[b]Tangent plane of a graph[/b][br][br]Given [math]z=f(x,y)[/math] and a point [math](x_0,y_0)[/math] in the domain of [math]f[/math]. Suppose [math]f[/math] is differentiable at [math](x_0,y_0)[/math]. How can we find the equation of the tangent plane to the graph of [math]z=f(x,y)[/math] at [math](x_0,y_0,f(x_0,y_0))[/math]?[br][br]We can use the following trick: Define the function of three variables [math]g(x,y,z)=z-f(x,y)[/math]. Then the graph of [math]z=f(x,y)[/math] can be regarded as the level surface [math]g(x,y,z)=0[/math]. Let [math]z_0=f(x_0,y_0)[/math]. In the previous section, we know that the equation of the tangent plane to the level surface [math]g(x,y,z)=0[/math] at [math](x_0,y_0,z_0)[/math] is as follows:[br][br][math]g_x(x_0,y_0,z_0)(x-x_0)+g_y(x_0,y_0,z_0)(y-y_0)+g_z(x_0,y_0,z_0)(z-z_0)=0[/math][br][br]By computation, we have[br][br][math]g_x=-f_x[/math], [math]g_y=-f_y[/math], [math]g_z=1[/math][br][br]i.e. [math]\nabla g(x_0,y_0,z_0)=\langle -f_x(x_0,y_0),-f_y(x_0,y_0),1 \rangle[/math] (normal vector to the tangent plane)[br][br]Therefore, we have[br][br][math]-f_x(x_0,y_0)(x-x_0)-f_y(x_0,y_0)(y-y_0)+(z-z_0)=0[/math][br][math]\implies z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)[/math][br][br][br][u]Example[/u]: Suppose [math]z=f(x,y)=x^2y[/math]. Find the equation of the tangent plane to the graph of [math]z=f(x,y)[/math] when [math](x,y)=(2,1)[/math]. Also, find the parametric equation of the normal line when [math](x,y)=(2,1)[/math].[br][br][u]Answer[/u]:[br][br][math]f_x=2xy[/math], [math]f_y=x^2[/math]. Then [math]f(2,1)=4[/math], [math]f_x(2,1)=4[/math] and [math]f_y(2,1)=4[/math]. Therefore, the equation of the tangent plane is [br][br][math]z=4+4(x-2)+4(y-1)[/math][br][math]\implies 4x+4y-z-8=0[/math][br][br]As for the normal line at [math](2,1,4)[/math], the direction vector is [math]\langle -f_x(2,1),-f_y(2,1),1\rangle=\langle -4,-4,1\rangle[/math]. Hence, its parametric equations are [br][br][math]\begin{cases} x = & 2-4t \\ y = & 1-4t \\ z = & 4+t \end{cases}[/math][br][br]where [math]t[/math] is any real number.[br][br][br][br]

[b]Linear approximation[/b][br][br]Let [math]z=f(x,y)[/math] be a function of two variable such that it is differentiable at [math](x_0,y_0)[/math]. Then we have following linear approximation:[br][br][math]f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0) \approx f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y[/math][br][br]Let [math](x,y)=(x_0+\Delta x,y_0+\Delta y)[/math], we can rewrite the above as follows:[br][br][math]f(x,y) \approx L(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)[/math][br][br][math]L(x,y)[/math] is the [b]linear approximation[/b] of [math]f(x,y)[/math] when [math](x,y)[/math] is near [math](x_0,y_0)[/math] (or equivalently, [math](\Delta x,\Delta y)[/math] is near [math](0,0)[/math]).[br][br][br][u]Example[/u]: Let [math]f(x,y)=\sqrt{x^2+y^2}[/math]. Find its linear approximation at [math](3,4)[/math] and then use it to approximate [math]f(3.04,3.98)[/math].[br][br][u]Answer[/u]:[br][br][math]f(3,4)=\sqrt{3^2+4^2}=5[/math], [math]f_x=\frac x{\sqrt{x^2+y^2}}[/math] and [math]f_y=\frac y{\sqrt{x^2+y^2}}[/math]. Hence [math]f_x(3,4)=\frac 35[/math] and [math]f_y(3,4)=\frac 45[/math].[br][br][math]L(x,y)=5+\frac 35(x-3)+\frac 45(y-4)[/math][br][br]We can use this to approximate the value of [math]f(3.04,3.98)[/math]:[br][br][math]f(3.04,3.98)\approx 5+\frac 35(3.04-4)+\frac 45(3.98-4)=5.008[/math][br][br]([u]Note[/u]: The approximation is justified only when [math]\Delta x[/math] and [math]\Delta y[/math] are small. In this example, [math]\Delta x=0.04[/math] and [math]\Delta y=-0.02[/math].)[br][br][br][u]Remark[/u]: For [math]w=g(x,y,z)[/math] a function of three variables such that it is differentiable at [math](x_0,y_0,z_0)[/math], we can define its linear approximation at [math](x_0,y_0,z_0)[/math] in a similar way:[br][br][math]g(x,y,z)\approx L(x,y,z)=f(x_0,y_0,z_0)+f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)[/math][br][br]where [math](x,y,z)[/math] is near [math](x_0,y_0,z_0)[/math].[br][br][br][br][br][br]