Answer:
30(x-4)-16(y-1)+18(z-3)
OR
30x-16y+18z=158
Step-by-step explanation:
In order to find the tangent plan equation at point P,you know that r₁(t) and r₂(u) lie on surface S, Find the vectors B₁(t)=[tex]\frac{d}{dt} \\[/tex]r₁(t) and B₂(u)=[tex]\frac{d}{du}[/tex]r₂(u)
B₁(t)=[tex]\frac{d}{dt}[/tex](4+3t, 1-[tex]t^{2}[/tex],3-5t+[tex]t^{2}[/tex]
B₁(t)=(3, -2t, -5+2t)
B₂(u)=[tex]\frac{d}{du}[/tex](3+[tex]u^{2}[/tex], 2[tex]u^{3}[/tex]-1, 2u+1)
B₂(u)=(2u, 6[tex]u^{2}[/tex], 2)
Put t=0 in r₁(t), we will get:
r₁(t)=(4, 1, 3), it means it is on point P
Put u=1 in r₂(u), we will get:
r₂(u)=(4, 1, 3), it means it is on point P
Put t=0 in B₁(t), we will get:
B₁(t)=(3,0,-5)
Put u=1 in B₂(u), we will get:
B₂(u)=(2,6,2)
So Plane Contains two vectors B₁(t) and B₂(u), For Normal Vectors to the plane Cross Product is:
B₁(t)xB₂(u)= (3,0,-5)x(2,6,2) [Cross Product]
B₁(t)xB₂(u)=(30,-16,18)
Equation will become:
30(x-4)-16(y-1)+18(z-3)
OR
30x-16y+18z=158
