Question

Suppose you need to know an equation of the tangent plane to a surface S at the point P(4, 1, 4). You don't have an equation for S but you know that the curves r1(t) = 4 + 3t, 1 − t2, 4 − 5t + t2 r2(u) = 3 + u2, 2u3 − 1, 2u + 2 both lie on S. Find an equation of the tangent plane at P.

Answer 1

Answer:

$15x-8y+9z=88$

Step-by-step explanation:

We are given that

Equation of curves

$r_1(t)=<4+3t,1-t^2,4-5t+t^2>$

$r_2(u)=<3+u^2, 2u^3-1, 2u+2>$

Both curves lie on S.

We have to find the equation of tangent plane at P(4,1,4).

$r'_1(t)=<3,-2t,-5+2t>,r'_2(u)=<2u,6u^2,2>$

$r_1(0)=<4,1,4>, r_2(1)=<4,1,4>$

Hence, t=0 and u then it satisfied the given point.

Substitute the values in the derivatives

$r'_1(0)=<3,0,-5>,r'_2(1)=<2,6,2>$

$r'_1(0)\times r'_2(1)=<3,0,-5>\times <2,6,2>=<30,-16,18>$

The equation of tangent at point P(4,1,4) is given by

$30(x-4)-16(y-1)+18(z-4)=0$

$30x-120-16y+16+18z-72=0$

$30x-16y+18z=176$

$15x-8y+9z=88$