Docsity
Docsity

Prepara tus exámenes
Prepara tus exámenes

Prepara tus exámenes y mejora tus resultados gracias a la gran cantidad de recursos disponibles en Docsity


Consigue puntos base para descargar
Consigue puntos base para descargar

Gana puntos ayudando a otros estudiantes o consíguelos activando un Plan Premium


Orientación Universidad
Orientación Universidad


EJERCICIOS DE PEARSON, Guías, Proyectos, Investigaciones de Ingeniería

EJERCICIOS DE PEARSON DE LA GUIA 4

Tipo: Guías, Proyectos, Investigaciones

2021/2022

Subido el 20/09/2022

laura-cortes-43
laura-cortes-43 🇨🇴

8 documentos

1 / 2

Toggle sidebar

Esta página no es visible en la vista previa

¡No te pierdas las partes importantes!

bg1
15/9/22, 15:25
Actividad 1 - Guía 4-Laura Cortes
https://xlitemprod.pearsoncmg.com/api/v1/print/highered
1/2
Student: Laura Cortes
Date: 09/15/22
Instructor: JOSE ALEXANDER
FUENTES MONTOYA
Course: Análisis Multivariado y Funcional
G1C3
Assignment: Actividad 1 - Guía 4
Find an equation for the plane that is tangent to the given surface at the given point.
z
= y x (1,2,1)
The tangent plane at the point on the level surface of a differentiable function f is the plane through
normal to . The tangent plane to at is given by the following equation.
P0x0,y0,z0f(x,y,z) = c
P0
f |P0f(x,y,z) = c P0
f + f + f = 0
xP0x x0 y P0y y0 z P0z z0
Begin by setting the equation for the given surface equal to zero.
z = y x
zy x = 0 Rewrite the equation.
Find .
f
x
f(x,y,z) = zy x
f
x=1
2 y x Differentiate.
Now find by substituting the values of the given point and simplifying.fxP0
f
x=1
2 y x
fxP0=1
2 2 1 Substitute.
fxP0=1
2Simplify.
Next find .
f
y
f(x,y,z) = zy x
f
y=1
2 y x Differentiate.
Now find by substituting the values of the given point and simplifying.fyP0
pf2

Vista previa parcial del texto

¡Descarga EJERCICIOS DE PEARSON y más Guías, Proyectos, Investigaciones en PDF de Ingeniería solo en Docsity!

15/9/22, 15:25 Actividad 1 - Guía 4-Laura Cortes

https://xlitemprod.pearsoncmg.com/api/v1/print/highered 1/

Student: Laura Cortes Date: 09/15/

Instructor: JOSE ALEXANDER FUENTES MONTOYA Course: Análisis Multivariado y Funcional G1C

Assignment: Actividad 1 - Guía 4

Find an equation for the plane that is tangent to the given surface at the given point.

z = y − x (1,2,1)

The tangent plane at the point on the level surface of a differentiable function f is the plane through normal to. The tangent plane to at is given by the following equation.

P 0 x 0 ,y 0 ,z 0 f(x,y,z) = c

P 0 ∇f |P 0 f(x,y,z) = c P 0

f (^) x P 0 x − x 0 + f (^) y P 0 y − y 0 + f (^) z P 0 z − z 0 = 0

Begin by setting the equation for the given surface equal to zero.

z = (^) y − x y − x − z= 0 Rewrite the equation.

Find.

∂f

∂x

f(x,y,z) = y − x −z

∂f

∂x =^

2 y − x

Differentiate.

Now find fx P 0 by substituting the values of the given point and simplifying.

∂f

∂x

= −^

2 y − x

fx P 0 = − 1 2 2 − 1

Substitute.

fx P 0 = (^) −^1 2

Simplify.

Next find.

∂f

∂y

f(x,y,z) = (^) y − x −z

∂f

∂y

2 y − x

Differentiate.

Now find fy P 0 by substituting the values of the given point and simplifying.

15/9/22, 15:25 Actividad 1 - Guía 4-Laura Cortes

https://xlitemprod.pearsoncmg.com/api/v1/print/highered 2/

∂f

∂y =^

2 y − x

fy P 0 =

Substitute.

fy P 0 = 1 2

Simplify.

Finally, find.

∂f

∂z

f(x,y,z) = (^) y − x −z

∂f

∂z

= − 1 Differentiate.

Find fz P 0 by substituting the values of the given point and simplifying.

∂f

∂z

fz P 0 = − 1

Substitute the values found above and the point P 0 into the equation for the tangent plane.

f (^) x P 0 x − x 0 + f (^) y P 0 y − y 0 + fz P 0 z − z 0 = 0

− ( ) + ( ) − 1( )

2 x − 1^

2 y − 2^ z − 1^

= 0 Substitute.

Clear the equation of fractions, and simplify the expression to obtain the equation for the plane that is tangent to the given surface at the given point.

2 x − 1^

2 y − 2^ z − 1^

(x − 1) − (y − 2) + 2(z − 1) = 0 Clear the fractions. x − 1 − y + 2 + 2z − 2 = 0 Distribute. x − y + 2z − 1 = 0 Simplify.

Therefore, the equation for the plane tangent to the surface z = y − x at the point (1,2,1) is x − y + 2z − 1 =0.