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.

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)=r₁(t) and B₂(u)=r₂(u)

B₁(t)=(4+3t, 1-,3-5t+

B₁(t)=(3, -2t, -5+2t)

B₂(u)=(3+, 2-1, 2u+1)

B₂(u)=(2u, 6, 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

45

step-by-step explanation:

no this is a nonlinear negative