Calculus Archive: Questions from July 20, 2022
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(3) \( \sum_{n=0}^{\infty} a_{n} x^{n} \quad \begin{array}{l}a_{0}=7 \\ a_{1}=5\end{array} \quad y^{\prime \prime}+y=0 \) fard the \( { }^{{ }^{4} 4}= \) ?1 answer -
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\( 2 . \) Compute derivatives \( d y / d x \). (a) \( y=\frac{3 x^{2}-5}{2 x+3} \) (b) \( y=\sqrt{1+\sqrt{x}} \) (c) \( x^{2} y-y^{2 / 3}-3=0 \)3 answers -
3. Compute total differentials \( d y \). (a) \( y=\left(x_{1}-1\right) /\left(x_{2}+1\right) \) (b) \( y=x_{1} x_{2}^{2}+\frac{x_{1}^{2}-x_{2}^{2}}{x_{1}+1} \)1 answer -
Find y' and y" y' = y" =
Find \( y^{\prime} \) and \( y^{\prime \prime} \) \[ y=\cos (\sin (4 \theta)) \] \[ y^{\prime}= \] \[ y^{\prime \prime}= \]2 answers -
Find the orthogonal trajectories of the family of curves \( y^{6}=k x^{4} \). (A) \( \frac{5}{2} y^{3}+\frac{7}{2} x^{3}=C \) (B) \( 3 y^{3}+4 x^{2}=C \) (C) \( 2 y^{2}+3 x^{2}=C \) (D) \( 2 y^{2}+\fr1 answer -
Calculate the following expression.
(1) \( x^{2} y+x^{2}-2 x y^{2}+5 x^{2} y-3 x y^{2} \) (2) \( (9 x y) \times\left(2 x y^{2}\right) \div 6(y)^{3} \) (3) \( \frac{1}{x+1}+\frac{1}{x-1} \) (4) \( \frac{\sqrt{2}}{2}+\frac{3}{\sqrt{2}} \)1 answer -
12. Solve the initial value problem \[ (x \ln x) y^{\prime}+y=x^{2} e^{x} \quad y(e)=1 \] (a) \( y=\frac{x e^{x}+1-e^{e+1}}{\ln x} \) (b) \( y=\frac{e^{x}(x+1)+1-e^{e}(e+1)}{x} \) (c) \( y=\frac{x e^{1 answer -
pls solve
Find the gradient of the function \( f(x, y)=x^{2} y+y^{3} \) \[ \begin{array}{l|l} A & \nabla f(x, y)=x \hat{i}+\left(x+3 y^{2}\right) \hat{j} \\ B & \nabla f(x, y)=x y \hat{i}+\left(x^{2}+3 y^{2}\ri1 answer -
51,57 only please.
In Problems 35-64 solve the given differential equation by undetermined coefficients. 51. \( y^{\prime \prime}-y=x^{2} e^{x}+5 \) 52. \( y^{\prime \prime}+2 y^{\prime}+y-x^{2} e^{x} \) 53. \( y^{\pri2 answers -
If I = .... where C is the curve with parametrization a(t) = (sin(2t), cos(2t)), with 0<= t <= pi/8. The value of I would be? [Line integrals (Scalar Fields)]
\( 0 \leq t \leq \frac{\pi}{8} \) En referencia al valor de \( I \), es correcto afirmar que: Seleccione una. \( I=\sqrt{2}-1 \) \( I=\frac{1}{\sqrt{2}}-\frac{1}{2} \) \( I=\frac{\pi}{4} \) \( I=1 \)3 answers -
By evaluating, by definition, the line integral I = ...., where C is the segment of the line that goes from (0,0,0) to (0,1,2) you get:
Al evaluar, mediante la definiciรณn, la integral de lรญnea \( I=\int_{C}(y+z) d x+x d y+(x+y) d z \), donde \( C \) es el segmento de recta que va desde \( (0,0,0) \) a \( (0,1,2) \) se obtiene: Selec1 answer -
Consider I = ...., where C is parameterized by a(t) = (2t -6, 4-2t), where t belongs to [0,1] the value of I is equal to:
Considere \( I=\int_{\mathcal{C}}(x+y) d x+(x-y) d y \), donde \( \mathcal{C} \) estรก parametrizada por \( \alpha(t)=(2 t-6,4-2 t) \), donde \( t \in[0,1] \). El valor de \( I \) es igual a: Seleccio1 answer -
Let ๐ผ (๐ผ is in the image below), where C is the curve with parameterization ๐ผ(๐ก)=(cos(3๐ก),sin(3๐ก)), with 0 โค ๐ก โค ๐/12. In reference to the value of ๐ผ, it is correct to
Sea \( I=\int_{\mathcal{C}} \frac{y}{x^{3}} d s \), donde \( \mathcal{C} \) es la curva con parametrizaciรณn \( \alpha(t)=(\cos (3 t), \sin (3 t)) \), con \( 0 \leq t \leq \frac{\pi}{12} \). En refere1 answer -
Let F(x,y,z) = ..., where n is a positive whole number. We can assure that: a) F is not conservative for any value of n b) F is conservative for any value of n c) none of the options d) F is conservat
Sea \( \vec{F}(x, y, z)=\left(9 e^{\cos (n x+z)}, y^{n} z^{n+1}, y^{n+1} z^{n}+n e^{\cos (n x+z)}\right) \), donde \( n \) es un entero positivo. Se puede afirmar lo siguiente: Seleccione una: \( \vec1 answer -
If C is the circle x^2 + y^2 = 9 clockwise. Use Green's theorem to calculate: .... the value of I is:
Sea \( \mathcal{C} \) el circulo \( x^{2}+y^{2}=9 \) recorrido de forma horaria. Al usar el teorema de Green para calcular : \[ I=\oint_{C}\left(96 y-\cos \left(x^{2}\right)\right) d x+\left(100 x+\sq1 answer -
Let S be a surface that is parameterized by the function r(u, v) = ...., where 0 <= u <= 2; and 0 <= v <= 1. If A representa the area of the surface S then we can ensure:
Sea \( S \) una superficie la cual posee parametrizaciรณn dada por la funciรณn \( r(u, v)=\left(2 u,-\frac{v}{2}, \frac{v}{2}\right) \), donde \( 0 \leq u \leq 2 ; 0 \leq v \leq 1 \) Si \( A \) repres1 answer -
The surface area of the hemisphere z = ... that is left between the planes z= 0 and z= 2 is equal to:
El รกrea superficial de un hemisferio \( z=\sqrt{16-x^{2}-y^{2}} \) que queda entre los planos \( z=0 \) y \( z=2 \) es igual a: Seleccione una: \( 16 \pi \) \( 16 \pi / 3 \) Ninguna de las otras opci1 answer -
The value of the integral (Integral in the image below), where ๐:โ3โโ, is defined by ๐(๐ฅ,๐ฆ,๐ง)= ๐ฅ2/17 + ๐ฆ2, and the curve ๐ is given by the intersection of the surfaces ๐
El valor de la integral \( \int_{\zeta} f d s \), donde \( f: \mathbb{R}^{3} \rightarrow \mathbb{R} \), estรก definida por \( f(x, y, z)=\frac{x^{2}}{17}+y^{2} \), y la curva \( \zeta \) estรก dada po1 answer -
Let F(x,y,z) = ... be a vectorial field and let G be the closed surface oriented with a normal vector pointing outside like the figure shown below: By using the Divergence theorem, we have that I = .
Sea \( \vec{F}(x, y, z)=\left(e^{y^{2}}, \cos \left(z^{9}+7\right), z+1\right) \) un campo vectorial, y sea \( G \) la superfice cerrada y orientada con vector normal apuntando hacia afuera que se mue1 answer -
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Let C be the curve resulting from the intersection of the sphere x^2 + y^2 + z^2 = 1 with the cone z= ..., oriented in a way that seen from above looks counterclockwise. Using Stokes theorem the calc
Sea \( \mathcal{C} \) la curva que resulta de la intersecciรณn de la esfera \( x^{2}+y^{2}+z^{2}=1 \) con el cono \( z=\sqrt{x^{2}+y^{2}} \), orientada de tal forma que cuando se ve desde arriba es re1 answer -
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Consider ๐ผ (๐ผ is given in the image below), where C is parametrized by ๐ผ(๐ก)=(๐ก+1,โ๐กโ1), where ๐กโ[0,1]. The value of ๐ผ is equal to: Select one: - 20 - none of the othe
Considere \( I=\int_{C}(x+y) d x+(x-y) d y \), donde \( C \) estรก parametrizada por \( \alpha(t)=(t+1,-t-1) \), donde \( t \in[0,1] \). El valor de \( I \) es igual a: Seleccione una: 20 Ninguna de l1 answer -
Consider ๐ผ=โซ๎ฏ(๐ฅ+๐ฆ)๐๐ฅ+(๐ฅโ๐ฆ)๐๐ฆ, where ๎ฏ is parametrized by ๐ผ(๐ก)=(4๐กโ4,2โ2๐ก), where ๐กโ[0,1]. The value of ๐ผ is equal to:
Considere \( I=\int_{C}(x+y) d x+(x-y) d y \), donde \( \mathcal{C} \) estรก parametrizada por \( \alpha(t)=(4 t-4,2-2 t) \), donde \( t \in[0,1] \). El valor de \( I \) es igual a: Seleccione una: 21 answer -
Let ๐น(๐ฅ,๐ฆ,๐ง)=(๐ฆ2๐ง3)๐โ +(2๐ฅ๐ฆ๐ง3)๐โ +(๐ผโพโพโ๐ฅ๐ฆ2๐ง2)๐โ be a vector field. The value, or values, of ๐ผ for ๐น(๐ฅ,๐ฆ,๐ง) to be a conservat
Sea \( F(x, y, z)=\left(y^{2} z^{3}\right) \vec{i}+\left(2 x y z^{3}\right) \vec{j}+\left(\sqrt{\alpha} x y^{2} z^{2}\right) \vec{k} \) un campo vectorial. El valor, o valores, de \( \alpha \) para qu1 answer -
The value of ๐ผ=โซ๐๐ โ ๐๐ซโ where ๐ :=(๐ฆ+6๐ฅ5,๐ฅ+8๐ฆ7,sin๐ง), and the curve is parameterized by ๐(๐ก):=(1+๐ก,2+sin(17๐๐ก),9 +cos(4๐๐ก2)), with 0โค๐กโค
Sea \( F(x, y, z)=\left(y^{2} z^{3}\right) \vec{i}+\left(2 x y z^{3}\right) \vec{j}+\left(\sqrt{\alpha} x y^{2} z^{2}\right) \vec{k} \) un campo vectorial. El valor, o valores, de \( \alpha \) para qu1 answer -
Let ๐นโ (๐ฅ,๐ฆ,๐ง)=(9๐cos(๐๐ฅ+๐ง),๐ฆ๐๐ง๐+1,๐ฆ๐+1๐ง๐+๐๐cos(๐๐ฅ+๐ง)), where ๐ is a constant. The following can be stated: Select one: - none o
Sea \( \vec{F}(x, y, z)=\left(9 e^{\cos (n x+z)}, y^{n} z^{n+1}, y^{n+1} z^{n}+n e^{\cos (n x+z)}\right) \), donde \( n \) es una constante. Se puede afirmar lo siguiente: Seleccione una: Ninguna de l1 answer -
Let ๎ฏ be the circle ๐ฅ2+๐ฆ2=9 traversed counterclockwise. By using Green's theorem to calculate: ๐ผ=โฎ๎ฏ(6๐ฆโsin(๐ฅ4))๐๐ฅ+(10๐ฅ+๐ฆ2022+๐ฆ+25โพโพโพโพโพโพโพโพโพโพโ
Sea \( C \) el cรญrculo \( x^{2}+y^{2}=9 \) recorrido de forma antihoraria. Al usar el teorema de Green para calcular : \[ I=\oint_{C}\left(6 y-\sin \left(x^{4}\right)\right) d x+\left(10 x+\sqrt{y^{23 answers -
Let ๐ be a surface which has parameterization given by the function ๐(๐ข,๐ฃ)=(2๐ข,โ๐ฃ2,๐ฃ2), where 0โค๐ขโค2; 0โค๐ฃโค1. If ๐ด represents the area of โโthe surface ๐
Sea \( S \) una superficie la cual posee parametrizaciรณn dada por la funciรณn \( r(u, v)=\left(2 u,-\frac{v}{2}, \frac{v}{2}\right) \), donde \( 0 \leq u \leq 2 ; 0 \leq v \leq 1 \) Si \( A \) repres1 answer -
The surface area of โโa hemisphere ๐ง=sqrt(16โ๐ฅ2โ๐ฆ2) lying between the planes ๐ง=0 and ๐ง=2 is equal to:
El รกrea superficial de un hemisferio \( z=\sqrt{16-x^{2}-y^{2}} \) que queda entre los planos \( z=0 \) y \( z=2 \) es igual a: Seleccione una: \( 16 \pi / 3 \) \( 16 \pi \) \( 12 \pi \) \( 8 \pi \)3 answers -
The value of ๐ผ=โซ๐๐ โ ๐๐ซโ where ๐ :=(๐ฅ3+๐ฆ,๐ฆ3+๐ฅ,2๐ง+๐ง5), and ๐(๐ก):=(๐ก,2+sin(15๐๐ก),cos(2๐๐ก2)), with 0 โค ๐ก โค 1 is equal to: Hint: think c
El valor de \( I=\int_{\mathbf{C}} \mathbf{F} \cdot d \overrightarrow{\mathbf{r}} \) donde \( \mathbf{F}:=\left(x^{3}+y, y^{3}+x, 2 z+z^{5}\right) \), y \( \mathbf{C}(t):=\left(t, 2+\sin (15 \pi t), \3 answers -
The value of the integral โซฮถ fds, where f:R3โR, is defined by f(x,y,z)=x^2/17+y2, and the curve ฮถ is given by the intersection of the surfaces x^2+y^2+z^2=1 ; 4y+z=0, is equal to: Hint: you must
El valor de la integral \( \int_{\zeta} f d s \), donde \( f: \mathbb{R}^{3} \rightarrow \mathbb{R} \), estรก definida por \( f(x, y, z)=\frac{x^{2}}{17}+y^{2} \), y la curva \( \zeta \) estรก dada po1 answer -
The value of the surface integral ๐ผ:=โฌ๐1/sqrt(18๐งโ16๐ฆ+10) ๐๐, where ๐ is defined by ๐ง=๐ฅ2+2๐ฆ,0โค๐ฅโคsqrt(2) ,0โค๐ฆโค1, is equal to:
El valor de la integral de superficie \( I:=\iint_{S} \frac{1}{\sqrt{8 z-16 y+10}} d S \), donde \( S \) estรก definida por \( z=x^{2}+2 y, 0 \leq x \leq \sqrt{2}, 0 \leq y \leq 1 \), es igual a: Sele1 answer -
Let ๐ be the area given by the parameterization ๐โ (๐ข,๐ฃ)=(๐ข,๐ฃ,35(๐ข5/3+๐ฃ5/3)), where 0โค๐ขโค1;0โค๐ฃโค1. The value of ๐ผ=โฌ๐1/ sqrt(1+๐ฅ4/3+๐ฆ4/3) look image
Sea \( S \) la superficie dada por la parametrizaciรณn \( \vec{r}(u, v)=\left(u, v, \frac{3}{5}\left(u^{5 / 3}+v^{5 / 3}\right)\right) \), donde \( 0 \leq u \leq 1 ; 0 \leq v \leq 1 \) El valor de \(1 answer -
Let C be the circle ๐ฅ2+๐ฆ2=9 traveled clockwise. By using Green's theorem to calculate: ๐ผ (๐ผ is given in the image below), the following value is obtained: Select one: - โ36๐ - 3๐
Sea \( C \) el cรญrculo \( x^{2}+y^{2}=9 \) recorrido de forma horaria. Al usar el teorema de Green para calcular : \[ I=\oint_{C}\left(y-\cos (x)-e^{x^{2}}\right) d x+\left(5 x+\sqrt{y^{2022}+y+25}\r1 answer -
Let S be the part of the cylinder x^2+y^2=a^2 that is between the planes z=โ1;z=1, and which includes the caps: If Fโ (x,y,z)=(x,โy,z^4) is a vector field, then the value of โฌSFโ โ dSโ u
Sea \( S \) la parte del cilindro \( x^{2}+y^{2}=a^{2} \) que se encuentra entre los planos \( z=-1 ; z=1 \), y la cual incluye las tapas: Si \( \vec{F}(x, y, z)=\left(x,-y, z^{4}\right) \) es un camp1 answer -
(\#3) [4 pts.] If \( f(x, y, z)=\ln (x+y+z) \), then evaluate \( f_{x y z}(1,1,1) \). Show all algebra!1 answer -
Let ๐นโ (๐ฅ,๐ฆ,๐ง)=(๐๐ฆ2,cos(๐ง9+7),๐ง+1) be a vector field, and let ๐บ be the closed and oriented surface with normal vector pointing out shown in the following figure: Using the
Sea \( \vec{F}(x, y, z)=\left(e^{y^{2}}, \cos \left(z^{9}+7\right), z+1\right) \) un campo vectorial, y sea \( G \) la superfice cerrada y orientada con vector normal apuntando hacia afuera que se mue1 answer -
Let ๎ฏ be the curve resulting from the intersection of the sphere ๐ฅ2+๐ฆ2+๐ง2=1 with the cone ๐ง=๐ฅ2+๐ฆ2โพโพโพโพโพโพโพโ, oriented in such a way that when seen from above it is tr
Sea \( C \) la curva que resulta de la intersecciรณn de la esfera \( x^{2}+y^{2}+z^{2}=1 \) con el cono \( z=\sqrt{x^{2}+y^{2}} \), orientada de tal forma que cuando se ve desde arriba es recorrida de1 answer -
The surface area of โโa hemisphere ๐ง= (16โ๐ฅ2โ๐ฆ2)1/2 lying between the planes ๐ง=0 and ๐ง=2 is equal to: Select one: - None of the other options - 16๐ - 8๐ - 12๐ - 16๐
El รกrea superficial de un hemisferio \( z=\sqrt{16-x^{2}-y^{2}} \) que queda entre los planos \( z=0 \) y \( z=2 \) es igual a: Seleccione una: NInguna de las otras opciones \( 16 \pi \) \( 8 \pi \)1 answer -
Let S be the surface given by the parameterization rโ (u,v)=(u,v,3/5(u^5/3+v^5/3)), where 0โคuโค1;0โคvโค1. The value of I=โฌS 1/1+x^4/3+y^4/3โโโโโโโโโโโโโdS is equal
Sea \( S \) la superficie dada por la parametrizaciรณn \( \vec{r}(u, v)=\left(u, v, \frac{3}{5}\left(u^{5 / 3}+v^{5 / 3}\right)\right) \), donde \( 0 \leq u \leq 1 ; 0 \leq v \leq 1 \) El valor de \(1 answer -
Let Fโ (x,y,z)=(ycosx,x+sinx,cosz) and let C be the curve that is the intersection of the surfaces: x+y+z=3;(xโ1)^2+(yโ 1)^2=1, and oriented in such a way that when viewed from above it is trave
Sea \( \vec{F}(x, y, z)=(y \cos x, x+\sin x, \cos z) \) y sea \( C \) la curva que es la intersecciรณn de las superficies: \( x+y+z=3 ;(x-1)^{2}+(y-1)^{2}=1 \), y orientada de tal forma que cuando se1 answer -
The value of the surface integral ๐ผ (๐ผ is given in the image below), where ๐ is defined by ๐ง=๐ฅ2+2๐ฆ, 0โค๐ฅโค1 , 0โค๐ฆโค1, is equal to: Select one: - โ2/2 - โ2 - 1 - none
El valor de la integral de superficie \( I:=\iint_{S} \frac{1}{\sqrt{8 z-16 y+10}} d S \) donde \( S \) estรก definida por \( z=x^{2}+2 y, 0 \leq x \leq 1,0 \leq y \leq 1 \), es igual a: Seleccione un1 answer -
The value of ๐ผ=โซ๐๐ โ ๐๐ซโ where ๐ :=(๐ฆ+6๐ฅ5,๐ฅ+8๐ฆ7,sin๐ง), and the curve is parameterized by ๐(๐ก):=(1+๐ก,2+sin(17๐๐ก),9 +cos(4๐๐ก2)), with 0โค๐กโค
\begin{tabular}{l|l} \begin{tabular}{l|l} Pregunta 6 \\ Sin responder aรบn \end{tabular} & El valor de \( I=\int_{\mathbf{C}} \mathbf{F} \cdot d \overrightarrow{\mathbf{r}} \) donde \( \mathbf{F}:=\le3 answers -
Let ๐ be the surface given by the parameterization ๐โ (๐ข,๐ฃ)=(๐ข,๐ฃ,2/3(๐ข3/2+๐ฃ3/2)), where 0โค๐ขโค1;0โค๐ฃโค1. The value of ๐ผ (๐ผ is given in the image below) is eq
Sea \( S \) la superficie dada por la parametrizaciรณn \( \vec{r}(u, v)=\left(u, v, \frac{2}{3}\left(u^{3 / 2}+v^{3 / 2}\right)\right) \), donde \( 0 \leq u \leq 1 ; 0 \leq v \leq 1 \) El valor de \(1 answer -
The value of I=โซC Fโ drโ where F:=(y+6x^5,x+8y^7,sinz), and the curve is parametrized by C(t):=(1+t,2+sin(17ฯt),9 +cos(4ฯt^2)), with 0โคtโค1, is equal to: Hint: think conservative fields
El valor de \( I=\int_{\mathbf{C}} \mathbf{F} \cdot d \overrightarrow{\mathbf{r}} \) donde \( \mathbf{F}:=\left(y+6 x^{5}, x+8 y^{7}, \sin z\right) \), y la curva es parametrizada por \( \mathbf{C}(t)1 answer -
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Let C be the curve resulting from the intersection of the sphere ๐ฅ2+๐ฆ2+๐ง2=1 with the cone ๐ง=โ(๐ฅ2+๐ฆ2), oriented in such a way that when seen from above it is traversed in the form c
Sea \( C \) la curva que resulta de la intersecciรณn de la esfera \( x^{2}+y^{2}+z^{2}=1 \) con el cono \( z=\sqrt{x^{2}+y^{2}} \), orientada de tal forma que cuando se ve desde arriba es recorrida de1 answer -
The value of the surface integral I:=โฌS 1/zโy+1โโโโโโโโ dS, where S is defined by 2z=x^2+2y,0โคxโค2โโ,0โคyโค2โโ, is equal to:
El valor de la integral de superficie \( I:=\iint_{S} \frac{1}{\sqrt{z-y+1}} d S \) donde \( S \) estรก definida por \( 2 z=x^{2}+2 y, 0 \leq x \leq \sqrt{2}, 0 \leq y \leq \sqrt{2} \), es igual a: Se1 answer -
By evaluating, by definition, the line integral I=โซC (y+z)dx+xdy+(x+y)dz, where C is the line segment from (0,0,0) to (0 ,1,2) we get:
Al evaluar, mediante la definiciรณn, la integral de lรญnea \( I=\int_{C}(y+z) d x+x d y+(x+y) d z \), donde \( C \) es el segmento de recta que va desde \( (0,0,0) \) a \( (0,1,2) \) se obtiene: Selec1 answer -
plzz helppp By evaluating, by definition, the line integral ๐ผ=โซ๐ถ๐ง๐๐ฅ+(๐ฅ+๐ง)๐๐ฆ+(๐ฅ+๐ฆ)๐๐ง, where ๐ถ is the line segment from (1,0,1) to (1 ,0,2) we get:
Al evaluar, mediante la definiciรณn, la integral de lรญnea \( I=\int_{C} z d x+(x+z) d y+(x+y) d z \), donde \( C \) es el segmento de recta que va desde \( (1,0,1) \) a \( (1,0,2) \) se obtiene: Sele3 answers -
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Find the derivative of all orders of the function. \[ y=\frac{x^{4}}{4}+\frac{8}{9} x^{3}-x^{2}+4 x-5 \] \( \mathrm{y}^{\prime}= \) \( \mathrm{y}^{\prime \prime}= \) \( \mathrm{y}^{\prime \prime \prim3 answers -
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Find the equation of normal line to a function \( y=3 \sqrt[4]{x}-\sqrt{x} \) if \( x_{0}=1 \) a. \( y=4 x-6 \) b. \( y=\frac{7}{4}-\frac{x}{4} \) c. \( y=6-4 x \) d. \( y=\frac{x-7}{4} \) e. \( y=\fr3 answers -
Find the derivative of the following functions: (a) \( y=e^{2 x} \operatorname{Tan}^{-1}\left(x^{3}\right) \) (b) \( y=\frac{x^{2}+2}{e^{4 x}} \) (c) \( y=\cos \left(\sin \left(x^{2}-3\right)\right) \1 answer -
Solve the initial value problem \[ y^{\prime \prime}-6 y^{\prime}+9 y=0, y(0)=0, y^{\prime}(0)=11 \] \[ \begin{array}{l} y=11 e^{3 t} \\ y=11 t e^{3 t} \\ y=e^{3 t} \\ y=t e^{3 t} \\ y=11 e^{3 t}+11 e1 answer -
Find \( \operatorname{div}(F \times \mathbf{G})=\nabla \cdot(\mathbf{F} \times \mathbf{G}) \). \[ \begin{array}{l} \mathbf{F}(x, y, z)=\mathbf{i}+6 x \mathbf{j}+8 y \mathbf{k} \\ \mathbf{G}(x, y, z)=x1 answer