- Today we discussed cylindrical. The most important concepts here is the volume element in cylindrical coordinates: $dV = r \, d\theta \, d r \, dz$.
-
To understand the volume element in spherical coordinates one needs to understand the spherical coordinates first. The first step in understanding spherical coordinates is to understand the parametrization of the unit vectors in $\mathbb R^3$ using angles $\theta$ and $\phi$. I hope that the animation below will be helpfull:
-
The image below should help you to understand the element of volume in spherical coordinates formula $dV = \rho^2 (\sin \phi) \, d\theta\, d\phi \, d\rho$.
- Assigned problems for Section 16.5 are 1-20 (all, but in particular 9, 10, 11, 19, 20), 19-20 (and evaluate the integrals), 21-23, 24-28, 29-34, 36, 37, 38, 39, 42, 43, 48, 49, 52, 56
- Some relevant Chapter 16 Review exercises and problems are 17, 21, 23, 63, 64.
-
Today we discussed Problems 34 and 35 in Section 16.4.
- The most important information in these two problems is the fact that the polar equation for the circle of radus $1$ centered at $(1,0)$ is $r = 2 \cos \theta$ with $\pi/2 \leq \theta \leq \pi/2.$ One way to see this is to write the equation of this circle in Cartesian coordinates \[ (x-1)^2 + y^2 = 1, \] which simplisies to \[ x^2 - 2 x + y^2 = 0. \] The last equation implies that $x \geq 0$ and in polar coordinates the last equation reads: \[ r^2 - 2 r \cos \theta = 0. \] For $r \gt 0$ the last equation is equivalent to $r = 2 \cos\theta$ and since $x \geq 0$ we have $-\pi/2 \leq \theta \leq \pi/2.$
- The above picture shows two circles, the unit circle $r=1$ in dark green and the circle $r - 2 \cos \theta$ with $-\pi/2 \leq \theta \leq \pi/2$ in navy blue. The origin is the red point. The red lines are the lines $\theta = -\pi/3$ and $\theta = -\pi/3.$ The teal and cyan colored regions are described in polar coordinates as follows: The teal region consists of all points whose polar coordinates satisfy: \[ -\pi/3 \leq \theta \leq \pi/3 \quad \text{and} \quad 1 \leq r \leq 2 \cos \theta; \] The cyan region consists of all points whose polar coordinates satisfy: \[ \pi/3 \leq \theta \leq 5\pi/3 \quad \text{and} \quad 2 \cos \theta \leq r \leq 1. \] The description of the yellow region is a little more complicated: The yellow region consists of all points whose polar coordinates satisfy: \[ 0 \leq \theta \leq 2\pi \quad \text{and} \quad 0 \leq r \leq \min\bigl\{ 1, 2 \cos \theta \bigr\}. \] Without using the minimum function, the only way to describe the yellow region is as a union of two regions: The yellow region consists of all points whose polar coordinates satisfy: \[ -\pi/3 \leq \theta \leq \pi/3 \quad \text{and} \quad 0 \leq r \leq 1 \] or \[ \pi/3 \leq \theta \leq 5 \pi/3 \quad \text{and} \quad 0 \leq r \leq 2 \cos \theta . \]
- It is important to notice that the areas of the teal, yellow and cyan regions can be easily calculated by using the formula for the area of a circular segment: The area of a circula segment whose central angle is $\theta$ and the radius of the circle is $R$ is given as \[ \frac{1}{2} R^2 \theta - \frac{1}{2} R^2 \sin \theta. \] This formula is obtained as the difference of the area of the circular sector whose central angle is $\theta$ and the radius of the circle is $R$ \[ \frac{1}{2} R^2 \theta \] and the area of the isosceles triangle whose equal sides (legs) have length $R$ and the angle between them (apex angle) is $\theta$: \[ \frac{1}{2} R^2 \sin \theta. \] Thus, the yellow area is \[ \frac{2 \pi}{3} - \frac{\sqrt{3}}{2}. \] Consequently, the cyan area and the teal area are both equl to \[ \frac{\pi}{3} + \frac{\sqrt{3}}{2}. \] However, it is always nice to confirm this by calculationg the double integral \[ \iint_T dA = \int_{-\pi/3}^{\pi/3} \int_1^{2 \cos \theta} r \, dr \, d\theta = \cdots = \frac{\pi}{3} + \frac{\sqrt{3}}{2}. \] In the above double integral $T$ denotes the teal region.
-
The explorations in the previous item unable us to calculate the following six double integrals
$\displaystyle I_1 = \iint_{D_1} x\, dA$
$\displaystyle I_2 = \iint_{D_1} \sqrt{x^2 + y^2} \, dA$
$\displaystyle I_3 = \iint_{D_1} \frac{1}{2} (x^2 + y^2) \, dA$
$\displaystyle I_4 = \iint_{T} x \, dA$
$\displaystyle I_5 = \iint_{T} \sqrt{x^2 + y^2} \, dA$
$\displaystyle I_6 = \iint_{T} \frac{1}{2} (x^2 + y^2) \, dA$
Here $D_1$ denotes the disk of radius $1$ centered at $(1,0)$ and $T$ denotes the teal region from the previous item.
The functions in all double integrals have the maximum $2$ in the region of integration.
- For Section 16.4 do problems 1-8, 9, 12-15, 16-22, 25, 27, 28, 31, 33, 34, 35 (but do evaluate the area), 36.
- For the next Section 16.5 the problems will be 1-20 (all, but in particular 9, 10, 11, 19, 20), 19-20 (and evaluate the integrals), 21-23, 24-28, 29-34, 36, 37, 38, 39, 42, 43, 48, 49, 52, 56
- Some relevant Chapter 16 Review exercises and problems are 17, 21, 23, 63, 64.
- I just could not resist producing a fine Cartesian and a fine polar grid.
- For Section 16.3 do problems 1-4, 5, 7, 8, 11, 12, 14-18, 19-27, 35, 37, 39, 43, 45, 49, 50, 51, 52, 60-62, 59, 63, 64.
- Today we discussed the concept of the average value of a function. Below I present several problems involving average distance.
-
What is the average distance between two points in the interval $[0,1]?$ Let $x, y \in [0,1].$ Let $S$ be the unit square $\bigl\{ (x,y) | x, y \in [0,1]\bigr\}$. Notice that the area of $S$ is $1.$ The average distance is $(1/1) \iint_S |x-y| dA.$ The double integral involved is the volume below.
-
What is the average distance of a point in a unit disk to disk's center? Let $D$ be the unit disk centered at the origin. The average distance is $(1/\pi) \iint_D \sqrt{x^2+y^2}dA.$ The double integral involved is the volume below.
-
In the previous item we used so called Euclidean distance to calculate the distance from the origin. Another popular distance, which is more alike what we experience in everyday life, is so called taxicab distance. The taxicab distance of the point $(x,y)$ to the origin is $|x|+|y|.$ Thus, the average taxicab distance of a point in a unit disk $D$ to the center of the unit disk is
$$
\frac{1}{\pi} \iint_D (|x|+|y|)dA
$$
The preceding double integral is the volume below.
-
The calculation in the previous item becomes simpler if we replace the unit disk with the square with the corners: $(-1,-1)$, $(-1,1)$, $(1,1)$, $(-1,1).$ Call this square $S.$ Using the taxicab distance find the average distance of the point $(x,y)$ in the square $S$ to the origin. The average distance of a point in the square $S$ to the origin is
$$
\frac{1}{4} \iint_S (|x|+|y|)dA
$$
The double integral involved is the volume below.
- We started Section 16.2 today. Do problems 1-27 (do most), 32, 33, 34, 35, 41, 42, 43, 44, 53, 54, 56, 58, 59, 60.
-
Problem 42 in Section 15.3 reads as follows: Find the minimum distance from the point $(1, 2, 10)$ to the paraboloid given by the equation $z = x^2 + y^2.$ Give a geometric justification for your answer.
It turns out that the choice of the point $(1, 2, 10)$ is somewhat awkward since the resulting equations cannot be solved explicitly. One has to use approximate solutions. The point $(5, -15, 8)$ leads to much simpler equation; so the solution is clearer, in my opinion. Or, try the point $(5, -10, 8).$
It is important to understand the abstract setting for this problem. Instead of the specific function whose graph is a paraboloid we could have used any function $z = g(x,y)$ which is defined for all $x$ and $y$ and an arbitrary point $(a,b,c).$ Then, instead of minimizing the distance of this point to the graph of $z = g(x,y)$ we can minimize the square of the distance: \[ \operatorname{ds}(x,y) = (x-a)^2 + (y-a)^2 + \bigl(g(x,y) - c\bigr)^2. \] The minimum of this function will occur at its critical points which we obtain by solving the system of equations: \begin{align*} (x-a) + \bigl(g(x,y) - c \bigr) g_x(x,y) & = 0 \\ (y-b) + \bigl(g(x,y) - c \bigr) g_y(x,y) & = 0. \end{align*} Assume that we have found a critical point $(x_0,y_0)$, that is this point satisfies the above system. What is the meaning of the above system for the point $(x_0,y_0)?$ We rewrite the above system as two dot products. \begin{align*} \bigl\langle x_0 - a , y_0 -b, g(x_0,y_0) - c \bigr\rangle \cdot \bigl\langle 1 , 0, g_x(x_0,y_0) \bigr\rangle & = 0 \\ \bigl\langle x_0 - a , y_0 -b, g(x_0,y_0) - c \bigr\rangle \cdot \bigl\langle 0 , 1 , g_x(x_0,y_0) \bigr\rangle & = 0. \end{align*} The vector \[ \bigl\langle x_0 - a , y_0 -b, g(x_0,y_0) - c \bigr\rangle \] is the displacement vector with the tail at the point $(a,b,c)$ and the head at the point $\bigl(x_0,y_0,g(x_0,y_0)\bigr).$ The vectors \[ \bigl\langle 1 , 0, g_x(x_0,y_0) \bigr\rangle \quad \text{and} \quad \bigl\langle 0 , 1 , g_x(x_0,y_0) \bigr\rangle \] occurred in my post on November 1. The first of these vectors is tangent to the cross section $z = g(x,y_0)$ while the second one is tangent to the cross section $z = g(x_0,y).$ Therefore, the meaning of the above system is that the vector \[ \bigl\langle x_0 - a , y_0 -b, g(x_0,y_0) - c \bigr\rangle \] which connects the point $(a,b,c)$ to its closest point $\bigl(x_0,y_0,g(x_0,y_0)\bigr)$ on the graph $z=g(x,y)$ is orthogonal to the graph of $z=g(x,y).$
Notice that this problem is in Section 15.3 which deals with constrained optimization and that in our discussion above we did not use the concept of Lagrange multipliers. Therefore we should present a solution which involves Lagrange multipliers. Here we deal with two functions of three variables. We want to minimize the square of the distance: \[ F(x,y,z) = (x-a)^2 + (y-a)^2 + (z - c)^2 \] under the constraint \[ G(x,y,z) = z - g(x,y) = 0. \] The minimum of the function $F$ under the constraint $G(x,y,z) = 0$ will occur at the solution of the following system: \begin{align*} (\vec{\nabla} F)(x,y,z) & = \lambda (\vec{\nabla} G)(x,y,z) \\ G(x,y,z) & = 0, \end{align*} or, more detailed \begin{align*} x - a & = \lambda -g_x(x,y) \\ y - b & = \lambda -g_y(x,y) \\ z - c & = \lambda \\ z & = g(x,y). \end{align*} Substituting the last two equations in the first two, you can see that this system has the same solutions as the previous system which was obtained without using Lagrange multipliers.
Going back to the specific numbers given in Problem 42. \begin{align*} x - 1 & = \lambda -x \\ y - 2 & = \lambda -y \\ z - 10 & = \lambda/2 \\ z & = x^2 + y^2. \end{align*} From the first two equations we obtain that $y = 2 x$ and $\lambda = -1 + 1/x$. Substituting these two and the fourth equation into the third one and simplifying we obtain \[ 10 x^3 - 19 x - 1 = 0. \] The approximate solutions of this equation are \[ x_1 \approx -1.35129, \quad x_2 \approx -0.0527087, \quad x_3 \approx 1.404003. \] Since only the point \[ \bigl(x_3, 2 x_3, x_3^2 + 4 x_3^2 \bigr) \approx (1.404003, 2.808006, 9.85612) \] is in the same octant as the point $(1,2,10)$, this is the closest point point on the paraboloid. It is difficult to asses whether the line segments in the picture below are really orthogonal to the paraboloid. However, since this paraboloid is a surface of revolution of the parabola all four points \[ (1,2,10), \quad \bigl(x_1, 2 x_1, x_1^2 + 4 x_1^2 \bigr), \bigl(x_2, 2 x_1, x_2^2 + 4 x_2^2 \bigr) , \quad \bigl(x_3, 2 x_3, x_3^2 + 4 x_3^2 \bigr), \] are in the vertical plane $y = 2 x$. Below I also show the cross section of the paraboloid and this plane.
- For Section 16.1 do problems 1, 3, 6-11, 13, 15.
- For Section 15.3 do the problems 2, 3, 4, 6, 9, 10, 11, 14, 17, 21, 22, 25, 27, 30, 42, 45. Review problems for Chapter 15: 3, 6, 12, 14, 19, 23-26, 30, 35, 39.
-
In Review problem 6 find the minimum and the maximum value of the function $f(x,y)= \sin x + \sin y +\sin(x+y)$ on the square $-\pi \leq x, y \leq \pi.$ To help you, I provide the contour plot of the function $f.$ The contour plot below consists of 99 contour lines. It includes the contour line at level $0$ and the contours are the levels which are approximately the integer multiples of $0.0519615$
- Updated list of problems for Section 15.1: 2, 3, 8, 9, 13, 15, 19, 21, 23, 24, 28, 29, 30, 31, 39.
- For Section 15.2 do problems 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 19, 20, 21, 26, 27, 28, 29, 30.
- In Problem 1 on the exam I assigned vectors $\vec{a},$ $\vec{b},$ and $\vec{c}$ in very special position. I expected that while doing algebra with these vectors you will notice that they are mutually orthogonal. In this pdf picture I plotted $\vec{a}$ in red $\vec{b}$ in green and $\vec{c}$ in blue with the origin marked by a black dot.
- While grading the exam I came back to Problem 2. In the picture that I provided on the exam the units on the $x$-axis and the $y$-axis were not the same. When I created the contour plot with the aspect ratio of the graph 1, that is with the sam units on the axes, I realized that a part of the contour at level -11 is a circle. The algebraic explanation of that is the following identity \[ f(x,y) = x^3 + x y^2 -12 x - y^2 = (x - 1)*\left(\left(x + \frac{1}{2}\right)^2 + y^2 - \frac{45}{4} \right) - 11. \] Thus, the teal circle that you see in the contour plot of the function $f(x,y)$ is the circle centered at $(-1/2,0)$ with radius $3\sqrt{5}/2.$
-
The pattern that I observed in Problem 2 on the exam inspired me to look at some other products of a circle. The following function gave a really interesting contour plot:
\[
\left( x^2 - y \right)*\left(x^2 + y^2 - 1 \right) = x^4 +x^2 y^2 - y^3 - x^2 y - x^2 + y
\]
This function has six critical points. The critical points are colored navy blue point is the local maximum, two teal points are saddle points, one saddle point is in purple and two local minimums are in red. For each critical point, I colored the corresponding contour line in the same color.
- We will partially cover Section 15.1 on the exam on Wednesday. Just do the problems 2, 3, 8, 9, 13, 15, 19.
-
The exam of Wednesday will cover:
-
Section 13.4
The cross product: using the cross product to calculate areas of triangles and parallelograms and volumes of parallelepipeds, using the cross product to determine an equation of a plane in special circumstances. -
Chapter 14
Conceptual understanding of the concept of a partial derivative, a directional derivative and a gradient vector, ability to estimate and compute these objects and ability to use them to solve related problems. Conceptual and geometric understanding of the concepts of a linear and a quadratic approximation and ability to compute them. Understanding of the chain rule for functions with two and three variables. Understanding of the concept of differential and ability to use it to estimate change of a function of two and three variables. - few problems from Section 15.1.
-
Section 13.4
- For Section 14.7 do problems 5, 7, 8, 11, 15, 16, 20, 21, 22-31, 34, 38, 39, 41, 44, 51, 52, 53.
-
I want to emphasize an important concept that appeared in Section 14.5: the concept of a normal vector to a surface. In the picture below the surface is the graph of a function $z = f(x,y).$ I selected a point $\bigl( x_0,y_0,z_0 \bigr),$ where $z_0 = f(x_0,y_0),$ on this surface and placed the blue vector orthogonal to the surface at that point.
There are two ways to calculate the blue vector. The first way is explained in the textbook. It goes as follows: We view the graph of $z = f(x,y)$ as the contour surface at level $0$ of the function $F(x,y,z) = z -f(x,y).$ Then we calculate the gradient vector of the function $F$ at the point $\bigl( x_0,y_0,z_0 \bigr):$ \[ \bigl( \vec{\nabla} F \bigr)(x_0,y_0,z_0) = \bigl\langle -f_x(x_0,y_0), -f_y(x_0,y_0), 1 \bigr\rangle. \] However, there is another way to calculate the blue vector in the picture below. First we determine the red vector in the direction of the tangent line to the cross section (this is the maroon curve) of the graph of $z = f(x,y)$ and the plane $y = y_0$. Since this tangent line to the maroon curve at the point $\bigl( x_0,y_0,z_0 \bigr)$ has the slope $f_x(x_0,y_0)$, the red vector is given by \[ \bigl\langle 1, 0, f_x(x_0,y_0) \bigr\rangle. \] Similarly, we determine the green vector in the direction of the tangent line to the cross section (this is the dark green curve) of the graph of $z = f(x,y)$ and the plane $x = x_0$. Since this tangent line to the dark green curve at the point $\bigl( x_0,y_0,z_0 \bigr)$ has the slope $f_y(x_0,y_0)$, the green vector is given by \[ \bigl\langle 0, 1, f_y(x_0,y_0) \bigr\rangle. \] Next we observe that both, the red vector, and the green vector are in the tangent plane to the graph of $z=f(x,y)$ at the point $\bigl( x_0,y_0,z_0 \bigr)$. Therefore, the cross product of the red vector and the green vector is normal to this tangent plane. That is, it is normal to the graph of $z=f(x,y).$ Let us calculate this cross product: \[ \bigl\langle 1, 0, f_x(x_0,y_0) \bigr\rangle \times \bigl\langle 0, 1, f_y(x_0,y_0) \bigr\rangle = \left| \begin{array}{ccc} \vec{\imath} & \vec{\jmath} & \vec{k} \\ 1 & 0 & f_x(x_0,y_0) \\ 0 & 1 & f_y(x_0,y_0) \end{array} \right| = \bigl\langle -f_x(x_0,y_0), -f_y(x_0,y_0), 1 \bigr\rangle, \] which is the same formula that was obtained earlier.
-
In the previous item I explained how to calculate the normal vector to a surface. In two pictures below I use this formula to place $81$ normal vectors to this surface.
- For Section 14.6 do problems 18, 20, 24, 31, 32, 33.
- For Section 14.5 do problems 3, 4, 5, 13, 15, 16, 20, 21, 23, 24, 25, 26, 31, 41, 42, 44, 46, 47, 49, 51, 53, 54, 59, 60, 61, 67, 68.
- For Section 14.4 do problems 3, 9, 12, 18, 21, 23, 24, 25, 32-37, 38, 42, 43, 45, 47, 49, 51, 53, 56-60, 61, 65, 67, 68, 70, 72, 73, 74, 75, 76, 77, 79, 80, 81, 83, 86, 87.
-
The concept of the gradient is the most important concept of multivariable calculus. Here is my small contribution to help you understand it better.
In the animation below, a linear function \[ z = L(x,y) = m x + n y + p \] is represented by its graph, represented by a transparent plane. In this plane I emphasized navy blue contour lines at equidistant levels. The projections of these countour lines in $xy$-plane are teal lines. The value $z_0 = L(x_0,y_0)$ of the function $L$ at the point $(x_0,y_0)$ is represented by the navy point $P = (x_0,y_0,z_0)$ where $z_0 = L(x_0,y_0).$Place the cursor over the image to start the animation.
Notice that the animation pauses at four important directions.
The pause is longest in the direction of the gradient.
The above animation illustrates directional derivatives of the function $L$. The purple unit vector in $xy$-plane goes through all the possible directions in $xy$-plane. The point $Q$ represents the change in the value of $L$ in the direction of the purple unit vector. Since the purple vector is a unit vector, the directional derivative, that is the slope of the $PQ$ line, is the rise which is represented by the green line segment (if positive) or by red line segment (if negative).
- For Section 14.3 do problems 1, 2, 5, 6, 18, 20, 21, 23, 24, 27, 36.
-
It is interesting to find the tangent plane to a half-sphere of radius $5$ given by
\[
z = f(x,y) = \sqrt{25 - x^2 - y^2}.
\]
Let us find the tangent plane to this half-sphere at the point $(2,1,2\sqrt{5}).$ First we calculate s
\[
\frac{\partial}{\partial x} \sqrt{25 - x^2 - y^2} = -\frac{x}{\sqrt{25 - x^2 - y^2}}, \quad \frac{\partial}{\partial y} \sqrt{25 - x^2 - y^2} = -\frac{y}{\sqrt{25 - x^2 - y^2}}.
\]
At the point $(2,1)$ we have
\[
f(2,1) = 2\sqrt{5}, \quad f_x(2,1) = - \frac{1}{\sqrt{5}}, \quad f_y(2,1) = - \frac{1}{2\sqrt{5}} .
\]
Therefore the equation of the tangent plane to the half-shere which is the graph of the function $f$ is
\[
z = - \frac{1}{\sqrt{5}} (x-2) - \frac{1}{2 \sqrt{5}} (y-1) + 2 \sqrt{5} = \frac{1}{2 \sqrt{5}} \bigl(25 - 2x - y \bigr).
\]
- For Section 14.2 do problems 2, 3, 4, 8, 13, 25, 41, 43, 44, 46, .
- Below is an oscillating string. Its equation is $y = F(x,t) = (\sin x) (\cos t)$. In the figure below $y$ is the vertical axis, $x$ is the horizontal axis and the time $t$ is displayed above the frame. The string at time $t$ is black. To indicate the motion of the string, I added several previous positions of the string in various shade of gray. In fact, the previous positions fade away. At each point in time $t_0 \in [0, 2 \pi)$ and for each point $x_0 \in [0,\pi]$ you should be able to determine the sign of the following quantities: \[ F(x_0,t_0), \ F_x(x_0,t_0), \ F_t(x_0,t_0), \ F_{xx}(x_0,t_0), \ F_{xt}(x_0,t_0), \ F_{tx}(x_0,t_0), \ F_{tt}(x_0,t_0). \] The best way to achieve this is to understand the meaning of each of these quantities. In fact, you should be able to determine the signs of the quantities listed above just based on one scene in the movie below and just based on a point emphasized on the graph of the string (even without explicitly given the values for $x_0$ and $t_0$).
Clicking on the image will cycle through 8 different individual scenes of this movie with various values of $x_0 \in [0,\pi]$.
- For Section 14.1 do problems 5, 7, 10-13, 17, 18, 22, 23, 25, 26, 27, 30.
-
In the post of Thursday I complained about the picture illustrating Problem 53 in Section 13.4. Here is a better picture. In Probelm 53 we we given a light blue plane $z=mx+ny+p$. In that plane we are given a navy blue parallelogram spaned by a maroon vector and a dark green vector. This parallelogram is projected onto $xy$-plane. This plane is light cyan plane in the picture below. The projection is the light blue parallelogram in the light cyan plane. In the problem we are asked to express the dark blue area in terms of the light blue area and the quantities $m$, $n$ and $p.$
A better picture to accompany Problem 53.
- For Section 13.4 do problems 5, 6, 8, 9, 12, 13, 14, 15, 16, 24, 27, 28, 29, 30, 33, 42, 44, 45, 49, 50, 51, 52, 53. Please note that there are much simpler ways of solving 14 and 15 than using the cross product. Please note that the picture accompanying 49 is missing three arrows to make sense. Please note that the picture accompanying 53 does not make much sense. in this picture the plane $z = mx+ny+c$ is shown as parallel to $xy$-plane. In this case the areas $R$ and $S$ are clearly equal, so, this case is trivial. Also, the label for the vector $\vec{v}$ is misplaced in this picture.
-
Finally we are able to solve the following problem:
Problem. The temperature in $xyz$-space is given by the linear function in three variables: \begin{equation*} T(x,y,z) = 3 x+2y-z+5. \end{equation*} This means that the temperature at the point $(x,y,z)$ in this space is $3 x + 2 y - z+5.$ Answer the following questions:
-
A miniature drone is hovering at the origin, that is at the point $(0,0,0).$ Answer the following questions.
- What is the temperature experienced by the drone?
- At which rate will drone's temperature change if it proceeds in the direction of the vector $\vec{\imath}?$ (Here we assume that $\vec{\imath}$ is the velocity vector of the drone.)
- At which rate will drone's temperature change if it proceeds in the direction of the vector $\vec{\jmath}?$ (Here we assume that $\vec{\jmath}$ is the velocity vector of the drone.)
- At which rate will drone's temperature change if it proceeds in the direction of the vector $\vec{k}?$ (Here we assume that $\vec{k}$ is the velocity vector of the drone.)
- In which direction should this drone move if it wants to increase its temperature at the largest rate? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to maintain its temperature and stay in the $xy$-plane? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to increase its temperature at the largest rate and stay in the $xy$-plane? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to maintain its temperature and stay in the $yz$-plane? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to increase its temperature at the largest rate and stay in the $yz$-plane? (Give your answer as a three dimensional vector.)
-
A miniature drone is hovering at the point $(1,2,3).$ Answer the following questions.
- What is the temperature experienced by the drone?
- At which rate will drone's temperature change if it proceeds in the direction of the vector $\vec{\imath}?$ (Here we assume that $\vec{\imath}$ is the velocity vector of the drone.)
- At which rate will drone's temperature change if it proceeds in the direction of the vector $\vec{\jmath}?$ (Here we assume that $\vec{\jmath}$ is the velocity vector of the drone.)
- At which rate will drone's temperature change if it proceeds in the direction of the vector $\vec{k}?$ (Here we assume that $\vec{k}$ is the velocity vector of the drone.)
- In which direction should this drone move if it wants to increase its temperature at the largest rate? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to maintain its temperature and stay in the plane which is parallel to $xy$-plane? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to increase its temperature at the largest rate and stay in in the plane which is parallel to the $xy$-plane? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to maintain its temperature and stay in the plane which is parallel to the $yz$-plane? (Give your answer as a three dimensional vector.)
- In which direction should this drone move if it wants to increase its temperature at the largest rate and stay in the plane which is parallel to the $yz$-plane? (Give your answer as a three dimensional vector.)
-
A miniature drone is hovering at the origin, that is at the point $(0,0,0).$ Answer the following questions.
- Today we started Section 13.3. The essential ingredient for the understanding of the geometric meaning of the dot product is the law of cosines. Wikipedia's page Law of cosines gives several proofs. I wrote a page which contains just a proof of the law of cosines and its connection with the dot product.
- For Section 13.3 do problems 1-9, 10, 11, 13, 15, 19, 22, 23, 25, 27, 28, 30, 32, 34, 35, 36, 37, 38, 41, 48, 56, 57, 58, 59, 62, 65, 70.
- Today we discussed the material that will be on the exam, but is not in the book. I summarized this material in the last item in yesterday's post. However, notice that to understand this fully you need to work through the post on October 3 and the last two items in yesterday's post. But, certainly, working through the entire post of yesterday is a good idea.
- We also did few problems in Section 13.2. For this section do problems 8, 9, 10, 11, 12, 15, 16, 20, 25, 26.
- Today we discussed the importance of the unit vectors.
-
The picture below shows the unit circle in $xy$-plane. Each point on the unit circle is given by $\bigl(\cos \alpha, \sin \alpha\bigr)$ where alpha is a number in the interval $[0, 2\pi)$. This is what you learned in a trigonometry class. For us this is important since we can think of the point $\bigl(\cos \alpha, \sin \alpha\bigr)$ as a head of a unit vector $\overrightarrow{u}$ whose tail is at the origin. Then
\[
\overrightarrow{u} = (\cos \alpha) \overrightarrow{\imath} + (\sin \alpha) \overrightarrow{\jmath}.
\]
It is important to know the components of unit vectors in standard directions. Below it the unit circle with the most important directions indicated and labeled by the corresponding angles. The corresponding cosines and sines are given. The picture below can be viewed as a "cheat sheet" of components for some of the most important unit vectors.
Click here to get a pdf file.
Mouse-over the image for a simplified version.
-
I created the above image by modifying an image that I found at the Wikipedia page about the unit circle.
- You learned how to calculate the coordinates of the blue points in the above picture in a trigonometry class. (That is just an application of the Pythagorean theorem.)
- To calculate the coordinates of the red points in the above picture I used what we learned about vectors in this class. For example, denote by $\overrightarrow{u}$ the unit vector which forms the angle $\pi/8$ with $\overrightarrow{\imath}$, that is \[ \overrightarrow{u} = \Big(\cos \frac{\pi}{8}\Big) \overrightarrow{\imath} + \Big(\sin \frac{\pi}{8}\Big) \overrightarrow{\jmath}. \] Since both vectors \[ \overrightarrow{v} = \Big(\cos \frac{\pi}{4}\Big) \overrightarrow{\imath} + \Big(\sin \frac{\pi}{4}\Big) \overrightarrow{\jmath} = \frac{\sqrt{2}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}}{2} \overrightarrow{\jmath} \qquad \text{and} \qquad \overrightarrow{\imath} \] are unit vectors, their sum \[ \overrightarrow{v} + \overrightarrow{\imath} =\frac{\sqrt{2}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}}{2} \overrightarrow{\jmath} + \overrightarrow{\imath} = \frac{2+\sqrt{2}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}}{2} \overrightarrow{\jmath} \] bisects the angle formed by $\overrightarrow{v}$ and $\overrightarrow{\imath}$. Therefore the vector $\overrightarrow{u}$ is the unit vector in the direction of the sum $\overrightarrow{v} + \overrightarrow{\imath}$. That is \[ \overrightarrow{u} = \frac{1}{\big\|\overrightarrow{v} + \overrightarrow{\imath}\big\|} \big( \overrightarrow{v} + \overrightarrow{\imath} \big) \] Next we calculate the magnitude of this sum: \[ \big\|\overrightarrow{v} + \overrightarrow{\imath}\big\| = \sqrt{\frac{1}{2} + \sqrt{2} + 1 + \frac{1}{2}} = \sqrt{2+\sqrt{2}}. \] Hence, we calculated the vector $\overrightarrow{u}$ to be \begin{align*} \overrightarrow{u} &= \frac{1}{\sqrt{2+\sqrt{2}}} \left( \frac{2+\sqrt{2}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}}{2} \overrightarrow{\jmath} \right) \\ & = \frac{\sqrt{2+\sqrt{2}}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}}{2\sqrt{2+\sqrt{2}}} \overrightarrow{\jmath} \\ & = \frac{\sqrt{2+\sqrt{2}}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}\sqrt{2-\sqrt{2}}}{2\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}} \overrightarrow{\jmath} \\ & = \frac{\sqrt{2+\sqrt{2}}}{2} \overrightarrow{\imath} + \frac{\sqrt{2}\sqrt{2-\sqrt{2}}}{2\sqrt{4-2}} \overrightarrow{\jmath} \\ & = \frac{\sqrt{2+\sqrt{2}}}{2} \overrightarrow{\imath} + \frac{\sqrt{2-\sqrt{2}}}{2} \overrightarrow{\jmath} \end{align*}
- As an exercise, use the above method to calculate the exact values of the cosine and the sine of the angle $\pi/12$.
- With the expressions for $\cos\bigl(\pi/8\bigr),\sin\bigl(\pi/8\bigr)$ and $\cos\bigl(\pi/12\bigr),\sin\bigl(\pi/12\bigr)$ all other expressions in the figure above can be obtained using the symmetries of the cosine and sine expressed by the following formulas \begin{alignat*}{2} \cos\Big(\frac{\pi}{2} - \alpha \Big) & = \sin \alpha & \qquad \qquad \sin\Big(\frac{\pi}{2} - \alpha \Big) & = \cos \alpha \\ \cos\Big(\frac{\pi}{2} + \alpha \Big) & = -\sin \alpha & \qquad \qquad \sin\Big(\frac{\pi}{2} + \alpha \Big) & = \cos \alpha \\ \cos\big(\pi - \alpha \big) & = -\cos \alpha & \qquad \qquad \sin\big(\pi - \alpha \big) & = \sin \alpha \\ \cos\big(\pi + \alpha \big) & = -\cos \alpha & \qquad \qquad \sin\big(\pi + \alpha \big) & = -\sin \alpha \\ \cos\Big(\frac{3\pi}{2} - \alpha \Big) & = -\sin \alpha & \qquad \qquad \sin\Big(\frac{3\pi}{2} - \alpha \Big) & = -\cos \alpha \\ \cos\Big(\frac{3\pi}{2} + \alpha \Big) & = \sin \alpha & \qquad \qquad \sin\Big(\frac{3\pi}{2} + \alpha \Big) & = -\cos \alpha \\ \cos\big(2\pi - \alpha \big) & = \cos \alpha & \qquad \qquad \sin\big(2\pi - \alpha \big) & = -\sin \alpha. \end{alignat*}
-
I mention this property in class but I have not posted it. Given a vector $\vec{a} = x \vec{\imath} + y \vec{\jmath},$ very often it is important to identify a vector which is orthogonal to it. The pictures below show a simple rule that leads to an orthogonal vector. In the pictures below the dark green vector is the given vector $\vec{a} = x \vec{\imath} + y \vec{\jmath}.$ Notice that the navy blue vector $-y \vec{\imath} + x \vec{\jmath}.$ Because of the congruence of the triangles with colored sides, the navy blue vector is orthogonal to the dark green vector.
-
With the fact established in the previous item, we go back to the contour diagram of the linear function posted on October 3. If a bug moves in a heated plate in which the temperature is given by this linear function in the direction of the vector $-\vec{\imath} + 2 \vec{\jmath},$ then the bug will stay on a contour line and experience no change in temperature. Now we have the language to express the direction in which a bug moving in this heated plate will experience the largest rate of change of temperature:
- For Section 13.1 do Exercises 1-6, 7, 8, 12, 14, 15-17, 21. 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43.
- For Section 12.6 do Exercises 1-6, 13, 18, 19, 23.
- Motivated with the discussion of the direction of largest slope on a heated plate in which the temperature is given by a linear function of two variables we briefly introduced the concept of a displacement vector, or simply vector. This is done in more detail in Section 13.1.
- For Section 12.5 do Exercises 1, 2, 5, 7, 8-15, 19, 20, 21, 24, 27, 28, 2, 30, 31, 32, 33, 35.
-
Here is an animation which shows level surfaces of the function $w = x^2 + y^2 - z^2$ studied in Example 3 in Section 12.5.
Place the cursor over the image to start the animation.
Five of the above level surfaces.
-
Here I present animations of cross sections of a graph of $z = y^2-x^2$ by planes parallel to $xy$-plane, $zx$-plane and $yz$-plane, respectively.
Place the cursor over an image to start the animation.
-
Here I present animations of cross sections of a little bit more interesting graph of $z = y^3 + xy.$ As above, the cross sections are by planes parallel to $xy$-plane, $zx$-plane and $yz$-plane, respectively.
Place the cursor over an image to start the animation.
-
Today we talked about linear functions of two variables. The general form for a linear function in two variables is
\[
z = mx+ny+c.
\]
The graph this function is a plane. The number $m$ is the slope of this plane in the direction of $x$-axis and $n$ is the slope of this plane in the direction of $y$-axis. Notice that the point $(0,0,c)$ is on this plane; it is the $z$-intercept of this plane. The above form could be called "slopes-intercept" form of the linear function.
Another important form of a linear function is $slopes-point$ form: Given slopes $m$ in $x$-direction $n$ in $y$-direction and a point $x_0,y_0,z_0)$ on the plane, the equation of that plane is \[ z = m (x-x_0) + n (y-y_0) + z_0 \] -
Let us consider a specific example. Consider the plane whose axes intercepts are $(1,0,0)$, $(0,2,0)$ and $(0,0,3).$ The equation of this plane is
\[
z = -3 x - \frac{3}{2} y + 3.
\]
A contour plot of this plane is given below by navy-blue lines. Contour lines are given at levels -9,-6,-3,0,3,6,9,12,15, 18.
In class we discussed the direction in $xy$-plane in which the corresponding slope in the given plane is the largest. We concluded that that is the direction which is orthogonal to the contour lines of the given plane. That is, the direction of the motion along the maroon line from the navy-blue point to the maroon point. The corresponding slope of the line in the given plane directly above the maroon line is "rise/run" where the rise is from $0$ to $15$ (thus $15$) and the run is the distance from the navy-blue point to the maroon point, that is $\sqrt{20} = 4\sqrt{5}.$ Thus the slope is $3\sqrt{5}/4.$
This slope is bigger than any other slope in this plane. Why? For example if we move in along the teal line from the navy-blue point to the teal point then the corresponding slope of the line in the given plane directly above the teal line is "rise/run" where the rise is again $15$ but now the run is the distance from the navy-blue point to the teal point. Since the rise is the same and the run is larger, the slope is smaller. - For Section 12.4 do Exercises 1-6, 7, 8, 9, 11, 21-24, 25-28, 30, 31, 32.
-
We encountered two kinds of hyperbolic paraboloids: $z = 2 xy$ and $z = x^2 - y^2$. It is remarkable that these two formulas lead to the essentially same looking surface. That can be observed just looking at these plots:
Another way to see that the surfaces representing graphs of $z = 2 xy$ and $z = x^2 - y^2$ is to look at their contour diagrams.
- Later we will use vectors to prove that the the counterclockwise rotation by the angle of $\pi/4$ of the graph of $z = x^2 - y^2$ is exactly the graph of $z = 2 xy$.
-
Here I present a geometric proof that the maroon graph of $y = \sqrt{x^2-1}$ with $x \geq 1$ when rotated counterclockwise by $\pi/4$ coincides with the navy-blue graph of $y=1/(2x)$ with $0\lt x \leq \sqrt{2}/2.$
In the above reasoning, the point $P$ is an arbitrary point on the navy-blue graph. That is the number $s$ is an arbitrary real number such that $0\lt x \leq \sqrt{2}/2.$ The point $Q$ is the point on the maroon graph which is at the distance $\frac{\sqrt{1+4s^4}}{2s}$ from the origin. This point is determined by finding the positive solution $t$ of the equation \[ \sqrt{t^2 + t^2 - 1} = \frac{\sqrt{1+4s^4}}{2s}. \] The solution is $t = \frac{1+2 s^2}{2\sqrt{2} s}.$ Notice that \[ \sqrt{\left(\frac{1+2 s^2}{2\sqrt{2} s}\right)^2 - 1} = \frac{1-2 s^2}{2\sqrt{2} s}. \] To calculate the distance between $P$ and $Q$ is a lengthy calculation: \[ d(P,Q) = \sqrt{\left(s - \frac{1+2 s^2}{2\sqrt{2} s} \right)^2 + \left(\frac{1}{2s} - \frac{1-2 s^2}{2\sqrt{2} s} \right)^2 } = \sqrt{2-\sqrt{2}} \, \frac{\sqrt{1+4s^4}}{2s}. \]
The navy-blue graph is the graph of $y=1/(2x)$ with $0\lt x \leq \sqrt{2}/2.$ The maroon graph is the graph $y = \sqrt{x^2-1}$ with $x \geq 1.$ I claim that the maroon graph when rotated counterclockwise by $\pi/4$ coincides with the navy-blue graph.
To prove this claim first notice that the vertices of the teal triangle are \[ (0,0), \quad \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right), \quad (1,0). \] Thus, the teal triangle is an isosceles triangle with the sides, $1$, $1$ and $\sqrt{2-\sqrt{2}}$ and its angle at the origin is $\pi/4.$
The purple triangle is similar to the teal triangle. This follows by calculating the vertices of the purple triangle: \[ O = (0,0), \quad P = \left(s,\frac{1}{2s} \right), \quad Q = \left(\frac{1+2 s^2}{2\sqrt{2} s}, \frac{1-2 s^2}{2\sqrt{2} s} \right). \] and then calculating the sides of the purple triangle: \[ d(O,P) = d(O,Q) = \frac{\sqrt{1+4s^4}}{2s}, \] \[ d(P,Q) = \sqrt{2-\sqrt{2}} \, \frac{\sqrt{1+4s^4}}{2s}. \]
- For Section 12.3 do Exercises 1-4, 8, 10, 11, 12, 13, 14, 16, 17, 20, 21, 22, 23, 28, 30, 36, 37, 38, 39, 40, 41, 42.
- For Section 12.2 do Exercises 2, 3, 4-11, 14, 15, 16-18, 19, 20, 28, 30, 31.
-
Our first illustration of a multivariable function was the function of multiplication of two numbers. We all remember the multiplication table from the elementary school. In class I created this
Excel file with a standard multiplication table from elementary school and one with 1681 products of real numebrs between -2 and 2. What do we learn from this large Excel table? For example we can use the Excel manu item
Notice that the curves formed by the heighlighted cells in the above Excel file are in fact Excel approximations of contour lines, see Section 12.3 and an item below. - Next we utilize the three-dimensional space to visualize the function $z = xy.$ As the last equation indicates, we plot all the points $(x,y,z)$ in the three-dimensional coordinate space for which $z = xy.$ Below is a Mathematica plot of the multiplication function from two different viewpoints.
- The curves formed by the highlighted lines on our Excel file are rough approximations of contour lines of the function $f(x,y) = xy$. In the plot below we show many more contour lines of this function. In fact we show all the countour lines with $xy = c$ where $c = k/4$ where $k \in \{-15,,\ldots,-1,0,1,\ldots,15\}.$ In addition to the countour lines, the plot below indicates the values of the function by colloring following the following color coding:
Hyperbolic Paraboloyd $z=xy$
|
|
- The information sheet
- Here is the Excel file that I created in class today. You can explore other functions by creating similar Excel files.
-
Since we will be working in the three dimensional space the following is must know terminology:
2-dimensional object: square rectangle parallelogram circle disk ellipse 3-dimensional object: cube cuboid parallelepiped sphere ball ellipsoid
In Section 12.1, Problem 36 in our textbook the geometric object which is above called cuboid is called rectangular solid. However, I do not think that this name is widely accepted. Since there is no well established name for this object I chose cuboid from several names that are used. The main reason that I decided to call this solid cuboid is the following picture
- I recommend that you always check Wikipedia and other web pages related to the topics that we study. Usually Wikipedia pages are very informative. However, the page for cuboid is strange. The definition given for cuboid on this page is far too technical and very difficult to understand. This Wikipedia entry lists the following alternative names for this concept: rectangular box, rectangular hexahedron, right rectangular prism, and rectangular parallelepiped.
-
The key concept that we discussed was the concept of Cartesian (or rectangular) coordinate system in three-dimensional space. The main ingredients of the Cartesian coordinate system in three-dimensional space are:
- The origin of the coordinate system which is denoted by $O$ (this is the capital letter O, not zero).
- The three coordinate axes: $x$-axis, $y$-axis, $z$-axis.
- The three coordinate planes: $xy$-plane, $yz$-plane, $zx$-plane.
-
One of the immediate benefits of a coordinate system is that when we have coordinates of two points, then we can calculate some related geometric objects. The best example is the distance. Given two points $P_1 = (x_1,y_1,z_1)$ and $P_2 = (x_2,y_2,z_2)$ the formula
\[
d(P_1,P_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 }
\]
gives the distance between these two given points.
To understand this formula assume that $x_1\neq x_2$, $y_1\neq y_2$ and $z_1\neq z_2$ and focus on the cuboid whose diagonal is the line segment $\overline{P_1P_2}$ and whose faces are parallel to the coordinate planes. The eight vertices of this cuboid are \[ (x_1,y_1,z_1), (x_1,y_1,z_2), (x_1,y_2,z_1), (x_2,y_1,z_1), (x_1,y_2,z_2), (x_2,y_1,z_2), (x_2,y_2,z_1), (x_2,y_2,z_2). \] Now choose a coordinate plane through the vertex $P_1$ and observe the face of the cuboid in this plane (this is a rectangle). This rectangle has two diagonals. One of its diagonals has an end point at $P_1$. Denote its other endpoint by $Q$. Calculate the length of the diagonal $P_1Q$. One uses the Pythagorean theorem for this. Next calculate the length of the line segment $QP_2$. This line segment is an edge of the cuboid that we study. The triangle $P_1QP_2$ is a right triangle with the right angle at $Q$. The diagonal $P_1P_2$ is the hypothenuse of this triangle. Now we calculate the length of $P_1P_2$ using the Pythagorean theorem. - The proof of the distance formula in space is based on the Pythagorean theorem. As you can see in this Wikipedia page, there are many different proofs of the Pythagorean theorem.
- Here you will find a visual proof of Pythagorean theorem that I designed.
- For Section 12.1 do Exercises 1-10, 27, 28, 29, 30, 32, 33, 35, 36, 38.
-
Problem.
Consider the cube with the following three properties:
- The cube contains the sphere $(x-2)^2 + (y-3)^2 + (z-4)^2 = 25$.
- Each face of the cube is parallel to a coordinate plane.
- Each face of the cube touches the sphere.
- Find the points where the faces of this cube touch the sphere.
- Find the eight vertices of this cube.
- Determine whether the origin inside or outside of the given sphere.
- Determine whether the origin inside or outside of the given cube?