Question

The curves r₁(t) = < t, 4, t²-9 > and r₂(s) = < 3, s², 4-2s >lie on surface of f and intersect at ( 3, 4, 0 ). Find a linear approximation for the # f(3.1, 4.1).

Answer 1

given r₁(t) = < t, 4, t²-9 > and r₂(s) = < 3, s², 4-2s >

at (3,4,0)
t=3
4-2s=0
=>2s=4
=>s=2

at the point (3,4,0) , correspoindind values of t and s on r₁(t) = < t, 4, t²-9 > and r₂(s) = < 3, s², 4-2s > are t=3,s=2

r₁(t) = < t, 4, t²-9 > and r₂(s) = < 3, s², 4-2s >
=>r'₁(t) = < 1, 0, 2t-0 > and r'₂(s) = < 0, 2s, 0-2 >
=>r'₁(t) = < 1, 0, 2t > and r'₂(s) = < 0, 2s, -2 >
=>r'₁(3) = < 1, 0, 2(3) > and r'₂(2) = < 0, 2(2), -2 >
=>r'₁(3) = < 1, 0, 6 > and r'₂(2) = < 0, 4, -2 >

r'₁(3) = < 1, 0, 6 >
r'₂(2) = < 0, 4, -2 >

(r'₁(3))X(r'₂(2)) =<(0*-2)-(4*6),(0*6)-(1*-2),(1*4)-(0*0)>
=>(r'₁(3))X(r'₂(2)) =<-24,2,4>

equation of tangent plane to the surface at (3,4,0):

<-24,2,4>.<x-3,y-4,z-0>=0
=>-24(x-3)+2(y-4)+4(z-0)=0
=>4(z-0)=24(x-3)-2(y-4)
=>(z-0)=6(x-3)-0.5(y-4)
=>z=6x-18-0.5y+2
=>z=6x-0.5y-16

linear approximation:

f(x,y)=6x-0.5y-16
=>f(3.1, 4.1)≈6(3.1)-0.5(4.1)-16
=>f(3.1, 4.1)≈18.6-2.05-16
=>f(3.1, 4.1)≈0.55