Linear approximation in two variables
The linear approximation formula for multivariable functions
We can use the linear approximation formula
???L(x,y)=f(a,b)+\frac{\partial{f}}{\partial{x}}(a,b)(x-a)+\frac{\partial{f}}{\partial{y}}(a,b)(y-b)???
???(a,b)??? is the given point
???f(a,b)??? is the value of the function at ???(a,b)???
???\frac{\partial{f}}{\partial{x}}(a,b)??? is the partial derivative of ???f??? with respect to ???x??? at ???(a,b)???
???\frac{\partial{f}}{\partial{y}}(a,b)??? is the partial derivative of ???f??? with respect to ???y??? at ???(a,b)???
to find an approximation of the function at the given point ???(a,b)???.
How to use the formula to build the linear approximation equation for an equation in two variables
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Find each piece of the linear approximation equation, then plug every piece into the formula
Example
Find the linear approximation of the multivariable function at at ???(1,2)???.
???f(x,y)=6x^3-2xy^2???
The problem tells us that ???(a,b)=(1,2)???, so we need to find ???f(a,b)=f(1,2)???.
???f(1,2)=6(1)^3-2(1)(2)^2???
???f(1,2)=6-2(4)???
???f(1,2)=-2???
Then we need to find the partial derivatives of the function with respect to ???x??? and ???y???.
???\frac{\partial{f}}{\partial{x}}=18x^2-2y^2???
???\frac{\partial{f}}{\partial{x}}(1,2)=18(1)^2-2(2)^2???
???\frac{\partial{f}}{\partial{x}}(1,2)=10???
and
???\frac{\partial{f}}{\partial{y}}=-4xy???
???\frac{\partial{f}}{\partial{y}}(1,2)=-4(1)(2)???
???\frac{\partial{f}}{\partial{y}}(1,2)=-8???
Plugging the slope in each direction, ???(a,b)???, and ???f(a,b)??? into the linear approximation formula, we get
???L(x,y)=-2+(10)(x-1)+(-8)(y-2)???
???L(x,y)=-2+10x-10-8y+16???
???L(x,y)=10x-8y+4???
This is the linear approximation of the given function at the given point.