Calculus Archive: Questions from November 21, 2022
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Find the Laplace Transformation of \[ \begin{array}{l} y^{\prime \prime}-3 y^{\prime}+2 y=4 e^{-2 x} \\ \text { when } y(0)=1, y^{\prime}(0)=4 \end{array} \] \[ \begin{array}{l} Y(s)=\frac{s^{2}+4 s-62 answers -
2 answers
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\( y "-3 y^{\prime}+2 y=2 e^{-2 x} \) when \( y(0)=6, y^{\prime}(0)=2 \) \( Y(s)=\frac{6 s^{2}-8 s+30}{s^{3}-s^{2}-4 s+4} \) \( Y(s)=\frac{6 s^{2}-4 s+28}{s^{3}-s^{2}-4 s+4} \) \( Y(s)=\frac{6 s^{2}-42 answers -
Sabiendo que (1/(3)^1/2)=pi/6, use la serie de Mclaurin de (x) para aproximar con un error menor que 10^-6.
(10 puntos) Use la serie de Mclaurin de \( e^{x} \) para aproximar \( \int_{0}^{1} e^{-x^{2}} d x \) con un error menor que \( 10^{-6} \) :2 answers -
Use la serie de Mclaurin de e^x para aproximar la integral de 0 a 1 de e^-x^2dx con un error menor que 10^-6.
(10 puntos) Sabiendo que \( \arctan (1 / \sqrt{3})=\pi / 6 \), use la serie de Mclaurin de \( \arctan (x) \) para aproximar \( \pi \) con un error menor que \( 10^{-6} \).2 answers -
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A. \( e^{-s} \frac{1}{(s-3)} \) B. \( e^{-3 s} \frac{1}{(s+3)} \). C. \( e^{-3 s} \frac{1}{(s+2)} \). D. \( e^{-s} \frac{1}{(s+3)} \)2 answers -
nts) \( \mathcal{L}(u(t-\pi)) \sin (t)= \) A. \( e^{-\pi s}\left(\frac{1}{(s-\pi)^{2}+1}\right) \) B. \( e^{-\pi s}\left(\frac{1}{s^{2}+1}\right) \) C. \( e^{-\pi s}\left(\frac{s}{s^{2}+1}\right) \) D2 answers -
13. (10 points) Solve: \( y^{\prime \prime}+7 y^{\prime}+12 y=\delta(t-1) \) where \( y(0)=0, y^{\prime}(0)=0 \)2 answers -
help me
14) \( y=\frac{x^{2}}{2}+2 x+3 ;[-2 \) A) \( \{-2\} \) B) \( \left\{-\frac{3}{2}\right\} \) C) \( \{-1\} \) D) \( \left\{-\frac{1}{2}\right\} \)2 answers -
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15) \( y=2 x^{2}-12 x+20 \) A) \( \left\{\frac{5}{2}\right\} \) B) \( \left\{\frac{7}{2}\right\} \)2 answers -
(2) In each case, find the average value of the function over the corresponding region. (a) \( f(x, y)=4-x-y, R=\{(x, y) \mid 0 \leq x \leq 2,0 \leq y \leq 2\} \). (b) \( f(x, y)=x y \sin \left(x^{2}\2 answers -
Find the derivatix of ecen of the following \[ \Rightarrow y=\tan (\sin (x)) \quad b \neq y=7 \sec (\sqrt{x}) \] c) \( y=e^{x^{2}} \ln (x \sqrt{x}) \) d) \( y=5^{\tan (x)}\left(x^{3} \sqrt{x}\right) \2 answers -
Question 5
5. Differentiate Implicitly (4 marks each) a. \( x^{4}(x+y)=y^{2}(3 x-y) \) b. \( \tan (x-y)=\frac{y}{1+x^{2}} \)2 answers -
Sabiendo que arctan(1/(3)^1/2) = pi/6, use la serie de Mclaurin de arctan(x) para aproximar pi con un error menor que 10^-6
(10 puntos) Sabiendo que \( \arctan (1 / \sqrt{3})=\pi / 6 \), use la serie de Mclaurin de \( \arctan (x) \) para aproximar \( \pi \) con un error menor que \( 10^{-6} \).2 answers -
2 answers
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Question 12 1 pts \( L\{t \cdot \sin 3 t\}= \) \( \frac{6 s}{\left(s^{2}+9\right)^{2}} \) \( \frac{6 s}{\left(s^{2}-9\right)^{2}} \) \( \frac{-6 s}{\left(s^{2}+9\right)^{2}} \) \( \frac{-6 s}{\left(s^2 answers -
Use implicit differentation to find dy/dx a) x^2 + y^2 - x^2 y + xy^2 - 2 =0 b) x^2 sin y - y^2 sin x = 0
Use implicit Differentation to find \( d y / d x \) a) \( x^{2}+y^{2}-x^{2} y+x y^{2}-2=0 \) b) \( x^{2} \sin y-y^{2} \sin x=0 \)2 answers -
2 answers
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2 answers
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2 answers
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If \( z=x \sin y, x=\sin t, y=t \), what is \( d f / d t \) ? \[ \begin{array}{l}\sin y \cos x-\sin x \cos y \\ 2 \sin y \cos y \\ \sin y \cos x+\sin x \cos y \\ 2 \sin x \cos x\end{array} \]2 answers -
2 answers
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questions 24,26,28,30,32, and 34 please
Find derivatives of the functions defined as follows. 1. \( y=e^{A x} 2 \). 3. \( y=-8 e^{3} 4 \). \[ y=e^{-2 x} \] 5. \( y=-16 e^{2 x+1} 6 \). \( y=1.2 e^{5 x} \) 7. \( y=e^{2} \) 8. \( y \) \( y=-42 answers -
\[ f(x, y)=x e^{y}-\ln (x), P_{0}(2,0) \] A) \( \left(\frac{4}{\sqrt{17}}\right) i+\left(\frac{1}{\sqrt{17}}\right) j \) B) \( \left(-\frac{1}{\sqrt{17}}\right) i \) C) \( \left(-\frac{1}{\sqrt{17}}\r2 answers -
Find \( f_{x}(x, y) \) \[ f(x, y)=\frac{5 x}{y}-\frac{y}{5 x} \] \[ f_{x}(x, y)=-\frac{5}{y^{2}}-\frac{5}{y} \] \[ \begin{array}{l} f_{x}(x, y)=\frac{5}{y}+\frac{y}{5 x^{2}} \\ f_{x}(x, y)=\frac{5}{y}2 answers -
\( E=\left\{(x, y, z) \mid 0 \leqslant y \leqslant 1,0 \leqslant z \leqslant y^{2}, 1 \leqslant x \leqslant z+1\right\} \)2 answers -
If \( \frac{d y}{d x}=\frac{-2}{\sqrt{1-4 x^{2}}} \), then \( y= \) Find \( \frac{d y}{d x} \) given \( \sin (x y)=x \) If \( y=\ln \left(\frac{e^{x}}{e^{x}-1}\right) \) then \( y^{\prime}= \)2 answers -
Given \( f(x, y)=4 x^{5}+3 x^{2} y^{4}-y^{2} \) \[ f_{x}(x, y)= \] \[ f_{y}(x, y)= \] \[ f_{x x}(x, y)= \] \[ f_{x y}(x, y)= \] Question Help: \( D \) Video2 answers -
In problems 11-20, find the directional derivative of the given function at the point indicated in the indicated direction. do exercises 15 and 17
15. \( f(x, y)=(x y+1)^{2} ; \quad(3,2) \), en la dirección de \( (5,3) \) 16. \( f(x, y)=x^{2} \tan y ; \quad\left(\frac{1}{2}, \pi / 3\right) \), en la dirección del eje \( x \) negativo. 17. \( F2 answers -
In problems 23-26, find a vector that produces the direction in which the given function increases faster at the indicated point. Find the maximum rate. do 23 and 25 hacer el ejercicio #25 solam
En los problemas \( 23-26 \), encuentre un vector que produzca la dirección en la cual la función dada aumenta más rápidamente en el punto indicado. Encuentre la tasa máxima. 23. \( f(x, y)=e^{22 answers -
2. Find the exact value of the following expressions. A. \( \sin \left(75^{\circ}\right)=\sin \left(45^{\circ}+30^{\circ}\right) \) B. \( \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)+\cos2 answers -
1. Find the exact values. A. \( 2 \sin \left(22.5^{\circ}\right) \cos \left(22.5^{\circ}\right) \) B. \( \cos ^{2}\left(75^{\circ}\right)-\sin ^{2}\left(75^{\circ}\right) \) C. \( 1-2 \sin ^{2}\left(\2 answers -
2 answers
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10. Find \( \frac{d y}{d x} \). [A] \( \frac{y}{1-e^{2 y}} \) [B] \( \frac{y+1}{1-2 e^{2 y}} \) [C] \( \frac{1}{1-2 e^{2 y}} \) [D] \( \frac{y-1}{1+2 y} \) [E] \( \frac{1}{1-e^{y}} \)2 answers -
Find \( d y / d x \) by implicit differentiation. \[ 2+3 x=\sin \left(x y^{2}\right) \] Find \( y^{\prime} \) and \( y^{\prime \prime} \). \[ \begin{array}{c} y=e^{7 e^{x}} \\ y^{\prime}=(\ln (7)+1)2 answers -
La fuerza \( F \) (Newton) que actủa sobre un objeto está dada por \[ F=11 \frac{d v}{d t}+2 v+4 \] donde \( v \) es la velocidad \( (\mathrm{m} / \mathrm{seg}) \) y \( t \) es el tiempo \( (\mathr2 answers