Question

whats the difference between linear approximation and finding the equation of a tangent plane?

Answer 1

was currently taking a look at this

http://tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes.aspx

http://tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes_files/eq0058MP.gif

Answer 2

why is it only approximate? it seems to be the same thing

Answer 3

@zepdrix

Answer 4

the linear approximation (called sometimes the differential, total derivative, Jacobian, and just plain derivative) of a function \(f:\mathbb{R}^2\to\mathbb{R}\) at a point \(p\) is a linear function \(df_p\) in \(x,y\) such that: $$\lim_{(x,y)\to p}\frac{|f(x,y)-f(p)-df_p(x,y)|}{\|(x,y)-p\|}=0$$

Answer 5

in particular, if the derivative exists, because it's defined to be *linear* we have that: $$df_p(ax_1+bx_2,ay_1+by_2)=a\cdot df_p(x_1,y_1)+b\cdot df_p(x_2,y_2)$$

now consider that the derivative in a particular direction, say, in an offset of \((1,0)\) should give the partial derivative \(\frac{\partial f}{\partial x}\), and likewise for \((0,1)\) and the other partial derivative \(\frac{\partial f}{\partial y}\) so: $$df_p(1,0)=\frac{\partial f}{\partial x}\\df_p(0,1)=\frac{\partial f}{\partial y}$$

Answer 6

by linearity this tells us that: $$df_p(x,y)=df_p(x+0,0+y)=x\cdot df_p(1,0)+y\cdot df_p(0,1)=\frac{\partial f}{\partial x}x+\frac{\partial f}{\partial y}y$$

Answer 7

now, \(df_p\) captures the first-order or linear behavior of \(f\) around a point \(p\), giving us: $$f(p+(x,y))\approx f(p)+df_p(x,y)=f(p)+\frac{\partial f}{\partial x}\bigg|_{p}\cdot x+\frac{\partial f}{\partial y}\bigg|_{p}\cdot y$$ now notice that if we subsitute \(z\) for this approximate value for \(f(p+(x,y))\) we have: $$z=f(p)+\frac{\partial f}{\partial x}x+\frac{\partial f}{\partial y}\\\implies \frac{\partial f}{\partial x}x+\frac{\partial f}{\partial y}y-z=f(p)\\\implies ax+by+cz=d$$where \(a=\frac{\partial f}{\partial x},b=\frac{\partial f}{\partial y},c=-1,d=f(p)\) are all constants here (since we're evaluating the partial derivatives at the point \(p\))

Answer 8

which is the equation of a plane; in fact, the tangent plane to a surface given by \(z=f(x,y)\) at a point \(p=(x_0,y_0)\) is precisely the linear approximation in terms of the total derivative \(df_{(x_0,y_0)}\), so: $$f(x,y)\approx f(x_0,y_0)+\frac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}(x-x_0)+\frac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}(y-y_0)$$

Answer 9

at a point \(p\) the tangent plane there is locally what the function "looks like" to first-order, i.e. the best linear approximation around the point \(p\)