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Multivariable Linear Approximation – An expression with ln function in 2 variables – Exercise 3397

Exercise

Find an approximate value of

\ln(0.09^3+0.99^3)

Final Answer


-0.03

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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