   Chapter 14.4, Problem 42E

Chapter
Section
Textbook Problem

Suppose you need to know an equation of the tangent plane to a surface S at the point P(2, 1,3). You don't have an equation for S but you know that the curvesr1(t) = ⟨2 + 3t, 1 –t2, 3 – 4t + t2⟩r2(u) = ⟨1 + u2, 2u3 – 1, 2u + 1⟩both lie on S. Find an equation of the tangent plane at P.

To determine

To find: The equation of the tangent plane to a surface S at the point P(2,1,3) .

Explanation

Given:

Let S be the surface and the curves are,

r1(t)=2+3t,1t2,34t+t2r2(u)=1+u2,2u31,2u+1

Calculation:

Let r1(t)=2+3t,1t2,34t+t2 (1)

Differentiate the equation (1) with respect to t at the point P(2,1,3) and obtain the tangent vector r1'(t) ,

r1'(t)=0+3,0(2t),04(1)+(2t)=3,2t,4+2t

Let r2(u)=1+u2,2u31,2u+1 (2)

Differentiate the equation (2) with respect to u at the point P(2,1,3) and obtain the tangent vector r2'(u) ,

r2'(u)=0+(2u),2(3u2)0,2+0=2u,6u2,2

To find the curve passes through the point P(2,1,3) or not. By assuming the value of t=0 and u=1 .

Assume t=0 and u=1 in the equation (1) and (2),

r1(0)=2+3(0),1(0)2,34(0)+(0)2=2,1,3

Assume u=1 in the equation (2),

r2(1)=1+(1)2,2(1)31,2(1)+1=1+1,21,2+1=2,1,3

From this, it is clear that the curve satisfies the point P(2,1,3) when t=0 and u=1

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