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Thread: tangent-plane approximation

  1. #1
    Feb 2009

    tangent-plane approximation

    How do you show that the tangent plane at the point $\displaystyle P_{0}(x_{0},y_{0})$, $\displaystyle f(x_{0},y_{0})$ on the surface z=f(x,y) defined by a differentiable function f is the plane:

    $\displaystyle f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})-(z-f(x_{0},y_{0})=0$


    $\displaystyle z=f(x_{0},y_{0})+f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$
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  2. #2
    Senior Member Pinkk's Avatar
    Mar 2009
    Uptown Manhattan, NY, USA
    Let's look at the partial derivatives and see what we can understand from them. Given some function $\displaystyle f(x,y)$, $\displaystyle \frac{\partial f}{\partial x}(a,b)$ measures the rate of change of $\displaystyle f(x,y)$ at the point $\displaystyle (a,b)$ with respect to $\displaystyle x$, and $\displaystyle \frac{\partial f}{\partial y}(a,b)$ measures the rate of change of $\displaystyle f(x,y)$ at the point $\displaystyle (a,b)$ with respect to $\displaystyle y$. Now we will look for vector-valued functions of tangent lines at the point $\displaystyle (a,b)$ to the curves that are created when we intersect $\displaystyle f(x,y)$ with the plane $\displaystyle y=b$ for $\displaystyle \frac{\partial f}{\partial x}(a,b)$ and the plane $\displaystyle x=a$ for $\displaystyle \frac{\partial f}{\partial y}(a,b)$. Now, how can we write the vector formula for the tangent line at that point in each plane? Let's look at the $\displaystyle y=b$ intersection. Remember that the general vector-valued function is going to be $\displaystyle <x(t),y(t),z(t)>$. Let's just choose $\displaystyle x(t)=1$. We could pick any constant, but let's choose 1 for ease. This is just saying that the x-coordinate of the tangent line to the curve created by the $\displaystyle y=b$ intersection will constantly increase or decrease. Since y is fixed, $\displaystyle y(t)=0$ since y will never change. $\displaystyle z(t)$ will be the slope of the tangent at the point $\displaystyle (a,b)$, which is none other than $\displaystyle z(t)=\frac{\partial f}{\partial x}(a,b)$. Let's call this vector $\displaystyle A=<1,0,\frac{\partial f}{\partial x}(a,b)>$. Following the logic of the $\displaystyle y=b$ plane intersection, let's have the tangent line at the point $\displaystyle (a,b)$ to the curve that results from the intersection $\displaystyle x=a$ be the vector $\displaystyle B=<0,1,\frac{\partial f}{\partial y}(a,b)>$. These vectors are not parallel and they are not skew (they intersect at the point $\displaystyle (a,b,f(a,b))$. So they must lie on some plane that will be tangental to the point. To get the equation of the plane, we need the point and the normal vector. The normal vector is simply the cross product of those two tangent vectors.
    $\displaystyle B \times A=<\frac{\partial f}{\partial x}(a,b), \frac{\partial f}{\partial y}(a,b), -1>$. So we have our point $\displaystyle (a,b,f(a,b))$ and our normal vector. Plugging this into the general formula for a plane, we get:

    $\displaystyle 0=\frac{\partial f}{\partial x}(a,b)(x-a)+\frac{\partial f}{\partial y}(a,b)(y-b)-(z-f(a,b))$

    We can rewrite this as:

    $\displaystyle z=f(a,b)+\frac{\partial f}{\partial x}(a,b)(x-a)+\frac{\partial f}{\partial y}(a,b)(y-b)$
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