For 2-dimensional functions ( \(f:\R \to \R\)), differentiable is defined as the derivative existing. For higher dimensional functions simply extending this definition to all partial derivatives exist is insufficient.
Activity32.
The following steps review the geometric characterization of differentiable, 2D functions. Use a device to perform the zooming in on the graphs.
(a)
\(f(x)=x^2\) is differentiable at \(x=0.\) Graph \(f(x)\) for \(x \in [-1,1]\) starting with a view that shows \(x \in [-1,1]\) and \(y \in [0,1]\text{.}\) Zoom in about \(x=0\text{.}\)
(b)
What does \(f(x)\) look increasingly like as you zoom in?
(c)
\(g(x)=x^{2/3}\) is not differentiable at \(x=0.\) Graph \(g(x)\) for \(x \in [-1,1]\) starting with a view that shows \(x \in [-1,1]\) and \(y \in [0,1]\text{.}\) Zoom in about \(x=0\text{.}\)
(d)
How does \(g(x)\) differ from \(f(x)\) as you zoom in?
(e)
Calculate and graph the line tangent to \(f(x)\) at \(x=0.\) Compare the graphs of \(f(x)\) and its tangent line as you zoom in.
Subsection2.3.2Concept
The relationship between the tangent line and the differentiable curve provides a means to extend the definition of differentiable to surfaces and higher dimensions. First consider the extension of the concept of tangent to a surface. For a curve (one dimensional object) the tangent is a one dimensional, “flat” object that locally contacts the curve at only one point. This is a line. For a surface (two dimensional object) the tangent is a two dimensional, “flat” object, that locally contacts the surface at only one point. This is a plane.
Activity33.
The following illustrate how to calculate a tangent plane and how tangent planes help define differentiable.
(a)
Use the surface \(f(x,y)=x^2-y^2\) for the following.
(i)
If Guido is walking along this surface following the path \(p_1(t)=(t,0),\) what curve (\(P_1(t)=(x,y,z)\)) is he following?
(ii)
Calculate the line tangent to this curve above the point \((0,0).\)
(iii)
If Guido is walking along this surface following the path \(p_2(t)=(0,t),\) what curve (\(P_2(t)=(x,y,z)\)) is he following?
(iv)
Calculate the line tangent to this curve above the point \((0,0).\)
(v)
Both of these curves are in the surface. Does it make sense for their tangent lines to be in the tangent plane?
(vi)
Calculate the equation of the plane containing these two lines.
(vii)
If Guido is walking along this surface following the path \(p(t)=(t,t),\) what curve (\(P_3(t)=(x,y,z)\)) is he following?
(viii)
Calculate the tangent line to this curve above the point \((0,0).\)
(ix)
Check if this tangent line is also in the tangent plane.
(b)
Use the surface \(g(x,y)=\left\{ \begin{array}{ll} 0 & \mbox{if} (x,y)=(0,0) \\ \frac{xy}{x^2+y^2} & \mbox{elsewise} \end{array} \right.\) for the following.
(i)
If Guido is walking along this surface following the path \(p_1(t)=(t,0),\) what curve (\(P_1(t)=(x,y,z)\)) is he following?
(ii)
Calculate the line tangent to this curve above the point \((0,0).\)
(iii)
If Guido is walking along this surface following the path \(p_2(t)=(0,t),\) what curve (\(P_2(t)=(x,y,z)\)) is he following?
(iv)
Calculate the line tangent to this curve above the point \((0,0).\)
(v)
Using this information what is \(\frac{\partial f}{\partial x}\) at \((0,0).\)
(vi)
Using this information what is \(\frac{\partial f}{\partial y}\) at \((0,0).\)
(vii)
Write the equation of the plane defined by these two vectors.
(viii)
Compare this plane to the surface by graphing.
(c)
Which of these two surfaces appears to be differentiable at \((0,0)\text{?}\)
Subsection2.3.3Calculation
The next section explains how to conveniently calculate the tangent plane before formally defining differentiable.
Activity34.
Use the surface \(f(x,y)=x^2-y^2\) for the following.
(a)
Calculate \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}.\)
(b)
Use \(P(t)=(t,0,f(t,0))\text{,}\) from Activity 33. Calculate the slope of the line tangent to \(P(t)\) at the point \((x,0,f(x,0))\text{.}\)
(c)
Use \(P(t)=(0,t,f(t,0))\text{,}\) from Activity 33. Calculate the slope of the line tangent to \(P(t)\) at the point \((0,y,f(x,0))\text{.}\)
(d)
Where do the partial derivatives from the first step show up in the slopes of the tangent lines?
(e)
Calculate the vector orthogonal to these two tangent line slopes.
(f)
Write the equation for the plane tangent to \(f\) at \((x,y,f(x,y))\text{.}\)
(g)
Where do the partial derivatives from the first step show up in this tangent plane equation?
Definition2.3.1.Differentiable.
A surface \(z=f(x,y)\) is differentiable if and only if \(\partial z = f(a+\Delta x,b+\Delta y)-f(a,b) = \Delta x \frac{\partial f(a,b)}{\partial x} + \Delta y \frac{\partial f(a,b)}{\partial y} + \Delta x\epsilon_1 + \Delta y\epsilon_2 \) and \(\epsilon_1, \epsilon_2 \to 0\) as \((\Delta x,\Delta y) \to (0,0).\)
Note this means that the plane is a good approximation of the surface near the point of tangency. The following conditions are also sufficient for differentiability and are easier to check.
Theorem2.3.2.Differentiable.
For a surface \(f(x,y)\) if \(\frac{\partial f}{\partial x} \) and \(\frac{\partial f}{\partial y} \) exist in a neighborhood around \((a,b)\) and are continuous at \((a,b)\) then \(f\) is differentiable at \((a,b).\)
Checkpoint2.3.3.
Use the surface \(f(x,y)=x^2+xy+y^2-16x+2\) for the following.
(a)
Find the equation of the plane tangent to \(f\) at \((0,0).\)
(b)
Find the equation of the plane tangent to \(f\) that is parallel to the plane \(2x+3y-z+1=0.\)
