applications Archives - rweber.net https://www.rweber.net/tag/applications/ trying to be a mile wide AND a mile deep Thu, 12 Oct 2017 11:18:36 +0000 en-US hourly 1 https://wordpress.org/?v=6.9.4 37896774 Distinguishing related rates and optimization https://www.rweber.net/mathematics/calculus/distinguishing-related-rates-and-optimization/ https://www.rweber.net/mathematics/calculus/distinguishing-related-rates-and-optimization/#comments Mon, 11 Mar 2013 12:00:15 +0000 http://www.rweber.net/?p=272 One of the main things students struggle with on exams is identifying the method to use on each problem. When they are learned in class, they are segregated by type, and their distinguishing features are not always highlighted. Here’s a brief cheat sheet on related rates and optimization. optimization related rates sample keywords maximize, minimize, […]

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]]> One of the main things students struggle with on exams is identifying the method to use on each problem. When they are learned in class, they are segregated by type, and their distinguishing features are not always highlighted. Here’s a brief cheat sheet on related rates and optimization.

optimization related rates
sample keywords maximize, minimize, most, smallest how fast, rate of change/increase/decrease
sample problem What dimensions maximize the area enclosed in a rectangular fence built with 100 feet of fencing? A shape is deforming such that it is always a rectangle of perimeter 100 ft, but the width is increasing by 5 ft/min. How fast is the area changing when the width is 20 feet?
goal maximize area find rate of change of area
answer form dimensions in feet rate in ft2/min
equation A = l\cdot w = (50-w)\cdot w = 50w - w^2 likewise, A = 50w - w^2
equation, unabridged A(w) = 50w - w^2 (area viewed as function of width) A(t) = 50w(t) - [w(t)]^2 (area and width viewed as related functions of time)
final answer a 25′ square fence maximizes the area the area is changing by 50 ft2/min

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]]> https://www.rweber.net/mathematics/calculus/distinguishing-related-rates-and-optimization/feed/ 2 272 Optimization https://www.rweber.net/mathematics/calculus/optimization/ https://www.rweber.net/mathematics/calculus/optimization/#respond Mon, 04 Mar 2013 13:00:50 +0000 http://www.rweber.net/?p=264 As with related rates, optimization is sometimes approached with some dread. Fortunately, although the calculus has more steps than in related rates, the process is somewhat more cut and dried. A. Make two-word phrase of goal, where first word is minimize or maximize and second word is the name of a quantity. B. Name quantities […]

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]]> As with related rates, optimization is sometimes approached with some dread. Fortunately, although the calculus has more steps than in related rates, the process is somewhat more cut and dried.

A. Make two-word phrase of goal, where first word is minimize or maximize and second word is the name of a quantity.
B. Name quantities and state all relationships they have, as equations. Graphs or other pictures can be helpful.
C. Make the quantity to be optimized a function of a single variable, using other relationship(s) as needed to simplify.
D. Find critical points and extrema for the function from C, and pick out the point(s) at which the goal from A is met.
E. Unpack the information from D to answer the question as stated.

Example 1.
Of all possible pairs of numbers that sum to 100, find those with the largest product.
A. Goal: maximize product.
B. Call the product and the numbers P, x, and y, respectively. We have P = xy by definition and x+y = 100 from the problem statement.
C. Use the sum equation to simplify the product equation: P = (100-x)\cdot x = 100x - x^2.
D. P'(x) = 100 - 2x. CPs: 100 - 2x = 0 only when x = 50. Is this a maximum? P'(49) = 100 - 98 = 2, P'(51) = 100-102 = -2, so the first derivative test says yes.
E. The pair of numbers with maximum product out of all those that sum to 100 is 50 and 50.

Example 2.
A cylindrical drum must hold 10 m3 of sand. The material for the sides costs $10 per m2 and the material for the base and lid cost $15 per m2. Find the dimensions of the drum which minimize its cost.
A. Goal: minimize cost.
B. The volume of a cylinder is \pi r^2 h, which we know equals 10. The cost of the drum is 10*(side area) + 15*(top and bottom area), which we may also put in terms of radius r and height h. Let cost be C; then C = 10(2\pi r h) + 15(2)(\pi r^2).
C. To avoid square roots, put everything in terms of r. We have h = \frac{10}{\pi r^2} from the volume constraint, which means cost as a function of radius is C(r) = 10(2\pi r \frac{10}{\pi r^2}) + 30\pi r^2 = \frac{200}{r} + 30\pi r^2.
D. C'(r) = -\frac{200}{r^2} + 60\pi r. Critical points occur when 60\pi r = \frac{200}{r^2}; solving for r we obtain r^3 = \frac{200}{60\pi} = \frac{10}{3\pi}, so r = \sqrt[3]{10/(3\pi)}. To find whether this is a minimum we’ll do the second derivative test. C''(r) = 400r^{-3} + 60\pi. For positive r (such as our critical point) C'' is positive, so the graph is concave up and we have indeed found a minimum.
E. We must have a drum with a radius of \sqrt[3]{10/(3\pi)} m and a height of h = 10/\pi \cdot (10/(3\pi))^{-2/3} m, which simplifies to \sqrt[3]{90/\pi} m. In decimal form, r = 1.02m and h = 3.06m.

Example 3.
Find the dimensions of the largest-area rectangle that can be inscribed in a circle of positive radius r.
Note: inscribed means the corners of the rectangle are on the circle.
A. Goal: maximize area.
B. For simplicity, let us suppose the circle is centered at the origin. Any inscribed rectangle must have a corner in each quadrant, and any corner tells us the location of the other three, by symmetry. Call the location of the first quadrant corner (x,y); the lengths of the sides of the rectangle are 2x and 2y. Now we can wrote relationship equations: area A = 4xy. Since the corners are on the circle, the distance from the origin to (x,y) must be r: r = \sqrt{x^2 + y^2}.
C. We must put the area equation in terms of one variable since it is what we are trying to maximize. Not having a numerical value for the radius can make it confusing, but r is a parameter: a stand-in for a constant we don’t wish to actually specify. It does not matter which variable we substitute for since we have perfect symmetry, so solve for x = \sqrt{r^2 - y^2}. Then A(y) = 4y\sqrt{r^2-y^2}, where r is an unspecified constant.
D. A'(y) = 4y(1/2(r^2-y^2)^{-1/2}(-2y)) + 4\sqrt{r^2-y^2} = \frac{-4y^2}{\sqrt{r^2-y^2}} + 4\sqrt{r^2-y^2}. Setting this equal to zero, we obtain 4\sqrt{r^2-y^2} = \frac{4y^2}{\sqrt{r^2-y^2}}, which simplifies to r^2-y^2 = y^2 and finally y = r/\sqrt{2}. Is this a maximum? Use the first derivative test. \sqrt{2} is approximately 1.4; $latex 5/4 = 1.25 < \sqrt{2} < 2$, so test $latex r/2 < r/\sqrt{2} < 4r/5$. $latex A'(r/2) = \frac{-4(r^2/2)}{\sqrt{r^2 - r^2/4}} + 4\sqrt{r^2-r^2/4} = \frac{-r^2}{\sqrt{3r^2/4}} + 4\sqrt{3r^2/4} = \frac{-r^2}{\sqrt{3}/2 r} + 2\sqrt3 r$ $latex = 2\sqrt 3 r - 2r/\sqrt 3$.    $latex 2\sqrt 3 > 2/\sqrt 3$ so this is positive.
A'(4r/5) = \frac{-4(\frac{16 r^2}{25})}{\sqrt{9/25 r^2}} + 4\sqrt{9/25 r^2} = 12/5 r - 64/15 r = (36/15 - 64/15) r, which is negative, and hence y = r/\sqrt 2 is a maximum.
E. When y = r/\sqrt 2, so does x, by the equation relating both of them to r. Hence a square of side length 2x = r\sqrt 2 is the largest possible inscribed rectangle.

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]]> https://www.rweber.net/mathematics/calculus/optimization/feed/ 0 264 Related Rates https://www.rweber.net/mathematics/calculus/related-rates/ https://www.rweber.net/mathematics/calculus/related-rates/#respond Mon, 25 Feb 2013 13:00:05 +0000 http://www.rweber.net/?p=258 These are perhaps the epitome of the dreaded “word problem.” Fortunately there is a step by step method for solving them. A. Assign variables to your quantities and find relationship equations for them. B. If relevant, simplify the equations to remove extraneous variables. C. Express given and asked-for rates of change in terms of d/dt. […]

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]]> These are perhaps the epitome of the dreaded “word problem.” Fortunately there is a step by step method for solving them.

A. Assign variables to your quantities and find relationship equations for them.
B. If relevant, simplify the equations to remove extraneous variables.
C. Express given and asked-for rates of change in terms of d/dt. At what point(s) do you want to know the rate(s) of change?
D. Differentiate relationship equation with respect to t.
E. Plug in known values and solve for desired value.

In C I will be finding the value of all variables at the desired point, although frequently you don’t need them.

Example 1.
The area of a square is increasing by 3 in2/s. What is the rate of change of the length of the sides when the side length is 2 in?
A. Name quantities: area A, side length \ell (same for all sides since it’s a square). Relationship: A = \ell^2.
B. This is already as simple as it gets.
C. Rates of change: given: \frac{dA}{dt} = 3 in2/s. We want to know \frac{d\ell}{dt} at the point \ell = 2, which is A = 4.
D. \frac{d}{dt}[A = \ell^2] is \frac{dA}{dt} = 2\ell \frac{d\ell}{dt}.
E. Know \frac{dA}{dt} = 3 and want \frac{d\ell}{dt} when \ell = 2; 3 = 4 \frac{d\ell}{dt} so \frac{d\ell}{dt} = \frac34. When the side length is 2 in, the sides are changing at a rate of 3/4 in/s.

Example 2.
You are testing walkie-talkie range with a friend. The two of you start at the same point, and then you amble north at 3 mph and your friend strides east at 4 mph. How fast are you moving away from each other when your friend is 3 mi away from the starting point?
A. Since you are moving north, label your position x, and since your friend is moving east, label her position y. You are really interested in the distance between you; call it R for (walkie-talkie) range. That is related to your positions by the equation R = \sqrt{x^2 + y^2}.
B. Since \frac{dx}{dt} and \frac{dy}{dt} are both constant, and hence x and y are constant multiples of each other, we could simplify by putting one in terms of the other but won’t gain much by doing so.
C. We are given \frac{dx}{dt} = 3 and \frac{dy}{dt} = 4 and want \frac{dR}{dt}. The point at which we want \frac{dR}{dt} is y=3, which is x = (\frac34)(3) = 2.25 and R = \frac{15}{4} = 3.75.
D. \frac{d}{dt}[R = \sqrt{x^2 + y^2}] is \frac{dR}{dt} = \frac12 (x^2+y^2)^{-1/2}(2x \frac{dx}{dt} + 2y \frac{dy}{dt}).
E. We know \frac{dx}{dt} = 3 and \frac{dy}{dt} = 4, and we want \frac{dR}{dt} when x = 2.25 and y = 3; \frac{dR}{dt} = (x^2+y^2)^{-1/2}(x \frac{dx}{dt} + y \frac{dy}{dt}) = (x \frac{dx}{dt} + y \frac{dy}{dt})/R = (2.25\cdot 3 + 3\cdot 4)/3.75 = 5. When you have gone 2.25 mi and your friend 3 mi, the distance between you is changing at a rate of 5 mph.

Example 3.
25 cubic inches of pizza dough are being spun in the air. Assume the surface is smooth and the outline is circular at all times. If the radius increases at a constant rate of 0.5 in/s, what is the rate of change of the thickness of the dough when the radius is 4 inches?
A. We have thickness, radius, and volume. Since t is taken, we’ll call them h, r, and V, respectively. The relationship is V = \pi r^2h.
B. Since V is constant, we can plug in its value. Since V is by itself on one side of the equation it is roughly equal amounts of work to do that as to leave it as V and plug in 0 for \frac{dV}{dt}, but since we did not simplify Example 2 we’ll simplify this one: 25 = \pi r^2 h.
C. We are given \frac{dr}{dt} = 0.5 in/s, and we want \frac{dh}{dt} at the point r = 4, which is h = \frac{25}{16\pi}.
D. \frac{d}{dt}[25 = \pi r^2h] is 0 = \pi (2r \frac{dr}{dt} h + r^2 \frac{dh}{dt}).
E. We know \frac{dr}{dt} = 0.5, and we want to know \frac{dh}{dt} when r = 4 and h = \frac{25}{16\pi}. Hence 0 = \pi (8 \cdot 0.5 \cdot \frac{25}{16\pi} + 16 \frac{dh}{dt}); -16\pi \frac{dh}{dt} = 100/16 = 25/4; \frac{dh}{dt} = -25/4 \cdot \frac{1}{16\pi} = -\frac{25}{64\pi} = -0.124 in/s. Make sense? If the volume is constant an increase in one dimension must be accompanied by a decrease in the other.

Example 4.
A coffee filter is being held under a running tap. It is cone-shaped, with height 8 cm and top radius 3 cm. Water flows in at the top at a rate of 4 cm3/s and drips out the bottom at 1.5 cm3/s. How fast is the water level in the filter rising when the depth is 2 cm?
A. Let’s call depth h since d is taken, radius r, and volume V. The volume equation for a cone is V = \frac{\pi}{3} r^2 h.
B. Because the filter is a cone, the ratio between depth and radius is the same for any amount of water in the filter. That is, while r is not always 3 and h is not always 8, r/h is always 3/8: r = 3/8 h. Thus the volume equation simplifies to V = \frac{3\pi}{64}h^3.
C. We are not given \frac{dV}{dt} right off, but the increase and decrease of water volume are both constant, so we can compute it: \frac{dV}{dt} = 4 - 1.5 = 2.5 cm3/s. We want to find \frac{dh}{dt} at the point h = 2, which gives V = \frac{3\pi}{8}.
D. \frac{d}{dt}[V = \frac{3\pi}{64}h^3] is \frac{dV}{dt} = \frac{9\pi}{64}h^2 \frac{dh}{dt}.
E. We have \frac{dV}{dt} = 2.5 and h = 2, so we obtain 2.5 = \frac{9\pi}{16} \frac{dh}{dt}, for \frac{dh}{dt} = \frac{40}{9\pi} = 1.41 cm/s. When the water is 2 cm deep, it is rising at a rate of 1.41 cm/s.

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]]> https://www.rweber.net/mathematics/calculus/related-rates/feed/ 0 258 Linear Algebra Presentations https://www.rweber.net/mathematics/linear-algebra/linear-algebra-presentations/ https://www.rweber.net/mathematics/linear-algebra/linear-algebra-presentations/#respond Mon, 14 Jan 2013 13:00:15 +0000 http://www.rweber.net/?p=86 Here are slide presentations for a linear algebra class that claims to include applications, but really barely has enough time to cover the nonapplied material. I made two presentations that could be done in 20 minutes of class time. These are both incorporated to some degree into the math club linear algebra talk I gave, […]

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]]> Here are slide presentations for a linear algebra class that claims to include applications, but really barely has enough time to cover the nonapplied material. I made two presentations that could be done in 20 minutes of class time.

These are both incorporated to some degree into the math club linear algebra talk I gave, but are much shorter and self-contained.

A friend gave me an apocryphal quote for the reason to shift to raster graphics: “I have 8 pixels. How do I make it look like a cat?”

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]]> https://www.rweber.net/mathematics/linear-algebra/linear-algebra-presentations/feed/ 0 86 Linear Algebra and its Uses https://www.rweber.net/mathematics/linear-algebra/linear-algebra-and-its-uses/ https://www.rweber.net/mathematics/linear-algebra/linear-algebra-and-its-uses/#respond Mon, 05 Nov 2012 13:00:23 +0000 http://www.rweber.net/?p=40 In February of 2011, I gave a talk to the Dartmouth Math Club about linear algebra. This post replicates the page of resources I made for the club members related to that talk, and adds one new one. Slides from the talk. Pdf about uses of linear algebra. From Oliver Knill, Harvard Math, near the […]

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]]> In February of 2011, I gave a talk to the Dartmouth Math Club about linear algebra. This post replicates the page of resources I made for the club members related to that talk, and adds one new one.

  • Slides from the talk.
  • Pdf about uses of linear algebra. From Oliver Knill, Harvard Math, near the bottom of this page, which has a large number of other handouts about linear algebra in general and applications.

Vector graphics

  • Some Adobe Photoshop help. Unfortunately you have to go to the index and click V to find the vector graphics segment, but it is the source of the bicycle graphic in my slides and has a nice explanation.
  • Interactive Mathematics’ vector art page, with explanation of more of the math behind the pictures and links to lots of other pages.
  • Inkscape is a free drawing program that makes vector graphics natively. They have many useful tutorials and you can find even more elsewhere on the web.

Google PageRank

Markov chains

  • Fun With Markov Chains. Has examples where Alice in Wonderland was mixed together with either Hamlet or the Biblical books of Genesis and Revelation.
  • Generating Text, Jon Bentley. Has examples where the King James Bible was remixed, and where his own text Programming Pearls was remixed.
    Note: the friend I corralled into compiling the C code for me said the first link’s program gave immense amounts of output without seeming to be near an end, whereas the second program gave much more reasonable files. I suspect based on the files I received that the amount of output in the second case is proportional to the amount of input. Reading the first page indicates you can specify the number of words of output in the command line; I don’t know if my friend did that.
  • Random Word Generator
  • Mark V. Shaney text generator

More details on the example in the slides above:
The Robert Frost poems included were Birches, Come In, Desert Places, Mending Wall, Nothing Gold Can Stay, On Looking Up by Chance at the Constellations, Reluctance, and Stars. The Dr. Seuss books included were One Fish, Two Fish, Red Fish, Blue Fish, and The Sneetches. The full recombined text may be found here, though I only put line breaks into the first third or so. The rest is unmined for Seufrostian gems. I tried putting a combination of the Dartmouth Marching Band song lyrics (only the songs related to Dartmouth, of course), and the standard description of Daniel Webster’s closing in the Dartmouth Case, but it did not turn up much of anything. If you don’t want to compile the C code in one of the first two links, Mark V. Shaney could be the answer. If it turns up any pearls let me know!

Principal component analysis

  • Source of graphic in the slides.
  • Chapter on PCA. Long, but a good example and overview at the start. This links directly to the pdf; I was unable to find a descriptive page for it.

Miscellaneous Fun

We’ll close with the only linear algebra joke I found worth sharing, a limerick by Donald E. Simanek.

Null vectors have zero projection.
So you ask, “What can be their direction?”
They point any which way.
“That’s magic!” you say?
Not really; it’s just misdirection.

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