Exercise
Given the function
f(x,y)=\sqrt{5e^x+y^2}
Use linear approximation around the point
(x,y)=(0,2)
In order to find an approximate value of
\sqrt{5e^{0.02}+{2.03}^2}
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=0.02, y=2.03
because these values appear in the question. And we set
x_0=0, 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{5e^x+y^2}
In the formula above we see the function partial derivatives. Hence, we calculate them.
f'_x(x,y)=\frac{1}{2\sqrt{5e^x+y^2}}\cdot 5e^x
f'_y(x,y)=\frac{1}{2\sqrt{5e^x+y^2}}\cdot 2y
We put all the data in the formula and get
f(0.02,2.03)\approx f(0,2)+f'_x(0,2)\cdot(0.02-0)+f'_y(0,2)\cdot(2.03-2)=
=\sqrt{5e^0+2^2}+\frac{1}{2\sqrt{5e^0+2^2}}\cdot 5e^0\cdot 0.02+\frac{1}{2\sqrt{5e^0+2^2}}\cdot 2\cdot 2\cdot (0.03)=
=\sqrt{9}+\frac{1}{6}\cdot 5\cdot 0.02+\frac{1}{6}\cdot 4\cdot 0.03=
=3+\frac{5}{6}\cdot 0.02+\frac{4}{6}\cdot 0.03=
=3+0.0166+0.02=
=3.037
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