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Multivariable Linear Approximation – An expression with a power in 2 variables – Exercise 3390

Exercise

Find an approximate value of

1.02^{4.05}

Final Answer


1.08

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=4.05

because these values appear in the question. And we set

x_0=1, y_0=4

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)=x^y

In the formula above we see the function partial derivatives. Hence, we calculate them.

f'_x(x,y)=yx^{y-1}

f'_y(x,y)=x^y\ln x

We put all the data in the formula and get

f(1.02,4.05)\approx f(1,4)+f'_x(1,4)\cdot(1.02-1)+f'_y(1,4)\cdot(4.05-4)=

=1^4+4\cdot 1^3\cdot 0.02+1^4\ln 1\cdot 0.05=

=1+4\cdot 0.02+1\cdot 0\cdot 0.05=

=1+0.08+0=

=1.08

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