   Chapter 16.6, Problem 38E

Chapter
Section
Textbook Problem

Find an equation of the tangent plane to the given parametric surface at the specified point. Graph the surface and the tangent plane.38. r(u, v) = (1 − u2 − v2) i − v j − u k; (−1, −1, −1)

To determine

To find: An equation of the tangent plane to the vector function r(u,v)=(1u2v2)ivjuk at the point (1,1,1) .

Explanation

Given data:

The vector function is given as follows.

r(u,v)=(1u2v2)ivjuk

The specified point is (1,1,1) .

Formula used:

Write the expression to find tangent plane to the parametric surface with the normal vector n=a,b,c at the specified point (x0,y0,z0) .

a(xx0)+b(yy0)+c(zz0)=0 (1)

Write the expression to find normal vector from the tangent vectors of the parametric surface.

n=|ijka1b1c1a2b2c2| (2)

Here,

The vector a1,b1,c1 is a tangent vector ru of the parametric surface and

The vector a2,b2,c2 is a tangent vector rv of the parametric surface.

Write the expression to find the tangent vector ru of the parametric surface.

ru=xui+yuj+zuk (3)

Write the expression to find the tangent vector rv of the parametric surface.

rv=xvi+yvj+zvk (4)

Write the vector function as follows.

r(u,v)=(1u2v2)ivjuk

Calculation u and v at the point (1,1,1) :

Equate the x-, y-, and z-coordinates of vector function r(u,v) to the corresponding coordinates of the point (1,1,1) as follows.

1u2v2=1v=1u=1

From the equations, it is clear that the point (1,1,1) corresponds to u=1 and v=1 .

Therefore, u=1 and v=1 .

Calculation of tangent vector ru :

Substitute (1u2v2) for x , (v) for y , and (u) for z in equation (3),

ru=(1u2v2)ui+(v)uj+(u)uk=[u(1u2v2)]i+[u(v)]j+[u(u)]k=(02u0)i+(0)j+(1)k=2u,0,1

Substitute 1 for u ,

ru=2(1),0,1=2,0,1

Calculation of tangent vector rv :

Substitute (1u2v2) for x , (v) for y , and (u) for z in equation (4),

rv=(1u2v2)cvi+(v)vj+

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