Michael0x2a · December 25, 2015 01:29
diff --git a/webassign_6-7.ipynb b/webassign_6-7.ipynb
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      "# Problem 7\n",
      "\n",
      "## Part a\n",
      "\n",
      "### General approach\n",
      "\n",
      "The area of the triangle is basically equal to one-half the product of the x-intercept and the y-intercept. Therefore, we need to first find the two intercepts, based on what we know.\n",
      "\n",
      "### Definition and basic setup:\n",
      "\n",
      "To make discussing this problem easier, I'm going to define a few variables:\n",
      "\n",
      "1. $(p, q)$ be the coordinates of $P(t)$\n",
      "2. $(a, b)$ be the coordinates of the intercept (either one)\n",
      "3. $(x, y)$ be general coordinates\n",
      "\n",
      "We can then start with the following equations:\n",
      "\n",
      "$$\n",
      "\\begin{array}{ll}\n",
      "    P(t) = (\\cos{t}, \\sin{t}) = (p, q) & \\text{I don't want to keep writing sin and cosine} \\\\\n",
      "    p^2 + q^2 = 1     & \\text{Equation to describe a circle} \\\\\n",
      "    y = mx + k        & \\text{Equation for the tangent line} \\\\\n",
      "    m = -\\frac{p}{q}  & \\text{Slope of the tangent line} \\\\\n",
      "\\end{array}\n",
      "$$\n",
      "\n"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "### Solving for the x and y intercepts\n",
      "\n",
      "If we substitute in the slope, $(q, p)$ and $(a, b)$ into the linear equation, we get:\n",
      "\n",
      "$$\n",
      "q = -\\frac{p}{q}p + k \\\\\n",
      "b = -\\frac{p}{q}a + k \\\\\n",
      "$$\n",
      "\n",
      "Now, rearrange so that $k$ is on one side:\n",
      "\n",
      "$$\n",
      "q = -\\frac{p^2}{q} + k \\\\\n",
      "b + \\frac{p}{q}a = k \\\\\n",
      "$$\n",
      "\n",
      "Substitute:\n",
      "\n",
      "$$\n",
      "q = -\\frac{p^2}{q} + \\frac{p}{q}a + b\\\\\n",
      "$$\n",
      "\n",
      "In order to get the y-intercept, $a$ must equal 0. Therefore, $b$ equals:\n",
      "\n",
      "$$\n",
      "q = -\\frac{p^2}{q} + \\frac{p}{q}a + b\\\\\n",
      "b = q + \\frac{p^2}{q}\\\\\n",
      "$$\n",
      "\n",
      "In order to solve for the x-intercept, $b$ must equal 0. Therefore, $a$ equals:\n",
      "\n",
      "$$\n",
      "q = -\\frac{p^2}{q} + \\frac{p}{q}a + b\\\\\n",
      "a = \\left( q + \\frac{p^2}{q} \\right) \\cdot \\frac{q}{p} \\\\\n",
      "a = \\frac{q^2}{p} + p \\\\\n",
      "$$\n",
      "\n",
      "After substituting back in cosine and sine, we get:\n",
      "\n",
      "$$\n",
      "a = \\frac{\\sin{(t)}^2}{\\cos{t}} + \\cos{t} \\\\\n",
      "b = \\frac{\\cos{(t)}^2}{\\sin{t}} + \\sin{t} \\\\\n",
      "$$\n",
      "\n"
     ]
    },
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     "cell_type": "markdown",
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     "source": [
      "### Equation for the area\n",
      "\n",
      "So, the y-intercept and the x-intercepts will equal the sides of the triangle. \n",
      "\n",
      "Therefore, the area is:\n",
      "\n",
      "$$\n",
      "A(t) = \\frac{1}{2}ab \\\\\n",
      "A(t) = \\frac{1}{2} \\left( \\frac{\\sin{(t)}^2}{\\cos{t}} + \\cos{t} \\right) \\left(\\frac{\\cos{(t)}^2}{\\sin{t}} + \\sin{t} \\right) \\\\\n",
      "A(t) = \\frac{1}{2} \\left( \\frac{\\sin{(t)}^2}{\\cos{t}} + \\frac{\\cos{(t)}^2}{\\cos{t}} \\right) \\left(\\frac{\\cos{(t)}^2}{\\sin{t}} + \\frac{\\sin{(t)}^2}{\\sin{t}} \\right) \\\\\n",
      "A(t) = \\frac{1}{2} \\left( \\frac{\\sin{(t)}^2 + \\cos{(t)}^2}{\\cos{t}} \\right) \\left(\\frac{\\cos{(t)}^2 + \\sin{(t)}^2}{\\sin{t}} \\right) \\\\\n",
      "A(t) = \\frac{1}{2} \\left( \\frac{1}{\\cos{t}} \\right) \\left(\\frac{1}{\\sin{t}} \\right) \\\\\n",
      "A(t) = \\frac{1}{2\\cos{t}\\sin{t}}\n",
      "$$"
     ]
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     "source": [
      "---\n",
      "\n",
      "## Parts b - d\n",
      "\n",
      "Think about these geometrically. If $t$ goes to $\\frac{\\pi}{2}$ or to $0$, what kind of shape would the triangle make?\n",
      "\n",
      "As $t$ goes to $\\frac{\\pi}{2}$, it steadily makes the triangle bigger and bigger, until it's basically a rectangle. Therefore, the area is $\\infty$.\n",
      "\n",
      "The same principle applies as $t$ goes to $0$.\n",
      "\n",
      "In every other case, you can just plug in whatever value of $t$ is given to you."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "---\n",
      "\n",
      "## Parts e - f\n",
      "\n",
      "### Definition and setup\n",
      "\n",
      "Setup:\n",
      "\n",
      "$$\n",
      "2 = \\frac{1}{2 \\cos{t} \\sin{t}} \\\\\n",
      "\\cos{t} \\sin{t} = \\frac{1}{4}\n",
      "$$\n",
      "\n",
      "We need to use the double angle identity, which states that the following statement is true:\n",
      "\n",
      "$$\n",
      "\\sin{2t} = 2\\cos{t}\\sin{t}\n",
      "$$\n",
      "\n",
      "This is basically a formula that I guess we were just sort of \"expected\" to know, similar to the quadratic formula or the Pythagorean theorem. As far as I know, we were never taught this, and there's probably alternate ways to solve this problem, but it did appear to be the simplest method, so I just ran with it.\n",
      "\n",
      "### Application\n",
      "\n",
      "If we rearrange our setup, we can exploit the double angle identity:\n",
      "\n",
      "$$\n",
      "\\cos{t_0} \\sin{t_0} = \\frac{1}{4} \\\\\n",
      "2\\cos{(t_0)}\\sin{(t_0)} = \\frac{1}{2} \\\\\n",
      "\\sin{(2t_0)} = \\frac{1}{2} \\\\\n",
      "2t_0 = \\arcsin{\\frac{1}{2}} \\\\\n",
      "2t_0 = \\frac{\\pi}{6} \\\\\n",
      "t_0 = \\frac{\\pi}{12} \\\\\n",
      "$$\n",
      "\n",
      "Now, geometrically, we need to sort of \"rotate\" and \"flip\" the triangle so we get the same area: \n",
      "\n",
      "$$\n",
      "t_1 = \\frac{\\pi}{2} - t_0 \\\\\n",
      "t_1 = \\frac{5\\pi}{12}\n",
      "$$\n",
      "\n",
      "Try drawing it out, and see what the triangle looks like at both the times, and compare the shapes/areas of both."
     ]
    },
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     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Part g\n",
      "\n",
      "Don't overthink this one. How do you find the average? \n",
      "\n",
      "$$\n",
      "\\frac{A(t_1) - A(t_0)}{t_1 - t_0}\n",
      "$$\n",
      "\n",
      "Our function, in this case, is the formula for the area:\n",
      "\n",
      "$$\n",
      "A(t) = \\frac{1}{2 \\cos{t} \\sin{t}}\n",
      "$$\n",
      "\n",
      "## Parts h - i\n",
      "\n",
      "This is basically taking the limit of the area as the delta approaches from the left or right. It's basically the same thing as above, except now we're finding the average rate of change between two times that are only a fraction of a second apart. Basically, taking the derivative.\n",
      "\n",
      "For example, for part h, you would try and find the limit by using the formula:\n",
      "\n",
      "$$\n",
      "\\text{limiting value} = \\lim_{b \\rightarrow 0^+} \\frac{A(\\frac{\\pi}{6} + b) - A(\\frac{\\pi}{6})}{(\\frac{\\pi}{6} + b) - \\frac{\\pi}{6}}\n",
      "$$\n",
      "\n",
      "...and finding the limiting value by plugging in smaller and smaller numbers for $b$.\n",
      "\n",
      "Question i is the same, except that you approach the limiting value from the other side (so instead of adding the delta, you subtract) and use a number other then $\\frac{\\pi}{6}$"
     ]
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	"cell_type": "markdown",
	"metadata": {},
	"source": [
	"# Problem 7\n",
	"\n",
	"## Part a\n",
	"\n",
	"### General approach\n",
	"\n",
	"The area of the triangle is basically equal to one-half the product of the x-intercept and the y-intercept. Therefore, we need to first find the two intercepts, based on what we know.\n",
	"\n",
	"### Definition and basic setup:\n",
	"\n",
	"To make discussing this problem easier, I'm going to define a few variables:\n",
	"\n",
	"1. $(p, q)$ be the coordinates of $P(t)$\n",
	"2. $(a, b)$ be the coordinates of the intercept (either one)\n",
	"3. $(x, y)$ be general coordinates\n",
	"\n",
	"We can then start with the following equations:\n",
	"\n",
	"$$\n",
	"\\begin{array}{ll}\n",
	" P(t) = (\\cos{t}, \\sin{t}) = (p, q) & \\text{I don't want to keep writing sin and cosine} \\\\\n",
	" p^2 + q^2 = 1 & \\text{Equation to describe a circle} \\\\\n",
	" y = mx + k & \\text{Equation for the tangent line} \\\\\n",
	" m = -\\frac{p}{q} & \\text{Slope of the tangent line} \\\\\n",
	"\\end{array}\n",
	"$$\n",
	"\n"
	]
	},
	{
	"cell_type": "markdown",
	"metadata": {},
	"source": [
	"### Solving for the x and y intercepts\n",
	"\n",
	"If we substitute in the slope, $(q, p)$ and $(a, b)$ into the linear equation, we get:\n",
	"\n",
	"$$\n",
	"q = -\\frac{p}{q}p + k \\\\\n",
	"b = -\\frac{p}{q}a + k \\\\\n",
	"$$\n",
	"\n",
	"Now, rearrange so that $k$ is on one side:\n",
	"\n",
	"$$\n",
	"q = -\\frac{p^2}{q} + k \\\\\n",
	"b + \\frac{p}{q}a = k \\\\\n",
	"$$\n",
	"\n",
	"Substitute:\n",
	"\n",
	"$$\n",
	"q = -\\frac{p^2}{q} + \\frac{p}{q}a + b\\\\\n",
	"$$\n",
	"\n",
	"In order to get the y-intercept, $a$ must equal 0. Therefore, $b$ equals:\n",
	"\n",
	"$$\n",
	"q = -\\frac{p^2}{q} + \\frac{p}{q}a + b\\\\\n",
	"b = q + \\frac{p^2}{q}\\\\\n",
	"$$\n",
	"\n",
	"In order to solve for the x-intercept, $b$ must equal 0. Therefore, $a$ equals:\n",
	"\n",
	"$$\n",
	"q = -\\frac{p^2}{q} + \\frac{p}{q}a + b\\\\\n",
	"a = \\left( q + \\frac{p^2}{q} \\right) \\cdot \\frac{q}{p} \\\\\n",
	"a = \\frac{q^2}{p} + p \\\\\n",
	"$$\n",
	"\n",
	"After substituting back in cosine and sine, we get:\n",
	"\n",
	"$$\n",
	"a = \\frac{\\sin{(t)}^2}{\\cos{t}} + \\cos{t} \\\\\n",
	"b = \\frac{\\cos{(t)}^2}{\\sin{t}} + \\sin{t} \\\\\n",
	"$$\n",
	"\n"
	]
	},
	{
	"cell_type": "markdown",
	"metadata": {},
	"source": [
	"### Equation for the area\n",
	"\n",
	"So, the y-intercept and the x-intercepts will equal the sides of the triangle. \n",
	"\n",
	"Therefore, the area is:\n",
	"\n",
	"$$\n",
	"A(t) = \\frac{1}{2}ab \\\\\n",
	"A(t) = \\frac{1}{2} \\left( \\frac{\\sin{(t)}^2}{\\cos{t}} + \\cos{t} \\right) \\left(\\frac{\\cos{(t)}^2}{\\sin{t}} + \\sin{t} \\right) \\\\\n",
	"A(t) = \\frac{1}{2} \\left( \\frac{\\sin{(t)}^2}{\\cos{t}} + \\frac{\\cos{(t)}^2}{\\cos{t}} \\right) \\left(\\frac{\\cos{(t)}^2}{\\sin{t}} + \\frac{\\sin{(t)}^2}{\\sin{t}} \\right) \\\\\n",
	"A(t) = \\frac{1}{2} \\left( \\frac{\\sin{(t)}^2 + \\cos{(t)}^2}{\\cos{t}} \\right) \\left(\\frac{\\cos{(t)}^2 + \\sin{(t)}^2}{\\sin{t}} \\right) \\\\\n",
	"A(t) = \\frac{1}{2} \\left( \\frac{1}{\\cos{t}} \\right) \\left(\\frac{1}{\\sin{t}} \\right) \\\\\n",
	"A(t) = \\frac{1}{2\\cos{t}\\sin{t}}\n",
	"$$"
	]
	},
	{
	"cell_type": "markdown",
	"metadata": {},
	"source": [
	"---\n",
	"\n",
	"## Parts b - d\n",
	"\n",
	"Think about these geometrically. If $t$ goes to $\\frac{\\pi}{2}$ or to $0$, what kind of shape would the triangle make?\n",
	"\n",
	"As $t$ goes to $\\frac{\\pi}{2}$, it steadily makes the triangle bigger and bigger, until it's basically a rectangle. Therefore, the area is $\\infty$.\n",
	"\n",
	"The same principle applies as $t$ goes to $0$.\n",
	"\n",
	"In every other case, you can just plug in whatever value of $t$ is given to you."
	]
	},
	{
	"cell_type": "markdown",
	"metadata": {},
	"source": [
	"---\n",
	"\n",
	"## Parts e - f\n",
	"\n",
	"### Definition and setup\n",
	"\n",
	"Setup:\n",
	"\n",
	"$$\n",
	"2 = \\frac{1}{2 \\cos{t} \\sin{t}} \\\\\n",
	"\\cos{t} \\sin{t} = \\frac{1}{4}\n",
	"$$\n",
	"\n",
	"We need to use the double angle identity, which states that the following statement is true:\n",
	"\n",
	"$$\n",
	"\\sin{2t} = 2\\cos{t}\\sin{t}\n",
	"$$\n",
	"\n",
	"This is basically a formula that I guess we were just sort of \"expected\" to know, similar to the quadratic formula or the Pythagorean theorem. As far as I know, we were never taught this, and there's probably alternate ways to solve this problem, but it did appear to be the simplest method, so I just ran with it.\n",
	"\n",
	"### Application\n",
	"\n",
	"If we rearrange our setup, we can exploit the double angle identity:\n",
	"\n",
	"$$\n",
	"\\cos{t_0} \\sin{t_0} = \\frac{1}{4} \\\\\n",
	"2\\cos{(t_0)}\\sin{(t_0)} = \\frac{1}{2} \\\\\n",
	"\\sin{(2t_0)} = \\frac{1}{2} \\\\\n",
	"2t_0 = \\arcsin{\\frac{1}{2}} \\\\\n",
	"2t_0 = \\frac{\\pi}{6} \\\\\n",
	"t_0 = \\frac{\\pi}{12} \\\\\n",
	"$$\n",
	"\n",
	"Now, geometrically, we need to sort of \"rotate\" and \"flip\" the triangle so we get the same area: \n",
	"\n",
	"$$\n",
	"t_1 = \\frac{\\pi}{2} - t_0 \\\\\n",
	"t_1 = \\frac{5\\pi}{12}\n",
	"$$\n",
	"\n",
	"Try drawing it out, and see what the triangle looks like at both the times, and compare the shapes/areas of both."
	]
	},
	{
	"cell_type": "markdown",
	"metadata": {},
	"source": [
	"## Part g\n",
	"\n",
	"Don't overthink this one. How do you find the average? \n",
	"\n",
	"$$\n",
	"\\frac{A(t_1) - A(t_0)}{t_1 - t_0}\n",
	"$$\n",
	"\n",
	"Our function, in this case, is the formula for the area:\n",
	"\n",
	"$$\n",
	"A(t) = \\frac{1}{2 \\cos{t} \\sin{t}}\n",
	"$$\n",
	"\n",
	"## Parts h - i\n",
	"\n",
	"This is basically taking the limit of the area as the delta approaches from the left or right. It's basically the same thing as above, except now we're finding the average rate of change between two times that are only a fraction of a second apart. Basically, taking the derivative.\n",
	"\n",
	"For example, for part h, you would try and find the limit by using the formula:\n",
	"\n",
	"$$\n",
	"\\text{limiting value} = \\lim_{b \\rightarrow 0^+} \\frac{A(\\frac{\\pi}{6} + b) - A(\\frac{\\pi}{6})}{(\\frac{\\pi}{6} + b) - \\frac{\\pi}{6}}\n",
	"$$\n",
	"\n",
	"...and finding the limiting value by plugging in smaller and smaller numbers for $b$.\n",
	"\n",
	"Question i is the same, except that you approach the limiting value from the other side (so instead of adding the delta, you subtract) and use a number other then $\\frac{\\pi}{6}$"
	]
	}
	],
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