Exercise
Find an approximate value of
\ln(0.09^3+0.99^3)
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=0.09, y=0.99
because these values appear in the question. And we set
x_0=0, y_0=1
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)=\ln(x^3+y^3)
In the formula above we see the function partial derivatives. Hence, we calculate them.
f'_x(x,y)=\frac{1}{x^3+y^3}\cdot 3x^2
f'_y(x,y)=\frac{1}{x^3+y^3}\cdot 3y^2
We put all the data in the formula and get
f(0.09,0.99)\approx f(0,1)+f'_x(0,1)\cdot(0.09-0)+f'_y(0,1)\cdot(0.99-1)=
=\ln(0^3+1^3)+\frac{1}{0^3+1^3}\cdot 3\cdot 0^2\cdot 0.09+\frac{1}{0^3+1^3}\cdot 3\cdot 1^2\cdot (-0.01)=
=\ln 1+1\cdot 0\cdot 0.02+1\cdot 3\cdot (-0.01)=
=0+0-0.03=
=-0.03
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