menu
Identidades

IDENTIDADES TRIGONOMÉTRICAS

$$ \sin^2{\alpha} + \cos^2{\alpha} = 1 $$ $$ \sec^2{\alpha} - \tan^2{\alpha} = 1 $$ $$ \csc^2{\alpha} - \cot^2{\alpha} = 1 $$ $$ \sin{ \left(\alpha \pm \beta\right) } = \sin{\alpha}\cos{\beta} \pm \sin{\beta}\cos{\alpha}$$ $$ \cos{ \left(\alpha \pm \beta\right) } = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$$ $$ \tan{ \left(\alpha \pm \beta\right) } = \dfrac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}$$ $$ \cos^2{\alpha} = \dfrac{1}{2} + \dfrac{\cos{2\alpha}}{2} $$ $$ \sin^2{\alpha} = \dfrac{1}{2} - \dfrac{\cos{2\alpha}}{2} $$ $$ \tan^2{\alpha} = \dfrac{1-\cos{2\alpha}}{\sin{2\alpha}} = \dfrac{\sin{2\alpha}}{1+\cos{2\alpha}} $$ $$ 2\,\cos{\alpha}\,\sin{\beta} = \left[ \sin{\left(\alpha + \beta\right)} - \sin{\left(\alpha - \beta\right)}\right] $$ $$ 2\,\cos{\alpha}\,\cos{\beta} = \left[ \cos{\left(\alpha + \beta\right)} + \cos{\left(\alpha - \beta\right)} \right] $$ $$ 2\,\sin{\alpha}\,\sin{\beta} = \left[ \cos{\left(\alpha - \beta\right)} - \cos{\left(\alpha + \beta\right)} \right] $$ $$ \sin{\alpha} \pm \sin{\beta} = 2\, \sin{\left(\dfrac{\alpha \pm \beta}{2}\right)}\, \cos{\left(\dfrac{\alpha \mp \beta}{2}\right)} $$ $$ \cos{\alpha} + \cos{\beta} = 2\, \cos{\left(\dfrac{\alpha + \beta}{2}\right)}\, \cos{\left(\dfrac{\alpha - \beta}{2}\right)} $$ $$ \cos{\alpha} - \cos{\beta} = - 2\, \sin{\left(\dfrac{\alpha + \beta}{2}\right)}\, \sin{\left(\dfrac{\alpha - \beta}{2}\right)} $$

LÍMITES

$$ \lim_{x \to 0}{\dfrac{\sin{x}}{x}} = \lim_{x \to 0}{\dfrac{x}{\sin{x}}} = 1$$ $$ \lim_{x \to 0}{\dfrac{\tan{x}}{x}} = \lim_{x \to 0}{\dfrac{x}{\tan{x}}} = 1$$ $$ \lim_{x \to 0}{\dfrac{1-\cos{x}}{x}} = 0$$ $$ \lim_{x \to 0}{\dfrac{\ln{(1+x)}}{x}} = 1$$ $$ \lim_{x \to 0}{(1+x)^{\frac{1}{x}}} = \lim_{x \to \infty}{\left(1+\dfrac{1}{x}\right)^{x}} = e$$

LOGARITMOS

$$ \log{a^n} = n\,\log{a} $$ $$ \log{ab} = \log{a} + \log{b} $$ $$ \log{\left(\dfrac{a}{b}\right)} = \log{a} - \log{b} $$ $$ \log_b{a} = \dfrac{\log_n{a}}{\log_n{b}} $$ $$ \log{1} = 0 $$ $$ \log_a{a} = 1 $$ $$ \log_e{a} = \ln{a} $$ $$ \ln{e} = 1 $$ $$ \log_a{x} = b \quad \rightarrow \quad x = a^b $$ $$ \ln{x} = b \quad \rightarrow \quad x = e^b $$

EXPANSIÓN TAYLOR

Polinomio de Taylor / Polinomio de MacLaurin ($x_o=0$)

$$ P_n(x) = \sum_{i=0}^{n}{\dfrac{(x-x_0)^i}{i!}\,\dfrac{d^if(x_0)}{dx^i}} $$ $$ P_n(x) = f(x_0) + (x-x_0)\,\dfrac{df(x_0)}{dx} + \dfrac{(x-x_0)^2}{2!}\,\dfrac{d^2f(x_0)}{dx^2} + ... \dfrac{(x-x_0)^n}{n!}\,\dfrac{d^nf(x_0)}{dx^n} $$ $$ F(x,y) = f(x,y)_0 + (x-x_0)\, \left. \dfrac{F(x,y)}{dx} \right|_{(x,y)_0} + (y-y_0)\, \left. \dfrac{F(x,y)}{dy} \right|_{(x,y)_0} $$

SERIES DE FOURIER

$$ f(t) = \dfrac{a_0}{2} + \sum^{\infty}_{n=1} a_n\,\cos{\left( \omega n t\right)} + \sum^{\infty}_{n=1} b_n\,\sin{\left( \omega n t\right)}$$ $$ a_0 = \dfrac{2}{T}\,\int^{T/2}_{-T/2}{f(t)\,dt} $$ $$ a_n = \dfrac{2}{T}\,\int^{T/2}_{-T/2}{f(t)\,\cos{\left( \omega n t\right)}\,dt} $$ $$ b_n = \dfrac{2}{T}\,\int^{T/2}_{-T/2}{f(t)\,\sin{\left( \omega n t\right)}\,dt} $$

FORMULARIO - DERIVADAS

$$d(C) = 0$$ $$d(u) = du$$ $$d(u^n) = n\,u^{n-1}\,du$$ $$d(a^u) = a^u\,\ln{a}\,du$$ $$d(u^v) = v\,u^{v-1}\,du+u^v\,\ln{u}\,dv$$ $$ d(e^{u}) = e^{u}\,du $$ $$d(uv) = u\,dv+v\,du$$ $$ d \left( \dfrac{u}{v} \right) = \dfrac{v\,du-u\,dv}{v^2} $$ $$ d(\sqrt{u}) = \dfrac{du}{2\,\sqrt{u}} $$ $$ d(\log_{a}{u}) = \dfrac{du}{u\,\ln{a}} $$ $$ d(\ln{u}) = \dfrac{du}{u} $$
$$d(\sin{u}) = \cos{u}\,du$$ $$d(\cos{u}) = -\sin{u}\,du$$ $$d(\tan{u}) = \sec^2{u}\,du$$ $$d(\csc{u}) = -\csc{u}\,\cot{u}\,du$$ $$d(\sec{u}) = \sec{u}\,\tan{u}\,du$$ $$d(\cot{u}) = -\csc^2{u}\,du$$ $$d(\arcsin{u}) = \dfrac{du}{\sqrt{1-u^2}} $$ $$d(\arccos{u}) = -\dfrac{du}{\sqrt{1-u^2}}$$ $$d(\arctan{u}) = \dfrac{du}{1+u^2}$$ $$d(\mathrm{arcsc}\,{u}) = -\dfrac{du}{u\,\sqrt{u^2-1}}$$ $$d(\mathrm{arsec}\,{u}) = \dfrac{du}{u\,\sqrt{u^2-1}}$$ $$d(\mathrm{arcot}\,{u}) = -\dfrac{du}{1+u^2}$$
$$d(\sinh{u}) = \cosh{u}\,du$$ $$d(\cosh{u}) = -\sinh{u}\,du$$ $$d(\tanh{u}) = \mathrm{sech}^2{u}\,du$$ $$d(\mathrm{csch}\,{u}) = -\mathrm{csch}\,{u}\,\coth{u}\,du$$ $$d(\mathrm{sech}\,{u}) = -\mathrm{sech}\,{u}\,\tanh{u}\,du$$ $$d(\coth{u}) = -\mathrm{csch}^2{u}\,du$$ $$d(\mathrm{arsinh}\,{u}) = \dfrac{du}{\sqrt{1+u^2}} $$ $$d(\mathrm{arcosh}\,{u}) = -\dfrac{du}{\sqrt{u^2-1}}$$ $$d(\mathrm{artanh}\,{u}) = \dfrac{du}{1-u^2}$$ $$d(\mathrm{arcsch}\,{u}) = -\dfrac{du}{u\,\sqrt{1+u^2}}$$ $$d(\mathrm{arsech}\,{u}) = -\dfrac{du}{u\,\sqrt{1-u^2}}$$ $$d(\mathrm{arcoth}\,{u}) = \dfrac{du}{1-u^2}$$

FORMULARIO - INTEGRALES

$$\int{du} = u+C$$ $$\int{\dfrac{du}{u}} = \ln{u}+C$$ $$\int{e^u \,du} = e^u+C$$ $$\int{u^n \,du} = \dfrac{u^{n+1}}{n+1}+C$$ $$\int{a^u \,du} = \dfrac{a^u}{\ln{a}}+C$$ $$\int{\ln{u}\,du} = u\,\ln{u}-u+C$$ $$\int{u\,e^u\,du} = e^u\,(u-1)+C$$
$$\int{\sin{u}\,du} = -\cos{u}+C$$ $$\int{\cos{u}\,du} = \sin{u}+C$$ $$\int{\tan{u}\,du} = -\ln{\left| \cos{u} \right|}+C = \ln{\left| \sec{u} \right|}+C$$ $$\int{\csc{u}\,du} = -\ln{\left| \csc{u}+\cot{u} \right|}+C$$ $$\int{\sec{u}\,du} = \ln{\left| \sec{u}+\tan{u} \right|}+C$$ $$\int{\cot{u}\,du} = \ln{\left| \sin{u} \right|}+C$$
$$\int{\csc^2{u}\,du} = -\cot{u}+C$$ $$\int{\sec^2{u}\,du} = \tan{u}+C$$ $$\int{\tan^2{u}\,du} = \tan{u}-u+C$$
$$\int{\sec{u}\,\tan{u}\,du} = \sec{u}+C$$ $$\int{\csc{u}\,\cot{u}\,du} = -\csc{u}+C$$
$$\int{\dfrac{du}{a^2+u^2}} = \dfrac{1}{a}\,\arctan{\dfrac{u}{a}}+C$$ $$\int{\dfrac{du}{\sqrt{a^2-u^2}}} = \arcsin{\dfrac{u}{a}}+C$$ $$\int{\dfrac{du}{u\,\sqrt{u^2-a^2}}} = \dfrac{1}{a}\,\mathrm{arsec}{\dfrac{|u|}{a}}+C$$ $$\int{\sqrt{a^2-u^2}\,du} = \dfrac{1}{2}\,u\,\sqrt{a^2-u^2}+\dfrac{1}{2}\,a^2\,\arcsin{\dfrac{u}{a}}+C$$ $$\int{\sqrt{u^2 \pm a^2}\,du} = \dfrac{1}{2}\,u\,\sqrt{u^2 \pm a^2}+\dfrac{1}{2}\,a^2\,\ln{ \left| u+\sqrt{u^2 \pm a^2} \right|}+C$$

FORMULARIO - ECUACIONES DIFERENCIALES

ECUACIONES DIFERENCIALES LINEALES

$$y'+p(x)\,y=q(x)$$ $$u=e^{\int p(x)\,dx}$$ $$y\,u=\int q\,u\,dx$$

ECUACIONES DIFERENCIALES BERNOULLI

$$y' + P(x)\,y = Q(x)\,y_n \qquad \rightarrow \qquad u = y^{1-n}$$

Se lo lleva a la forma de la ecuacion diferencial lienal de primer orden

$$x y' + y = 3x\sqrt{y}$$ $$u = y^{1-1/2} = y^{1/2}$$ $$u^2 = y \qquad \rightarrow \qquad y' = 2u\,u'$$ $$2xu\,u' + u^2 = 3xu$$ $$u' + \frac{1}{2x}u = \frac{3}{2} \qquad EDL$$

ECUACIONES DIFERENCIALES HOMOGÉNEAS DE ORDEN SUPERIOR

$$a\,y'''+b\,y''+c\,y'+d\,y=0 \qquad \rightarrow \qquad y=e^{r\,x}$$ $$a\,r^3+b\,r^2+c\,r+d=0 $$

Si tiene un raíz compleja es

$$r=a+b\,i$$

Se usa la formula de Euler

$$e^{i\omega}=\cos{\omega}+i\,\sin{\omega}$$

Nos queda la solución genreal de la siguiente forma

$$y=e^{ax}\,(C_1\,\cos{bx}+C_2\,\sin{bx})$$

Ejemplo 1

$$y'''-7\,y'+6\,y=0$$ $$r^3-7\,r+6=0$$ $$(r-1)(r+3)(r-2)=0$$ $$y = C_1\,e^{x}+C_2\,e^{-3x}+C_3\,e^{2x} \qquad \mathrm{Solucion\,Genreal}$$

Ejemplo 2

$$r'''-6\,y'+13\,y=0$$ $$r^2-6\,r+13=0$$ $$r=3 \pm 2\,i$$ $$y=e^{3x}\,(C_1\,\cos{2x}+C_2\,\sin{2x}) \qquad \mathrm{Solucion\,Genreal}$$

Se trabaja con la EDO

$$R' = \alpha_1\,R - \beta_1\,R\,F$$ $$F' = -\alpha_2\,F + \beta_2\,R\,F $$ $$\mathrm{det} (\lambda I-A) = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} - \begin{bmatrix} \alpha_1 & -\beta_1 \\ -\alpha_2 & \beta_2 \end{bmatrix}$$ $$\mathrm{det} (\lambda I-A) = \begin{bmatrix} \lambda-\alpha_1 & \beta_1 \\ \alpha_2 & \lambda-\beta_2 \end{bmatrix}$$ $$ \mathrm{det} (\lambda I-A) = (\lambda-\alpha_1)\,(\lambda-\beta_2)-(\alpha_2)\,(\beta_1)$$ $$ \mathrm{det} (\lambda I-A) = \lambda^2 - (\alpha_1+\beta_2)\,\lambda+\alpha_1\,\beta_2-\alpha_2\,\beta_1$$ $$ \mathrm{det} (\lambda I-A) = 0$$ $$ \lambda = \dfrac{\alpha_1+\beta_2 \pm \sqrt{(\alpha_1+\beta_2)^2-4\,(\alpha_1\,\beta_2-\alpha_2\,\beta_1)}}{2}$$ $$ \lambda_1= 2.6758$$ $$ \lambda_2 = 0.2242$$

Autovectores

$$(\lambda I-A)V_1=0$$ $$\begin{bmatrix} \lambda-\alpha_1 & \beta_1 \\ \alpha_2 & \lambda-\beta_2 \end{bmatrix} \,\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ Para $\lambda_1$ $$\begin{bmatrix} \lambda-\alpha_1 & \beta_1 \\ \alpha_2 & \lambda-\beta_2 \end{bmatrix} \,\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ $$\begin{bmatrix} 0.6758 & 1.2 \\ 1 & 1.7758 \end{bmatrix} \,\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

Se realiza lo respectivos reemplazos.

$$ \left \{ \begin{array}{ll} 0.6758\,v_1+1.2\,v_2=0\\ v_1+1.7758\,v_2=0\\ \end{array}\right.$$ $$ \left \{ \begin{array}{ll} v_1+1.7758\,v_2=0\\ v_1+1.7758\,v_2=0\\ \end{array}\right.$$$$ \left \{ \begin{array}{ll} v_1=-1.7758\,v_2\\ v_1=-1.7758\,v_2\\ \end{array}\right.$$ Si $v_1=1 \qquad \rightarrow \qquad v_2=-0.5631$ $$V_1 = \begin{bmatrix} 1 \\ -0.5631 \end{bmatrix}$$ Para $\lambda_2$ $$\begin{bmatrix} \lambda-\alpha_1 & \beta_1 \\ \alpha_2 & \lambda-\beta_2 \end{bmatrix} \,\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ $$\begin{bmatrix} -1.7758 & 1.2 \\ 1 & -0.6758 \end{bmatrix} \,\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

Se realiza lo respectivos reemplazos.

$$ \left \{ \begin{array}{ll} -1.7758\,v_1+1.2\,v_2=0\\ v_1-0.6758\,v_2=0\\ \end{array}\right.$$ $$ \left \{ \begin{array}{ll} 1.7758\,v_1=1.2\,v_2\\ v_1=0.6758\,v_2\\ \end{array}\right.$$ Si $v_1=1 \qquad \rightarrow \qquad v_2=1.4797$ $$V_2 = \begin{bmatrix} 1 \\ 1.4797 \end{bmatrix}$$

Solución analítica

$$\begin{bmatrix} R \\ F \end{bmatrix}\,(t) = C_1\,\begin{bmatrix} 1 \\ -0.5631 \end{bmatrix}\,e^{2.6758\,t} + C_2\,\begin{bmatrix} 1 \\ 1.4797 \end{bmatrix}\,e^{0.2242\,t}$$

Se reemplaza condiciones iniciales

$$\begin{bmatrix} 200 \\ 400 \end{bmatrix} = C_1\,\begin{bmatrix} 1 \\ -0.5631 \end{bmatrix} + C_2\,\begin{bmatrix} 1 \\ 1.4797 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 1\\ -0.5631 & 1.4797 \end{bmatrix} \,\begin{bmatrix} C_1 \\ C_2 \end{bmatrix}=\begin{bmatrix} 200 \\ 400 \end{bmatrix}$$ $$\begin{bmatrix} C_1 \\ C_2 \end{bmatrix}=\begin{bmatrix} -50.9399 \\ 250.9399 \end{bmatrix}$$

La ecuación analítica queda entonces:

$$\begin{bmatrix} R \\ F \end{bmatrix}\,(t) = -50.9399\,\begin{bmatrix} 1 \\ -0.5631 \end{bmatrix}\,e^{2.6758\,t} + 250.9399\,\begin{bmatrix} 1 \\ 1.4797 \end{bmatrix}\,e^{0.2242\,t}$$

Se deriva la primera ecuación

$$R'' = \alpha_1\,R' - \beta_1\, \left[ R\,F' + F\,R' \right] $$

Reemplazamos la segunda ecuación del sistema

$$R'' = \alpha_1\,R' - \beta_1\, \left[ R\, \left( -\alpha_2\,F + \beta_2\,R\,F \right) + F\,R' \right] $$ $$R'' = \alpha_1\,R' - \beta_1\, \left( - R\,\alpha_2\,F + \beta_2\,R^2\,F + F\,R' \right) $$ $$R'' - \left( \alpha_1-\beta_1\,F \right)\,R' + \beta_1\,R\, \left( - \alpha_2\,F + \beta_2\,R\,F \right) = 0$$

VECTORES

$$ \vec{\mu_A} = \frac{\vec{A}}{A} $$ $$ \vec{A} \cdot \vec{B} = A \, B\, \cos{\theta} $$ $$ \vec{A} \times \vec{B} = A \, B\, \sin{\theta} $$

MOVIMIENTO RECTILINEO UNIFORME (MRU)

Escalar

$$ d = vt $$

Vector

$$ \vec{\Delta_r} = \vec{v}\,t $$

MOVIMIENTO RECTILINEO UNIFORMEMENTE VARIADO (MRUV)

Escalar

$$ v_f = v_o + at $$ $$ v_f^2 = v_o^2 + 2ad $$ $$ d = v_ot + \frac{1}{2}at^2 $$

Vector

$$ \vec{v}_f = \vec{v_o} + \vec{a}t $$ $$ \vec{v}_f^2 = \vec{v_o}^2 + 2\vec{a}\,\vec{\Delta_r} $$ $$ \vec{\Delta_r} = \vec{v_o}\,t + \frac{1}{2}\,\vec{a}\,t^2 $$

MOVIMIENTO PARABÓLICO

Movimiento Horizontal

$$ d_x = v_xt $$

Movimiento Vertical y Caida Libre

$$ v_y = v_{oy} + gt $$ $$ v_y^2 = v_{oy}^2 + 2gh $$ $$ h = v_{oy}t + \frac{1}{2}\,g\,t^2 $$ $$ \vec{a}= (\vec{g} \cdot \vec{\mu_v})\,\vec{\mu_v}$$ $$ \vec{a_c}= \vec{a_T} - \vec{a}$$

MOVIMIENTO CIRCULAR UNIFORMEMENTE VARIADO (MCUV)

$$ w = w_o + \alpha\,t $$ $$ w^2 = w_o^2 + 2\,\alpha\,\Delta\theta $$ $$ \Delta\theta = w_ot + \frac{1}{2}\,\alpha\,t^2 $$

Lineales o Tangenciales

$$ l = \Delta\theta\,r$$ $$ v = w\,r$$ $$ a = \alpha\,r$$

Complementos

$$ a_c = w^2\,r = \frac{v^2}{r}$$ $$ w = 2\,\pi\,f$$ $$ T = \frac{1}{f}$$ $$ \vec{a_T}= \vec{a} + \vec{a_c}$$
Imagen de fondo