Suppose you need to know an equation of the tangent plane to a surface S at the point P(4, 1, 3). You don't have an equation for S but you know that the curves r₁(t) = 4 + 3t, 1 − t², 3 − 5t + t²r₂(u) = 3 + u², 2u³ − 1, 2u + 1 both lie on S. Find an equation of the tangent plane at P.

Answer:30(x-4)-16(y-1)+18(z-3)OR30x-16y+18z=158Step-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 PPut u=1 in r₂(u), we will get: r₂(u)=(4, 1, 3),  it means it is on point PPut 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)OR30x-16y+18z=158