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Multivariable Linear Approximation – An expression with a square root in 2 variables – Exercise 3400

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


\sqrt{5e^{0.02}+{2.03}^2}\approx 3.037

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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