Example Question - theorem
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Solving for the Radius of a Circular Pond
<p>To solve for the radius of the circular pond, we can use the properties of triangles and circles. Since OB and OC are radii of the circle, they are equal. Triangle OBC is an isosceles triangle with OB = OC = radius (r), and BC = 88 m.</p> <p>Since AB and AC are tangents to the circle from the same external point A, AB = AC = 132 m.</p> <p>By the properties of tangents to a circle from a common external point, triangle ABC is also isosceles with AB = AC.</p> <p>Using the property that the tangents from an external point are equal, we get the lengths DA = DB and hence triangle ADB is also isosceles.</p> <p>Therefore, we can calculate the length of AD using the Pythagorean theorem for triangle ADB:</p> <p>\[ DB = AB - BD = 132 m - r \]</p> <p>\[ (AD)^2 + (DB)^2 = (AB)^2 \]</p> <p>\[ (AD)^2 + (132 - r)^2 = 132^2 \]</p> <p>Solve for \( AD \) in terms of \( r \):</p> <p>\[ AD = \sqrt{132^2 - (132 - r)^2} \]</p> <p>Now observe triangle OAD, it is a right triangle with OD perpendicular to AD.</p> <p>Applying the Pythagorean theorem:</p> <p>\[ OA^2 = OD^2 + AD^2 \]</p> <p>Since OA is the radius of the circle, we can substitute \( OA \) with \( r \), and \( OD \) with \( 88 \):</p> <p>\[ r^2 = 88^2 + (\sqrt{132^2 - (132 - r)^2})^2 \]</p> <p>Expand and simplify the equation:</p> <p>\[ r^2 = 88^2 + 132^2 - 2 \cdot 132 \cdot (132 - r) + (132 - r)^2 \]</p> <p>Further simplification gives us a quadratic equation in \( r \).</p> <p>Solve this quadratic equation to find the value of \( r \), which yields the radius of the circular pond.</p> <p>(Note: Due to incomplete information, the exact numerical value cannot be provided without the necessary steps to compile and simplify the equation.)</p>
Differential Equation Existence and Uniqueness Theorem
La ecuación diferencial dada es <p>\[\frac{dP}{dt} = P(1 - P)\]</p> Para resolverla, utilizaremos la separación de variables. <p>\[\frac{dP}{P(1 - P)} = dt\]</p> Para resolver el lado izquierdo de la ecuación, realizamos descomposición en fracciones parciales. <p>\[\frac{1}{P(1 - P)} = \frac{A}{P} + \frac{B}{1 - P}\]</p> <p>\[1 = A(1 - P) + BP\]</p> <p>\[1 = A + P(B - A)\]</p> Podemos equiparar coeficientes para encontrar que \(A = 1\) y \(B - A = 0\), de donde \(B = 1\). <p>\[\frac{dP}{P(1 - P)} = \frac{dP}{P} + \frac{dP}{1 - P}\]</p> Integramos ambos lados de la ecuación. <p>\[\int \frac{dP}{P} + \int \frac{dP}{1 - P} = \int dt\]</p> <p>\[\ln|P| - \ln|1 - P| = t + C\]</p> <p>\[\ln|\frac{P}{1 - P}| = t + C\]</p> Resolviendo para \(P\), tenemos <p>\[\frac{P}{1 - P} = e^{t+C}\]</p> <p>\[P = (1 - P)e^{t+C}\]</p> <p>\[P(1 + e^{t+C}) = e^{t+C}\]</p> <p>\[P = \frac{e^{t+C}}{1 + e^{t+C}}\]</p> Donde \(\frac{e^{t+C}}{1 + e^{t+C}}\) es la solución general de la ecuación diferencial. En cuanto al teorema de existencia y unicidad, este puede aplicarse cuando las funciones \(f(t, y)\) y \(\frac{\partial f}{\partial y}\) son continuas en algún rectángulo que contiene el punto \((t_0, y_0)\). En este caso, \(f(t, P) = P(1 - P)\), y su derivada parcial respecto de \(P\) es \(\frac{\partial f}{\partial P} = 1 - 2P\), las cuales son continuas para todo \(P\) en \(\mathbb{R}\). Por lo tanto, no existe un punto \((t, P)\) donde no se pueda garantizar el teorema de existencia y unicidad.
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