Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Let S be the surface defined by p(x, y, z) = 0 where z = f(x,y); i.e. z is a function of the variables x and y. (i) Let P be an arbitrary point on S with position vector p. Show that the equation of the tangent plane to the surface S at p is given by Vo(x-p) = 0 where x = xi+yj+zk is the position vector of an arbitrary point in R³. Hint: Start with the usual equation of the tangent plane at the point P = (a, b, f(a, b)) on the surface z = f(x, y): : - f(a, b) = = af -(a, b) (x − a) + ?x af ду -(a, b) (y — b)

Let S be the surface defined by p(x, y, z) = 0 where z = f(x,y); i.e. z is a function of the variables x and y. (i) Let P be an arbitrary point on S with position vector p. Show that the equation of the tangent plane to the surface S at p is given by Vo(x-p) = 0 where x = xi+yj+zk is the position vector of an arbitrary point in R³. Hint: Start with the usual equation of the tangent plane at the point P = (a, b, f(a, b)) on the surface z = f(x, y): : - f(a, b) = = af -(a, b) (x − a) + ?x af ду -(a, b) (y — b)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 18T
See similar textbooks
icon
Related questions
Question
(a) Let S be the surface defined by p(x, y, z) function of the variables x and y. (i) Let P be an arbitrary point on S with position vector p. Show that the equation of the tangent plane to the surface S at p is given by = 0 where z = f(x, y); i.e. z is a z — f(a, b) - Vo. (x − p) = = 0 where x = xi+yj+zk is the position vector of an arbitrary point in R³. = Hint: Start with the usual equation of the tangent plane at the point P = (a, b, ƒ (a, b)) on the surface z = f(x, y): af af (a, b) (x − a) + (a, b) (y - b) ду əx (ii) Lety: R→→→→ R³; be a curve in R³ that lies on the surface S and passes through the point P on S, i.e. there exists a to such that y(to) = P. Show that the tangent vector y'(to) to the curve at P lies in the tan- gent plane to the surface S at the point P.
Transcribed Image Text:(a) Let S be the surface defined by p(x, y, z) function of the variables x and y. (i) Let P be an arbitrary point on S with position vector p. Show that the equation of the tangent plane to the surface S at p is given by = 0 where z = f(x, y); i.e. z is a z — f(a, b) - Vo. (x − p) = = 0 where x = xi+yj+zk is the position vector of an arbitrary point in R³. = Hint: Start with the usual equation of the tangent plane at the point P = (a, b, ƒ (a, b)) on the surface z = f(x, y): af af (a, b) (x − a) + (a, b) (y - b) ду əx (ii) Lety: R→→→→ R³; be a curve in R³ that lies on the surface S and passes through the point P on S, i.e. there exists a to such that y(to) = P. Show that the tangent vector y'(to) to the curve at P lies in the tan- gent plane to the surface S at the point P.
Expert Solution
steps

Step by step

Solved in 3 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
SEE MORE TEXTBOOKS