- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
We still have some tasks from Day 17 to finish discussing in class.
Day 17 - Prep
Task 17.1
We'll focus this task on making sure we understand how differentials can help us approximate changes in a function.
A forest ranger needs to estimate the height of a tree. The ranger stands 50 feet from the base of tree and measures the angle of elevation to the top of the tree to be about 60$^\circ$.
- If this angle of 60$^\circ$ is correct, then what is the height of the tree? Explain in general why the height of the tree is $h(\theta) = 50 \tan \theta$.
- Compute the differential $dh$ in terms of $\theta$ and $d\theta$.
- The ranger's angle measurement is mostly likely off by some amount. If the error in the ranger's measurement could be as much as $d\theta = 5^\circ$ (so $\frac{5\pi}{180}$ radians), then use differentials to estimate how large the error in the height could be (so compute $dh$). If your answer here is quite large (much larger than the height of the tree), then look back at your work and see if using radians instead of degrees makes a difference. Why does it? Feel free to ask in class.
- Compute the height if the angle were exactly 65 instead of 60. What's the actual difference between these two heights?
The US mint creates coins that are roughly a cylindrical shape, with volume $V = \pi r^2h$. Unfortunately, not every coin is exactly the same size, and small errors in $r$ (given by $dr$) and small errors in $h$ (given by $dh$), affect the amount of material needed to mint these coins.
- Compute $dV$ to give an approximate for the change in volume given by the errors $dr$ and $dh$.
- The radius of a coin is much larger than the height. Will an error in the radius, or an error in the height, cause a larger change in volume? Explain using your differentials.
- A soda can company has a cylindrical shape that instead has a large $h$ with small $r$. Will an error in the radius, or an error in the height, cause a larger change in volume in this situation.
Task 17.2
Suppose that our rover is located at point $P=(x,y)$ on a hill whose elevation is given by $z=f(x,y)$. The rover will be moving in the direction parallel to $\vec u$.
- Explain why the slope of the hill at $P$ in the direction $\vec u = (dx,dy)$ is given by $$\frac{dz}{\sqrt{(dx)^2+(dy)^2}}.$$
- Prove that this slope can be written, using gradients, as $$\vec \nabla f(P) \cdot \frac{\vec u}{|\vec u|}.$$
- Use the above fact to compute the slope of a hill given by $f(x,y) = x^2+3xy$ at $P=(2,-1)$ in the direction $\vec u = (3,4)$. (We call this the directional derivative of $f$ at $P$ in the direction $\vec u$, written $D_{\vec u}f(P)$.
The directional derivative of $f$ in the direction of the vector $\vec u$ at a point $P$ is defined to be $$D_{\vec u} f(P)=\vec \nabla f \cdot \frac{\vec u}{|\vec u|}.$$ We can simplify the above to just $f(P)=\vec \nabla f \cdot \hat u$ if $\hat u$ is a unit vector. We dot the gradient of $f$ with a unit vector in the direction of $\vec u$.
- Show that the partial derivative of $f$ with respect to $x$ is precisely the directional derivative of $f$ in the $(1,0)$ direction.
- Show that the partial derivative of $f$ with respect to $y$ is precisely the directional derivative of $f$ in the $(0,1)$ direction.
Please watch this short 2 part video that discusses the gradient a bit more, and how you can connect the gradient to the slope in various directions.
Task 17.3
Suppose our rover is located at a point $P$ on a hill whose elevation is given by $z=f(x,y)$. Recall that the directional derivative of $\vec f$ at $P$ in the direction $\vec u$ is the dot product $D_{\vec u} f(P)=\vec \nabla f(P)\cdot \frac{\vec u}{|\vec u|}.$ Also recall that we can compute dot products using the law of cosines $\vec \nabla f(P)\cdot \vec u= |\vec \nabla f(P)| |\vec u|\cos\theta,$ where $\theta$ is the angle between $\vec \nabla f(P)$ and $\vec u$.
- Give a formula for the angle $\theta$ between the two vectors $\vec \nabla f$ and $\vec u$?
- Given a direction $\vec u$, the directional derivative will give the slope of $f$ at $P$ in the direction $\vec u$. We want to know which direction we should be pick to obtain the largest slope (directional derivative). Explain why the angle between $\vec u$ and $\vec \nabla f(P)$ must be 0, in order to obtain the largest slope.
- State a vector $\vec u$ that yields the largest directional derivative.
- When $\vec u$ is parallel to $\vec \nabla f(P)$, show that $D_{\vec u}f(P) = |\vec \nabla f(P)|$. In other words, explain why the length of the gradient is precisely the slope of $f$ in the direction of greatest increase (the slope in the steepest direction).
- Which direction points in the direction of greatest decrease? What is the slope in that direction?
- In your own words, summarize what facts this task helped you learn about the gradient.
Task 17.4
The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.
- Return to any of the previous day's OpenStax problems to locate extra practice.
Day 18 - Prep
Task 18.1
In first semester calculus, differential notation is $dy=f' dx$. At $x=c$, the tangent line passes through the point $P=(c,f(c))$. If $Q=(x,y)$ is any other point on the line, then the vector $\vec {PQ} = (x-c,y-f(c))$ tells us that when $dx=x-c$ we have $dy=y-f(c)$. Substitution give us an equation for the tangent line tangent line as $$\underbrace{y-f(c)}_{dy}={f'(c)}\underbrace{(x-c)}_{dx}.$$ This equation tells us that a change in the output ($y-f(c)$) equals the derivative times a change in the input ($x-c$). In this task, we'll repeat this process to obtain an equation of a tangent plane to a function $f(x,y)$, where differential notation gives $$dz = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$
Consider the function $z=f(x,y)=9-x^2-y^2$. We'll be finding an equation of the tangent plane to $f$ at $(x,y)=(2,1)$. Here is surface plot along with the tangent plane at $(2,1,f(2,1))$, together with a contour plot.

- Compute the partial derivatives $f_x$ and $f_y$, and the differential $dz$. At the point $(x,y) = (2,1)$, evaluate the partial derivatives and the function $z=f(x,y)$.
- One point on the tangent plane to the surface at $(2,1)$ is the point $P=(2,1,f(2,1))$. Let $Q=(x,y,z)$ be another point on this plane. Use the vector $\vec{PQ}$ to obtain $dz$ when $dx = x-2$ and $dy = y-1$.
- We'd like an equation of the tangent plane to $f(x,y)$ when $x=2$ and $y=1$. Differential notation tells us that $$\underbrace{z-?}_{dz}=(-4)\underbrace{(x-?)}_{dx}+(?)\underbrace{(y-?)}_{dy}.$$ Fill in the blanks to get an equation of the tangent plane.
- Rewrite the equation you got in the form $A(x-a)+B(y-b)+C(z-c)=0$ and state a normal vector to the plane.
- The level curve of $f$ that passes through $(2,1)$ has no change in height, so $dz=0$. Use this fact to give an equation of the tangent line to this level curve at $(2,1)$.
Now let $z=f(x,y)=x^2+4xy+y^2$. At the point $P=(x,y)=(3,-1)$, we'll give an equation of the tangent plane to the surface and an equation of the tangent line to the level curve of $f$ that passes through this point.
- Give an equation of the tangent plane at $P=(x,y)=(3,-1)$. [Hint: Find $f_x$, $f_y$, $dx$, $dy$, and then $dz$, all at $(x,y)=(3,-1)$. Then substitute, as done above.]
- The level curve of $f$ that passes through $P$ is a curve in the plane. Give an equation of the tangent line to this curve at $P$. [Hint: Since we're on a level curve, what does $dz$ equal? The equation is almost identical to the previous part, with one minor change based on what $dz$ equals.]
The tangent plane and tangent line you just found are shown below.

Task 18.2
A rover moves on a hill where elevation is given by $z=f(x,y)=9-x^2-y^2$. The rover's path is parametrized by $\vec r(t)=(2\cos t, 3\sin t)$.
- At time $t=0$, what is the rover's position $\vec r(0)$, and what is the elevation $f(\vec r(0))$ at that position? Then find the positions and elevations at $t=\pi/2$, $t=\pi$, and $t=3\pi/2$ as well.
- In the plane, draw the rover's path for $t\in [0,2\pi]$. Then, on the same 2D graph, include a contour plot of the elevation function $f$. Include the level curves that pass through the points in part 1. Along each level curve drawn, state the elevation of the curve. [If you end up with an ellipse and several concentric circles, then you've done this right.]
- As the rover follows its elliptical path, the elevation is rising and falling. At which $t$ does the elevation reach a maximum? A minimum? Explain, using your graph.
- As the rover moves past the point at $t=\pi/4$, is the elevation increasing or decreasing? In other words, is $\dfrac{df}{dt}$ positive or negative? Use your graph to explain.
Notice above that we wanted $\frac{df}{dt}$, the rate of change of elevation with respect to time, even though the function $f(x,y)$ does not explicitly have $t$ as an input. The proper notation would be $\frac{d(f\circ r)}{dt}$, but this is so cumbersome that it's generally avoided. The notation $\frac{df}{dt}$ requires the reader to infer from context that $x$ and $y$ depend on $t$.
- At the point $\vec r(t)$, we'd like a formula for the elevation $f(\vec r(t))$. What is the elevation of the rover at any time $t$? [In $f(x,y)$, replace $x$ and $y$ with what they are in terms of $t$.]
- Compute $df/dt$ (the derivative as you did in first-semester calculus).
Let's repeat the above, but first compute differentials before substitution. For reference, we let $f(x,y)=9-x^2-y^2$ and $(x,y)=\vec r(t)=(2\cos t, 3\sin t)$.
- Compute the differential $df$ in terms of $x$, $y$, $dx$, and $dy$.
- Compute $dx$ and $dy$ in terms of $t$ and $dt$.
- Use substitution to write $df$ in terms of $t$ and $dt$. Then divide by $dt$ to obtain $\frac{df}{dt}$. Did you get the same answer as the previous part?
- Use your work above to state both $\vec\nabla f(x,y)$ and $\frac{d\vec r}{dt}$. Show that $\frac{df}{dt} = \vec\nabla f(x,y)\cdot \frac{d\vec r}{dt}$.
Task 18.3
A second-order partial derivative of $f$ is a partial derivative of one of the partial derivatives of $f$. The second-order partial of $f$ with respect to $x$ and then $y$ is the quantity $\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]$, so we first compute the partial of $f$ with respect to $x$, and then compute the partial of the result with respect to $y$. Alternate notations exist, for example the same second-order partial above we can write as $$\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]=\left(f_{x}\right)_y=f_{xy}=\ds\frac{\partial}{\partial y}\frac{\partial}{\partial x}f = \frac{\partial}{\partial y}\frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial y \partial x}.$$ The subscript notation $f_{xy}$ is easiest to write. Sometimes we'll use subscript notation to mean something other than a partial derivative (like the $x$ or $y$ component of a vector), at which point we use the fractional partial derivative notation to avoid confusion.
Consider the functions $f(x,y,z) = xy^2z^3$ and $g(x,y)=x\cos(xy)$.
- First compute $\vec \nabla f$. Then compute $f_{xy}$ and $\frac{\partial^2 f}{\partial z\partial y}$. Explain how you got these. End by computing the entire second derivative $D\vec\nabla f(x,y,z)$ (it is a 3 by 3 matrix with all 9 second partials placed inside).
- Compute $g_x$ and then $g_{xy}$. Then compute $g_y$ followed by $g_{yx}$.
- Now let $f(x,y)=3xy^3+e^{x}.$ Compute the four second partials $$\ds \frac{\partial^2 f}{ \partial x^2},\quad \ds\frac{\partial^2 f}{\partial y \partial x},\quad \ds\frac{\partial^2 f}{\partial y^2}, \quad \text{ and }\ds\frac{\partial^2 f}{\partial x \partial y}.$$
- For $f(x,y)=x^2\sin(y)+y^3$, compute both $f_{xy}$ and $f_{yx}$.
- Make a conjecture about a relationship between $f_{xy}$ and $f_{yx}$. Then use your conjecture to quickly compute $f_{xy}$ if $$f(x,y)=3xy^2+\tan^{2}(\cos(x)) (x^{49}+x)^{1000}.$$
Task 18.4
The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.
- Return to any of the previous day's OpenStax problems to locate extra practice.
Day 18 - In class
Grad School Comment
NOTE: Graduate school should be free! (Via Tuition waivers and Teaching/Research assistantships, for most people in STEM this is true.)
Brain Gains (Rapid Recall, Jivin' Generation)

- Consider the function $f(x,y) = x^2y+7y$. We'll focus on what happens at the point $(1,2)$. Note that $f(1,2) = 16$.
- Compute the differential $df(1,2)$ and gradient $\vec \nabla f(1,2)$.
- Let $(x,y,z)$ be a point on the tangent plane to $f$ at $(1,2)$. The differentials $dx = x-1$, $dy = y-2$, and $dz = z-16$ represent changes in $x$, $y$, and $z$ on the tangent plane. Give an equation of the tangent plane to $f$ at $(1,2,16)$.
- Give an equation of the tangent line to the level curve of $f$ at $(1,2)$ (what does $dz$ equal on a level curve).
- Compute the directional derivative of $f$ at $(1,2)$ in the direction $(3,4)$.
- Let $f(x,y) = x^2\sin(y)+y^3$.
- Compute $f_x$ and then compute $\dfrac{\partial}{\partial x}(f_x)$ and $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)$
- Compute $f_y$ and then compute $\dfrac{\partial}{\partial x}(f_y)$ and $\dfrac{\partial^2f}{\partial y^2}$
- Find a number $c$ so that the vectors $(1+c,2)$ and $(4,6)$ lie on the same line (are parallel or antiparallel).
Group Problems
- Let $f(x,y) = 9-x^2-y^2$ with $\vec r(t) = (2\cos t,3\sin t)$.
- State $f(\vec r(t))$ and then compute $\frac{df}{dt}$.
- Compute the differential $df$ in terms of $x,y,dx,dy$, and then compute the differentials $dx$ and $dy$ in terms of $t$ and $dt$.
- Use substitution from your previous computations to obtain $df$ in terms of $t$ and $dt$. Then state $df/dt$.
- Give an equation of the tangent plane to $f(x,y) = 9-x^2-y^2$ at the point $(2,-3)$. [Find the differential, and then substitute $dx = x-2$, $dy = y-?$, $dz = ?$.]
- Give an equation of the tangent line to the level curve of $f(x,y) = 9-x^2-y^2$ at the point $(2,-3)$. [Your answer will be very similar to the one above. What changes?]
- Let $g(x,y)=x\cos(xy)$.
- Compute $g_x$ and $g_y$, and then state the first derivative $Dg(x,y)$.
- Compute the second partials $\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)$, $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)$, $\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)$, and $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right)$.
- State the second derivative $D^2g(x,y)$ (it should be a 2 by 2 matrix).
- Find the directional derivative of $f(x,y)=xy^2$ at $P=(4,-1)$ in the direction $(-3,4)$. [Check: $D_{(-3,4)}f(4,-1) = \vec\nabla f(4,-1)\cdot \frac{(-3,4)}{5}=(-8,16)\cdot \frac{(-3,4)}{5}=88/5$.]
- Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$. [Check: $(-2)(x-(-3))+(-3)(y-(-2))+2(1)(z-1)=0$. ]
- Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$. [Check: $z-4 = (-1)^2(x-4)+2(4)(-1)(y-(-1))$.]
- Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the surface at the point $P=(1,-3,2)$.
- If $dx=0.1$, $dy=0.2$ and $dz=0.3$, then what is $df$ at $P$.
- Find the directional derivative of $f$ at $P$ in the direction $(1,-2,2)$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $P$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
Day 19 - Prep
Task 19.1
In the first calculus books, there was no mention of the chain rule. This is because differentials were extremely common notation, and the chain rule, when working with differentials, is simply substitution. In this task, we'll develop some rules for how to compute derivatives when functions depend on other functions (so composite functions).
- Suppose that $f(x,y,z) = ax+by+cz$, and $x=mt$, $y=nt$, and $z = pt$, for some constants $a,b,c,m,n,p$. Compute the differentials $df$, $dx$, $dy$, and $dz$. Then use substitution to obtain the differential of $f$ in terms of $t$ and $dt$. Finish by stating $\frac{df}{dt}$.
- Suppose now that $g$ is a function of $x$ and $y$, but $x$ and $y$ are functions of $u$, $v$, and $w$. This means, by definition of the differential and partial derivatives, that $dg = g_xdx+g_ydy$, along with $dx = x_udu+x_vdv+x_wdw$ and $dy = y_udu+y_vdv+y_wdw$. Substitution gives $$\begin{align*} dg &= g_xdx+g_ydy\\ &= g_x(x_udu+x_vdv+x_wdw)+g_y(y_udu+y_vdv+y_wdw)\\ &= (?)du+(?)dv+ (?)dw. \end{align*}$$ Fill in the question marks above, and then use your answer to state the three partials $\dfrac{\partial g}{\partial u}$, $g_v$, and $D_w g$.
- Consider the function $h(x,y,z)$, where $x$, $y$, and $z$ are functions of $r$ and $\theta$. State the differentials of $h$, $x$, $y$, and $z$, and then use substitution to prove that $$\dfrac{\partial h}{\partial r} = \dfrac{\partial h}{\partial x}\dfrac{\partial x}{\partial r} +\dfrac{\partial h}{\partial y}\dfrac{\partial y}{\partial r} +\dfrac{\partial h}{\partial z}\dfrac{\partial z}{\partial r}.$$ Obtain a similar formula for $\dfrac{\partial h}{\partial \theta}$.
Feel free to ask me in class how this relates to matrix multiplication.
Task 19.2
Suppose a rover travels around the circle $g(x,y)=x^2+y^2=1$. The elevation of the surrounding terrain is $f(x,y) = x^2+y+4$. The plot below shows the rover's path (the constraint $g(x,y)=1$), placed on the same grid as a contour plot of the elevation (the function $f(x,y)$ we wish to optimize).

Each level curve above represents a difference in elevation of 0.25 m. Our goal is to find the maximum and minimum elevation reached by the rover as it travels around the circle. We will optimize $f(x,y)$ subject to the constraint $g(x,y)=1$.
- Label each level curve with its elevation. Print this page, or copy the curves down on your paper.
- At which $(x,y)$ point does the rover encounter the minimum elevation? What is the minimum elevation? Explain, using the plot.
- Suppose the rover is currently at the point $(0,1)$ on its circular path. As the rover moves left, will the elevation rise or fall? What if the rover moves right? Is $(0,1)$ the location of a local maximum or local minimum?
- On your graph, place a dot(s) where the rover reaches a maximum elevation. What is the maximum elevation? Explain.
- Rather than visually inspecting level curves, let's examine the gradients $\vec \nabla f$ and $\vec \nabla g$ to see how these gradients compare at maximums and minimums.
- On the graph above, draw $\vec \nabla f$ at lots of places on your contour plot.
- At lots of points on the circle, with a different color, draw $\vec \nabla g$.
- Make sure you draw both gradients at all the points corresponding to local maxes and mins.
- At the local maximums and minimums, Lagrange noticed that $\vec \nabla f = \lambda \vec \nabla g$.
- How would you interpret the equation $\vec \nabla f = \lambda \vec \nabla g$?
- Compute $\vec \nabla f$ and $\vec \nabla g$.
- Explain why the system of equations $\vec \nabla f = \lambda \vec \nabla g$ and $g(x,y)=c$ is equivalent to the system of equations $$2x = \lambda 2x,\quad 1=\lambda 2y,\quad x^2+y^2=1.$$
- Solve the system of equations above to obtain 4 ordered pairs $(x,y)$. You can use the Mathematica notebook LagrangeMultipliers.nb to check your work.
- At each ordered pair, find the elevation. What is the maximum elevation obtained, and where does it occur? What is the minimum elevation obtained, and where does it occur?
Suppose $f$ and $g$ are continuously differentiable functions. Suppose that we want to find the maximum and minimum values of $f(x,y)$ subject to the constraint $g(x,y)=c$ (where $c$ is some constant). If a local maximum or minimum occurs, it must occur at a spot where the gradient of $f$ and the gradient of $g$ point in the same, or opposite, directions. This means the gradient of $g$ must be a multiple of the gradient of $f$. To find the $(x,y)$ locations of the maximum and minimum values (if they exist), we solve the system of equations that result from $$\vec \nabla f = \lambda \vec \nabla g,\quad \text{and}\quad g(x,y)=c$$ where $\lambda$ is the proportionality constant. The locations of maximum and minimum values of $f$ will be among the solutions of this system of equations.
Task 19.3
This task will mostly involve reading through some definitions and an example, with a short example at the end.
Let $A$ be a square matrix, as $A=\begin{bmatrix} \begin{pmatrix}a\\b\end{pmatrix}& \begin{pmatrix}c\\d\end{pmatrix}\end{bmatrix} = \begin{bmatrix}a&c\\b&d\end{bmatrix}$. The eigenvalues $\lambda$ and eigenvectors $\vec x$ of $A$ are solutions $\lambda$ and $\vec x\neq \vec 0$ to the equation $A\vec x=\lambda \vec x$, effectively replacing the matrix product (linear combination) with scalar multiplication.
The identity matrix $I$ is a square matrix with 1's on the diagonal and zeros everywhere else, so in 2D we have $I = \begin{pmatrix} 1&0\\0&1 \end{pmatrix}$. To find the eigenvalues, we rewrite $A\vec x = \lambda\vec x$ in the form $A\vec x -\lambda\vec x=\vec 0$ or $A\vec x -\lambda I \vec x=\vec 0$, which becomes $(A-\lambda I) =\vec 0.$ We need to find the values $\lambda$ so that $\left(\begin{bmatrix} a&c\\b&d\end{bmatrix}-\lambda \begin{bmatrix} 1&0\\0&1 \end{bmatrix} \right)\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}0\\0\end{pmatrix} \quad\text{or}\quad \begin{bmatrix} a-\lambda &c\\b&d-\lambda \end{bmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}0\\0\end{pmatrix}.$ A linear algebra course will show that $\lambda$ satisfies $$(a-\lambda)(d-\lambda)-bc=0.$$
Let $f(x,y)$ be a function so that all the second partial derivatives exist and are continuous. The second derivative of $f$, written $D^2f$ and sometimes called the Hessian of $f$, is a square matrix. Suppose $P=(a,b)$ is a critical point of $f$, meaning $\vec\nabla f(a,b) = (0,0)$.
- Suppose all the eigenvalues of $D^2f(a,b)$ are positive. Then at all points $(x,y)$ sufficiently near $P$, the gradient $\vec \nabla f(x,y)$ points away from $P$. The function has a local minimum at $P$.
- Suppose all the eigenvalues of $D^2f(a,b)$ are negative. Then at all points $(x,y)$ sufficiently near $P$, the gradient $\vec \nabla f(x,y)$ points inwards towards $P$. The function has a local maximum at $P$.
- Suppose the eigenvalues of $D^2f(a,b)$ differ in sign. Then at some points $(x,y)$ near $P$, the gradient $\vec \nabla f(x,y)$ points inwards towards $P$. However, at other points $(x,y)$ near $P$, the gradient $\vec \nabla f(x,y)$ points away from $P$. The function has a saddle point at $P$.
- If the largest or smallest eigenvalue of $f$ equals 0, then the second derivative tests yields no information.
Let's look at an example. Consider $f(x,y)=x^2-2x+xy+y^2$. The gradient is $\vec \nabla f(x,y)=(2x-2+y,x+2y)$. The critical points of $f$ occur where the gradient is zero. We need to solve the system $2x-2+y=0$ and $x+2y=0$, which is equivalent to solving $2x+y=2$ and $x+2y=0$. Double the second equation, and then subtract it from the first to obtain $0x-3y=2$, or $y=-2/3$. The second equation says that $x=-2y$, or that $x=4/3$. So the only critical point is $(4/3,-2/3)$.
The second derivatives is $ D^2f = \begin{bmatrix}2&1 \\1&2\end{bmatrix}.$ The second derivative is constant, so $D^2 f(4/3,-2/3)$ is the same as $D^2f(x,y)$. (In general, the critical point may affect your matrix.) To find the eigenvalues we solve $$(2-\lambda)(2-\lambda)-(1)(1)=0.$$ Expanding the left hand side gives $4-4\lambda + \lambda^2 -1 = 0$. Simplifying and factoring gives us $\lambda^2-4\lambda +3 = (\lambda-3)(\lambda -1) = 0$. The eigenvalues are $\lambda = 1$ and $\lambda=3$. Since both numbers are positive, we know the gradient points outwards away from the critical point. The critical point $(4/3,-2/3)$ corresponds to a local minimum of the function. The local minimum is the output $f(4/3,-2/3) = (4/3)^2-2(4/3)+(4/3)(-2/3)+(-2/3)^2$.
Let's try this process on our own. Consider the function $f(x,y)=x^2+4xy+y^2$.
- Find the critical points of $f$ by finding when $Df(x,y)$ is the zero matrix.
- Find the eigenvalues of $D^2f$ at any critical points.
- Label each critical point as a local maximum, local minimum, or saddle point, and state the value of $f$ at the critical point.
Task 19.4
Pick some problems related to the topics we are discussing from the Text Book Practice page.
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