2 4 ⋅ 6 ⋅ ( y − 1 ) = =\sqrt{4}+\frac{1}{2\sqrt{4}}\cdot 3\cdot (x-1)+\frac{1}{2\sqrt{4}}\cdot 6\cdot (y-1)= = 4 + 2 4 1 ⋅ 3 ⋅ ( x − 1 ) + 2 4 1 ⋅ 6 ⋅ ( y − 1 ) =
= 2 + 3 4 ⋅ ( x − 1 ) + 6 4 ⋅ ( y − 1 ) = =2+\frac{3}{4}\cdot (x-1)+\frac{6}{4}\cdot (y-1)= = 2 + 4 3 ⋅ ( x − 1 ) + 4 6 ⋅ ( y − 1 ) =
= 2 + 3 4 x − 3 4 + 6 4 y − 6 4 = =2+\frac{3}{4}x-\frac{3}{4}+\frac{6}{4}y-\frac{6}{4}= = 2 + 4 3 x − 4 3 + 4 6 y − 4 6 =
= 3 4 x + 3 2 y − 1 4 =\frac{3}{4}x+\frac{3}{2}y-\frac{1}{4} = 4 3 x + 2 3 y − 4 1
Hence, we get
x 3 + 3 y 2 ≈ 3 4 x + 3 2 y − 1 4 \sqrt{x^3+3y^2}\approx \frac{3}{4}x+\frac{3}{2}y-\frac{1}{4} x 3 + 3 y 2 ≈ 4 3 x + 2 3 y − 4 1
As required.
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