Exercise
Find an approximate value of
\sqrt{1.02^3+1.97^3}
Final Answer
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
Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!