Multivariable Calculus Linear Approximation
I am having a problem with the question:
Use the linear approximation of $f(x, y) = e^{2x^2+3y}$ at $(0, 0)$ to estimate $f(0.01, -0.02)$.
I know how to take linear approximations with one variable by taking the derivative, but I am a bit lost on how to do this with two variables (partial differentiation I think). I have tried many things but I am unable to get the correct answer. Later on in this assignment, a third variable is introduced. Could somebody please help me through this problem?
Thanks so much!
$\endgroup$ 12 Answers
$\begingroup$$$ df = \frac{\partial f}{\partial x} \,dx+ \frac{\partial f}{\partial y}\, dy. \tag{This is a chain rule.} $$ $$ \Delta f \approx \frac{\partial f}{\partial x} \,\Delta x+ \frac{\partial f}{\partial y}\, \Delta y. $$ $$ f(0+\Delta x,0+\Delta y) = f(0,0) + \Delta f. $$ \begin{align} f(0,0) & =1 \\ \Delta x & = 0.01 \\ \Delta y & = -0.02. \end{align}
$\endgroup$ $\begingroup$With one dependent variable we use the tangent line to approximate, with two dependent variables we use the tangent plane to approximate. The tangent plane has a normal vector of $\langle 1,0,f_x \rangle \times \langle 0,1,f_y \rangle=\langle -f_x,-f_y,1 \rangle$.
Hence the equation of the tangent plane at a point $(a,b,c)$ is:
$$-f_x(a,b)(x-a)-f_y(a,b)(y-b)+1(z-c)=0$$
Dividing by $-1$ on both sides and solving for $z$ we get:
$$f_x(a,b)(x-a)+f_y(a,b)(y-b)+c=z$$
Substituting values of $(x,y)$ near $(a,b)$ will give us an approximation of $z(x,y)$.
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