Question

In: Math

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

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).

Solutions

Expert Solution

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


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