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Chapter 10 Limits and Derivatives

“Real knowledge is to know the extent of one's ignorance.” - Confucius

For the next two definitions, suppose that \(x, y,\) and \(z : \re \to \re\) are differentiable functions.

Definition 10.1.

The limit of the vector valued function of one variable \(\oa f(t) = (x(t), y(t), z(t)),\) as \(t\) approaches \(a\) is defined by

\begin{equation*} \lim_{t\to a} \ \oa f(t)= \left( \lim_{t\to a}\ x(t), \lim_{t\to a}\ y(t),\lim_{t\to a}\ z(t) \right) \end{equation*}

as long as each of these limits exists. If any one of the limits does not exist, then the limit of \(\oa f\) does not exist at \(a\text{.}\)

Definition 10.2.

The derivative of the vector valued function of one variable \(\oa f(t) = (x(t), y(t), z(t)),\) is defined by

\begin{equation*} \oa f'(t)=(x'(t), y'(t), z'(t)) \end{equation*}

as long as each of the derivatives exists. If any one of the derivatives does not exist, then the derivative of \(\oa f\) does not exist.

Problem 10.3.

Let \(\oa f(t)=\dsp{\left(t^2-4,\ \frac{\sin(t)}{t},\ \frac{4t^3}{e^t} \right) }\text{.}\) Compute \(\dsp{\lim_{t\to 0}\ \oa f(t)}\text{.}\)

Problem 10.4.

Compute \(\oa f'\) and \(\oa f'(0)\) for \(\oa f\) from the previous problem.

We now wish to develop the rules for limits and derivatives that parallel the rules from calculus in one dimension. Of course, we know you have not forgotten any of these rules, so the ones we develop should look familiar! The good news is that the really hard work in proving these was done in the first semester of calculus and thus the work here is more notational than mathematical!

Problem 10.5.

Let \(\oa f(t) = (t^2,t^3-1,\sqrt{t-1})\) and \(\oa g(t)= (2-t^2,t^3,\sqrt{t+1})\text{.}\)

  1. Compute \(\dsp \lim_{t\to 2}\ \oa f(t).\)

  2. Compute \(\dsp \lim_{t\to 2}\ \oa g(t).\)

  3. Compute \(\dsp \lim_{t\to 2}\ \oa f(t) + \lim_{t\to 2}\ \oa g(t).\)

  4. Compute \(\dsp \lim_{t\to 2}\ [\oa f(t)+ \oa g(t)].\)

  5. What can you conjecture about \(\dsp \lim_{t\to a}\ [\oa f(t)+ \oa g(t)]\) for arbitrary choices of \(a, \oa f,\) and \(\oa g\text{?}\)

Problem 10.6.

State 5 rules for limits of the vector valued functions, \(\oa f, \oa g: \re \to \re^3\) that parallel the limit rules from Calculus I and prove one of these conjectures. You may grab a book or look on the web to remind you of the rules from Calculus I.

Problem 10.7.

Compute \(\oa f'(2)\) and \(\oa g'(2)\) where \(\oa f\) and \(\oa g\) are from Problem 10.5. Compute \(\oa{(f+g)}'(2)\text{.}\) What can you conjecture about \(\oa{(f+g)}'(t)\) for arbitrary choices of \(\oa f\) and \(\oa g\text{?}\)

Problem 10.8.

State 5 rules for derivatives of vector valued functions that parallel the derivative rules from Calculus I. Prove one of these conjectures. You may grab a book or look on the net to remind you of the rules from Calculus I.

You may assume that which ever one you do not prove will end up on the next test. Yes, it is a well known fact that all calculus teachers can read students' minds. How else would we always be able to schedule our tests on the same days as your physics tests?

Limits of Functions of Several Variables

Recall from Calculus I the various ways in which you computed limits. If possible, you substituted a value into the function. If not, perhaps you simplified the function via some algebra or computed a limit table or graphed the function. Or you might have applied L`Hôpital's Rule. Your instructor probably used the Squeeze Theorem to obtain the result that \(\dsp \lim_{t \to 0} \frac{\sin(t)}{t} = 1.\) Recall that if the left hand limit equaled the right hand limit then the limit existed.

In three dimensions, the difficulty is that there are more paths to consider than merely left and right. For the limit to exist at a point \((a,b)\text{,}\) we need that the limit as \((x,y)\) approaches \((a,b)\) exists regardless of the path we take as we approach \((a,b)\text{.}\) We could approach \((a,b)\) along the x-axis for example, setting \(y=0\) and taking the limit as \(x \to 0\text{.}\) Or we could take the limit along the line, \(y=x\text{.}\) The limit exists if the limit as \((x,y) \to (a,b)\) along every possible path exists. We will see an example where the limit toward \((a,b)\) exists along every straight line, but does not exist along certain non-linear paths!

Definition 10.9.

If \((a,b)\in {\re}^2\) and \(L\in \re\) and \(f: \re^2 \to \re\) is a function, then we say that

\begin{equation*} \dsp{\lim_{(x,y)\to (a,b)} f(x,y) = L} \end{equation*}

if \(f(x,y)\) approaches \(L\) as \((x,y)\) approaches \((a,b)\) along every possible path.

Problem 10.10.

Sketch \(f(x,y) = x^2 + y^2,\) indicate the point \(\big( 2,3,f(2,3) \big)\) and compute \(\dsp{\lim_{(x,y)\to (2,3)} x^2+y^2}.\)

Problem 10.11.

Use any free web-based software to graph the function from the previous example near \((0,0).\) Print and use a high-lighter to mark the paths \(x=0\text{,}\) \(y=0\text{,}\) and \(y=x\text{.}\) (Google Chrome on a Windows box will graph functions like \(z=x^2+y^2\) just by typing it into the search bar!)

Problem 10.12.

Convert the previous problem to polar coordinates via the substitution \(x=r\,\cos(\theta)\) and \(y=r\,\sin(\theta)\) and then compute the limit as \(r \to 0.\)

Problem 10.13.

Let \(\dsp f(x,y) = \frac{x+y^2+2}{x-y+2}.\)

  1. Graph \(f\) using any software and state the domain.

  2. Compute \(\dsp{\lim_{(x,y) \to (-1,2)} f(x,y)}.\)

Recall your Calculus I definition of continuity for functions of one variable.

Definition 10.14.

A function \(f: \re \to \re\) is continuous at \(a \in \re\) if \(\lim_{x \to a} f(x) = f(a).\)

This definition says that for \(f\) to be continuous at \(a\) three things must happen. First, the function must be defined at \(a\text{.}\) This means that \(a\) must be in the domain of the function so that \(f(a)\) is a number. Second, the limit of the function as we approach \(a\) must exist. And third, \(f(a)\) must equal the limit of \(f\) at \(a\text{.}\) The same statement defines continuity for all functions.

Definition 10.15.

A function \(f: \re^n \to \re^m\) is continuous at \(a \in \re^n\) if \(\lim_{x \to a} f(x) = f(a).\)

Problem 10.16.

Consider \(\dsp{\lim_{(x,y)\to (0,0)} \frac{x^4-y^4}{x^2+y^2}}.\)

  1. Compute this limit along the lines: \(x=0\text{,}\) \(y=0\text{,}\) \(y=x\text{,}\) and \(y=-x\text{.}\)

  2. Convert to polar coordinates and check the limit.

  3. Graph using any software.

  4. Why isn't this function continuous at \((0,0)\text{?}\)

  5. How can you modify \(f\) in such a way as to make it continuous at \((0,0)\text{?}\)

Problem 10.17.

Determine whether \(\dsp{\lim_{(x,y)\to (0,0)} \frac{xy+y^3}{x^2+y^2}}\) exists and if so, state the limit.

Problem 10.18.

Determine whether the function \(f\) is continuous at \((x,y)=(0,0)\) by considering the paths \(y=kx^2\) for several choices of k. \(f(x,y)=\left\{ \begin{array}{ll} \dsp{\frac{x^2y}{x^4+y^2}} \amp (x,y)\neq (0,0)\\ 0 \amp (x,y)=(0,0) \end{array} \right.\)

Directional Derivatives

Suppose we have the function \(f(x,y)=x^2+y^2\) and we are sitting on that function at some point \((a,b,f(a,b))\) other than \((0,0,0)\text{.}\) Then there are many directions we can walk while remaining on the surface. Depending on the direction of our path, the rate of increase of our height, or slope of our path, may vary. Some paths will move us uphill and others downhill. Suppose while we sit at the point, \((a,b,f(a,b)),\) we decide to walk in a direction that will not change the \(y\) coordinate, but only changes the \(x\) coordinate. Thus, we are walking on the surface and staying within the plane, \(y=b.\) Walking in this way, we could go in one of two directions. Either we could go in the direction that increases \(x\) or decreases \(x.\) Let's go in the direction that increases \(x.\) Now, consider the tangent line to the curve at this point that lies in the plane, \(y=b.\) As we take our first step along the curve our rate of increase in height, \(z,\) will be the same as the slope of that tangent line. This slope is the directional derivative of the function at the point \((a,b)\) in the \(x\) direction. If we had decided to fix \(x=a\) and walk in the direction that increases \(y\text{,}\) then the slope of the line tangent to the function and in the plane \(x=a\) is the directional derivative of \(f\) at \((a,b)\) in the \(y\) direction.

The next definition formalizes this discussion and the problem immediately following it is an example that will make these notions of directional derivative precise!

Definition 10.19.

If f: \({\re}^2\to \re\) is a function and \((a,b)\) is in the domain of \(f\text{,}\) then the derivative of \(f\) in the \((1,0)\) direction at \((a,b)\) is the slope of the line tangent to \(f\) at the point \((a,b,f(a,b))\) and in the plane, \(y=b.\)

When it exists, the derivative of \(f\) with respect to \(x\) at \((a,b)\) can be computed via the limit:

\begin{equation*} f_{x}(a,b)=\dsp{\lim_{h\to 0} \frac{f(a+h,b)-f(a,b)}{h}} \end{equation*}

or, since \(y\) is being held constant and \(x\) is changing, one may just compute the derivative of \(f\) as if \(x\) is the variable and \(y\) is a constant.

Notation. Suppose that \(f: {\re}^2\to \re\) is a function as in the previous definition. There are many phrases and notations used to denote the derivative of \(f\) in the \((1,0)\) direction at \((a,b)\). For example,

  • \(f_{1}\) — the derivative of f with respect to the first variable

  • \(f_{x}\) — the the derivative of f with respect to x

  • \(\dsp{\frac{\partial f}{\partial x}}\) — the partial derivative of f with respect to x

Similarly, \(f_{2}\text{,}\) \(f_{y}\text{,}\) and \(\dsp{\frac{\partial f}{\partial y}}\) would denote the same concepts where the derivative was taken in the \((0,1)\) direction.

Problem 10.20.

Let \(f(x,y) = x^2 + y^2\) and \((a,b)=(2,-3).\)

  1. Sketch \(f\) and sketch the line tangent to \(f\) at the point \((2,-3,f(2,-3))\) that is in the plane, \(y=-3.\)

  2. Let \(g(x) = f(x,-3)\) and compute \(g'(2).\)

  3. What part of the graph of \(f\) is the graph of \(g?\)

Problem 10.21.

Let \(\dsp f(x,y) = \frac{y}{x}.\) Compute \(f_x(1,2)\) using the limit described following Definition 10.19.

Problem 10.22.

Compute \(f_{x}\) and \(f_{y}\) for each function.

  1. \(f(x,y)=x^3-4x^2\)

  2. \(f(x,y)=e^{xy^2}\)

  3. \(f(x,y)=\dsp{\frac{x^2}{\sin(xy)}}\)

  4. \(f(x,y)=e^{x^2y}\sin(x-y)\)

Definition 10.23.

Just as in Calculus I, “second derivatives” are merely derivatives of the first derivatives. Thus \(f_{xx}=(f_{x})_{x}\text{.}\) I.e. \(f_{xx}\) is the derivative of \(f_{x}\) with respect to x. Similarly, \(f_{yy}=(f_{y})_{y}\text{,}\) \(f_{xy}=(f_{x})_{y}\) and \(f_{yx}=(f_{y})_{x}\text{.}\)

Other standard notations are:

\begin{equation*} f_x = \frac{\partial f}{\partial x}, \ \ \ f_y = \frac{\partial f}{\partial y}, \ \ \ f_{xx} = \frac{\partial^2 f}{\partial x^2}, \ \ \ f_{yy} = \frac{\partial^2 f}{\partial y^2}, \ \ \ \mbox{ and } \ \ \ f_{xy} = \frac{\partial^2 f}{\partial y \partial x} \end{equation*}
Problem 10.25.

Find \(f_{xx}, f_{xy}, f_{yx}\text{,}\) and \(f_{yy}\) for each function.

  1. \(f(x,y)=e^{x^2+y^2}\)

  2. \(f(x,y)=\sin(xy+y^3)\)

  3. \(f(x,y)=\sqrt{3x^2-2y^3}\)

  4. \(f(x,y)=\dsp{\cot\left( \frac{x}{y}\right) }\)

The study of partial differential equations is the process of finding functions that satisfy some equation that has derivatives with respect to multiple variables. For example Laplace's Equation is the equation \(u_{xx}+u_{yy} +u_{zz}=0\) and solutions give us information about the steady state of heat flow in a three dimensional object.

Problem 10.26.

Which of these functions satisfy \(u_{xx}(x,y)+u_{yy}(x,y)=0\) for all \((x,y) \in \re^2\text{?}\)

  1. \(u(x,y)=x^2+y^2\)

  2. \(u(x,y)=x^2-y^2\)

  3. \(u(x,y)=x^3+3xy^2\)

  4. \(u(x,y)=\ln(\sqrt{x^2+y^2})\)

  5. \(u(x,y)=\sin(x)\cosh(y)+\cos(x)\sinh(y)\)

  6. \(u(x,y)=e^{-x}\cos(y)-e^{-y}\cos(x)\)

  7. Find a solution to this equation other than the ones listed above.

In the last two problems, we studied functions of two variables and we defined derivatives in each direction, the \(x\) direction and the \(y\) direction. If \(f\) were a function of three variables, then there would be partial derivatives with respect to each of \(x, y\text{,}\) and \(z.\) Let's extend the notion of the partial derivatives with respect to \(x\) and \(y\) to functions with domain \(\re^n\) where \(n>2.\) If \(f : \re^n \to \re\text{,}\) then the domain of \(f\) is \(\re^n\) so there is a partial derivative of \(f\) with respect to the first variable, the second variable, and so on, up to the partial derivative of \(f\) with respect to the \(n^{th}\) variable. We use the notation, \(f_1, f_2, f_3, \dots, f_n\) or \(\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \frac{\partial f}{\partial x_3}, \dots , \frac{\partial f}{\partial x_n}\) to denote the derivative of \(f\) with respect to each variable.

Problem 10.27.

Let \(f(x_1,x_2,x_3,x_4,x_5) = x_3\sqrt{(x_1)^3+(x_2)^2} + x_4 e^{x_5 x_3}.\) Compute the five partial derivatives, \(f_1, f_2, \dots, f_5.\)

This allows us to define the derivative for functions of \(n\) variables.

Definition 10.28.

If \(f:{\re}^n \to \re\) is a function and each partial of \(f\) exists, then the gradient of f is the function

\begin{equation*} \nabla f:{\re}^n \to {\re}^n \end{equation*}

and is defined by

\begin{equation*} \nabla f= (f_{1}, f_{2}, f_{3}, \ldots , f_{n}) =(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2}, \frac{\partial f}{\partial x_3}, \dots , \frac{\partial f}{\partial x_n}). \end{equation*}
Problem 10.29.

For each function below, state the domain of the function, compute the gradient and state the domain of the gradient.

  1. \(f(x,y)=x^3y^2-x^2y^3\)

  2. \(g(x,y,z)=zx^3-3xyz+\ln(x^2yz^3)\)

For any function of two variables, \(f,\) let's assume once again that we are at a point on the function, \((a,b,f(a,b)).\) We know that the slope of the line tangent to \(f\) at \((a,b,f(a,b))\) and above the line \(y=b\) is \(f_x(a,b)\text{.}\) And we know that the slope of the line tangent to \(f\) at \((a,b,f(a,b))\) and above the line \(x=a\) is \(f_y(a,b)\text{.}\) Now consider a line in the \(xy\)-plane passing through \((a,b)\) in some direction \(\oa{(c,d)}\) that is not parallel to either the \(x\) axis or the \(y\) axis. What would the slope of the line tangent to \(f\) at \((a,b,f(a,b))\) and above this line be? From the point \((a,b)\) there are infinitely many directions that we might travel, not just the directions parallel to the \(x\) and \(y\) axes. We can define such a direction from \((a,b)\) by a vector, \(\oa{(c,d)}.\) The slopes of the lines tangent to \(f\) at \((a,b,f(a,b))\) in the direction \(\oa{(c,d)}\) are called the directional derivatives of \(f\) at \((a,b)\) in the direction \(\oa{(c,d)}.\) The partial derivatives of \(f\) with respect to \(x\) and \(y\) are your first examples of directional derivatives where the directions were \(\oa i = (1,0)\) and \(\oa j = (0,1).\) Here is the formal definition of the directional derivative.

Definition 10.30.

Let \(f: \re^n \to \re\) be a function. The directional derivative of \(f\) at \(\oa{u}\) in the direction \(\oa{v}\) is given by:

\begin{equation*} D_{\oa{v}}f(\oa{u})= \dsp{\lim_{h \to 0} \frac{f(\oa{u}+h\oa{v})-f(\oa{u})}{h}}, \end{equation*}

where \(\oa{v}\) must be a unit vector.

Of course, not every limit exists, so directional derivatives may exist in some directions but not others.

Problem 10.31.

Using the definition just stated, compute the directional derivative of \(f(x,y)=4x^2+y\) at the point \(\oa u = (1,2)\) in the direction \(\oa v\) for each \(\oa v\) defined below.

  1. \(\oa{v}=(0,1)\)

  2. \(\oa{v}=(1,0)\)

  3. \(\oa{v}=(1,1)\)

  4. \(\oa{v}=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)

Important. In the previous problem, you got different answers for the directional derivatives in parts 3 and 4 even though those represent the same direction. For this reason, it is a convention to always use a unit vector for the direction when computing directional derivatives.

Non-Definition. An analytical definition of f is differentiable at \(u\) is beyond the scope of this course, but a geometrical definition is not. In two dimensions (Calculus I), f was differentiable at \(a\) if there was a tangent line to f at \((a,f(a))\text{.}\) In three dimensions (Calculus III), f is differentiable at \(\oa{u}\) if there is a tangent plane to f at \((u,f(u)).\)

Definition 10.33.

An \(\epsilon\)-neighborhood of \(u \in \re^n\) is the set of all points with a distance from \(u\) of less than \(\epsilon\text{.}\) I.e. \(N_{\epsilon}(u)=\{v \in \re^n:\ \mid u-v \mid \lt \epsilon \}\text{.}\)

How will we be able to tell when a function is “nice,” that is, when a function has a derivative?

Problem 10.35.

Is \(\dsp f(x,y) = y^3(x-\frac{1}{2})^\frac{2}{3}\) differentiable at \((1,2)?\)

Problem 10.36.

For each of the following problems, either prove it or give a counterexample by finding functions and variables for which it does not hold. Assume \(f:{\re}^2\to \re\) and \(g:{\re}^2\to \re\) are differentiable. Assume that each of \(x = (x_1,x_2)\) and \(y = (y_1,y_2)\) are elements of \(\re^2\text{,}\) and \(c \in \re.\) Assume \(x, y, \text{and} x + y\) are in the domain of \(f\) and \(g.\)

  1. \(\nabla f(x+y)= \nabla f(x)+\nabla f(y)\)

  2. \(\nabla (f+g)(x)=\nabla f(x)+\nabla g(x)\)

  3. \(\nabla (cf)(x)=c\nabla f(x)\)

  4. \(\nabla f(cx)=c\nabla f(x)\)

  5. \(\nabla f(cx)=\nabla f(c)\ \nabla f(x)\)

  6. \(\nabla (f\cdot g)(x)= \nabla f(x)g(x)+f(x) \nabla g(x)\)

  7. \(\dsp \nabla (\frac{f}{g})(x) = \frac{ \nabla f(x)g(x)-f(x) \nabla g(x) }{ g^2 }\)

As in Calculus I, it is very nice to know when and where a function is continuous. The following theorem answers that question in both cases.

The next theorem is a very important one. In Calculus I, we first computed derivatives using the definition and then proved rules to help us differentiate more complex functions; we do the same here. We won't prove this theorem, but it gives a very easy way to compute directional derivatives by using the dot product and the gradient of the function.

To date we have studied the derivatives of functions from \(\re\) to \(\re,\) from \(\re\) to \(\re^2,\) and from \(\re^2\) to \(\re.\) Of course there is nothing special about \(\re^2\) here. We might as well have studied \(\re^n\) as all the derivatives would follow the same rules. Now let's consider the derivative of a function, \(f : \re^2 \to \re^2.\)

Definition 10.40.

If \(f : \re^2 \to \re^2\) is any vector valued function of two variables defined by \(f(x,y) = \big( u(x,y) \ , \ v(x,y) \big)\text{,}\) then the derivative of \(f\) is given by

\begin{equation*} Df = \begin{pmatrix}u_x(x,y) \amp u_y(x,y) \cr v_x(x,y) \amp v_y(x,y). \end{pmatrix} \end{equation*}
Problem 10.41.

Compute the derivative of \(\dsp f(x,y) = \big( x^2\sin(xy) \ , \ \frac{e^y}{\tan(x)} \big).\)

Because of the number of different domains and ranges of functions we are studying, we have several variations of the chain rule. Before we begin, I would like to take this opportunity to apologize for the number of notations used by mathematicians, physicists, and engineers for derivatives, partial derivatives, total derivatives, gradients, Laplacians, etc. There are a number of notations and all are convenient at one time or another. I attempt to adhere for the most part to the functional notation for derivatives (\(f_1, f_x, f',\) etc.), but Leibniz notation, \(\frac{\partial f} {\partial x}\text{,}\) is a convenient notation as well. Table 10.42 illustrates my preferred notations, where \(L(\re^2, \re^2)\) denotes the set of all 2x2 matrices.

Function Derivative
\(f: \re \to \re\) \(f' : \re \to \re\)
\(f: \re^2 \to \re\) \(\nabla f : \re^2 \to \re^2\)
\(f: \re^2 \to \re^2\) \(Df : \re^2 \to L(\re^2, \re^2)\)
Table 10.42. Derivative Notation

The chain rule you learned in Calculus I, which is stated next, can be generalized to take the derivative of the composition of any of the functions we have been studying. That is, given any two functions with domains so that their composition actually makes sense and so that they are differentiable at the appropriate places, we can compute their derivative using the same chain rule that you learned in Calculus I with one minor change. When the domains and ranges of the functions change, the derivatives change. Thus, in the following statement, depending on the domain of \(f\text{,}\) sometimes \(f'\) means the derivative of a parametric curve, but sometimes it means the gradient of \(f\text{,}\) \(\nabla f\text{,}\) and sometimes it means the matrix, \(Df\text{,}\) of derivatives of \(f\text{.}\) The same holds for the derivative of \(g\text{.}\) And finally, the symbol \(\cdot\) might mean multiplication or the dot product or matrix multiplication. You'll know from context which one. The point is to realize that no matter how many different notational ways we have of writing the chain rule, it always boils down to this one.

Next we state the chain rule for differentiating functions that are the composition of a function of two variables with a planar curve. Notice that this theorem is exactly the same as the original theorem, but restated for functions with different domains. Because \(f : \re^2 \to \re\) we replace the \(f'\) from the previous theorem with \(\nabla f\) and because \(\oa g : \re \to \re^2\) we replace the \(g'\) in the previous theorem with \(\oa g'.\) Thus the theorem still says (in English) that “the derivative of (\(f\) composed with \(g\)) is the (derivative of \(f\)) evaluated at \(g\) times (the derivative of \(g\)).”

On the right hand side of the last line of the Chain Rule, we have the composition of \(\nabla f\) with \(g(t)\text{.}\) Because we are multiplying vectors, “\(\cdot\)” represents dot product and not multiplication.

Problem 10.45.

Let \(f(x,y) = x^2 - 3y^2\) and \(\oa g(t) = (2,3) + (4,5)t.\) Compute \((f \circ \oa g)'\) both by direct composition and by using Theorem 10.44.

Problem 10.46.

Compute \((w \circ \oa g)'\) where \(\dsp w(x,y) = e^x\sin(y) - e^y\sin(x)\) and \(\oa g(t) = (3,2)t.\) Write a complete sentence that says what line \(\big(w \circ \oa g\big)'(-1)\) is the slope of.

Next we state the chain rule for differentiating functions that are the composition of a function of several variables with a function from the plane into the plane. Notice that this theorem is exactly the same as the original theorem, but restated for functions with different domains. Because \(f : \re^2 \to \re\) we replace the \(f'\) from the original theorem with \(\nabla f\) and because \(\oa g : \re^2 \to \re^2\) we replace the \(g'\) in the previous theorem with \(D\oa g.\) Thus the theorem still says that “the derivative of (\(f\) composed with \(g\)) is the (derivative of \(f\)) evaluated at \(g\) times (the derivative of \(g\)).” Because \(D \oa g\) is a matrix, the right hand side of this is now a vector times a matrix.

Problem 10.49.

Let \(f(x,y) = 2x^2 - y^2\) and \(\oa g(s,t) = (2s+5t,3st).\) Compute \(\nabla (f \circ \oa g)\) in two ways. First, compute by composing and then taking the derivative. Second, apply Theorem 10.48.

Problem 10.50.

Let \(w(x,y) = \ln(x+y) - \ln(x-y)\) and \(g(s,t) = (te^s, - e^{st}).\) Compute \(\nabla (w \circ \oa g)\) in two ways. First, compute by composing and then taking the derivative. Second, apply Theorem 10.48.

Problem 10.51.

Prove or give a counterexample to the statement that \(f'(\oa l(t)) = (f(\oa l(t)))'\) for all differentiable functions \(f: \re^2 \to \re\) and \(\oa l: \re \to \re^2.\)

The next problem tells us something important. If you are sitting on some function in three-space and you are trying to decide what direction you should travel to go uphill at the steepest possible rate, then the gradient tells us this direction.

Problem 10.52.

Use Problem 8.30 and Theorem 10.38 to show that \(D_{\oa{v}}f(\oa{u})\) is greatest when \(\oa{v}=\nabla f(\oa{u})\) and least when \(\oa{v}=-\nabla f(\oa{u})\text{.}\)

Surfaces. Earlier we gave a list of the types of functions we have studied so far. Of course, there is only one definition for a function, so we are really talking about functions with different domains and ranges as was illustrated by the need for different chain rules for functions with different domains. In linear algebra, we see functions with domain the set of matrices (the determinant function) and \(\dsp T(f) = \int_0^1 f(x) \ dx\) is a function from the set of all continuous functions into the real numbers.

In earlier courses, you studied not only functions, but relations such as \(x^2 + y^2 = 1.\) What was the difference? Well, a function has a unique \(y\) for each \(x\) while a relation may have several \(y\) values for a given \(x\) value. In three dimensions a function will have a unique \(z\) for a given coordinate pair, \((x,y).\) When one has multiple \(z\) values for a given \((x,y)\) value, we call it a surface. We have thus far studied mostly functions from \(\re^2\) to \(\re\) such as \(f(x,y)=x^2+y^2\) or \(f(x,y)=ye^{x^2}\text{,}\) but surfaces are equally important. Surfaces are to functions in three-space as relations were to functions in two-space.

Problem 10.53.

Sketch these functions in \({\re}^3\text{.}\)

  1. \(f(x,y)=\mid x \mid - \mid y \mid\)

  2. \(h(x,y)=\sqrt{xy}\)

  3. \(g(x,y)=\sin(x)\)

  4. \(i(x,y)=2x^2-y^2\)

Problem 10.54.

Sketch these surfaces in \({\re}^3\text{.}\)

  1. \(x^2+y^2+z^2=1\)

  2. \(y^2+z^2=4\)

  3. \(x^2-y+z^2=0\)

  4. \(\mid y \mid =1\)

Surfaces may be expressed as \(F(x,y,z)=k\text{,}\) where \(F: \re^3 \to \re\) is a function and \(k\) is a real number. The first example above represents the set of all points in \(\re^3\) that satisfy \(F(x,y,z)=1\) where \(F\) is the function defined by \(F(x,y,z)= x^2+y^2+z^2.\) The mathematical convention is to rewrite these as follows:

  1. \(F(x,y,z)=1\) where \(F(x,y,z)= x^2+y^2+z^2.\)

  2. \(F(x,y,z)=4\) where \(F(x,y,z)= y^2+z^2\text{.}\)

  3. \(F(x,y,z)=0\) where \(F(x,y,z)= x^2-y+z^2\text{.}\)

  4. \(F(x,y,z)=1\) where \(F(x,y,z)= \mid y \mid\text{.}\)

Tangent Planes to Functions

Example 10.55.

Demonstrate how to compute the equation of the tangent plane to \(f(x,y) = 2 + x^2 + 3y^2\) at \((1,2,15)\) by taking the product of \((1,0,f_x(1,2))\) with \((0,1,f_y(1,2))\) to find the normal. Compute a tangent line or two at this point.

Problem 10.56.

Find the equation of the plane tangent to the function \(f(x,y) = 25 - x^2 - y^2\) at the point \((3,1,15).\) Sketch the graph of the function and the plane.

Problem 10.57.

Find the equation of the plane tangent to the function \(f(x,y) = \sqrt{x^2 + y^2}\) at the point \((3,4,5).\) Sketch the graph of the function and the plane.

Tangent Planes to Surfaces

Example 10.58.

Let \(F(x,y,z)=k\) denote some surface and show that \(\grad F(p)\) is perpendicular to the surface by differntiating \(F(r(t)) = k\) where \(r\) is some paramatric curve on the surface. Reconsider our previous example. Write \(f(x,y) = 2 + x^2 + 3y^2\) as a surface and then compute the tangent plane using this new orthogonal to demonstrate that we find the same plane.

Definition 10.59.

If \(F(x,y,z)=k\) is a surface, then the tangent plane to F at \(u=(x,y,z)\) is the plane passing through \(u\) with normal vector, \(\oa{\nabla F(u)}\text{.}\)

Problem 10.60.

Find the equation of the tangent plane to the surface \(x^2-2y^2-3z^2+xyz=4\) at the point \((3,-2,-1)\text{.}\)

Problem 10.61.

Find the equations of two lines perpendicular to the surface in the previous problem at the point \((3,-2,-1)\) on the surface.

Problem 10.62.

Find the equations of two lines perpendicular to the surface \(z+1=ye^{y}\cos(z)\) at the point \(\oa{p}=(1,0,0).\)