Question

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 + 2t, 1 − t2, 3 − 5t + t2 r2(u) = 3 + u2, 2u3 − 1, 2u + 1 both lie on S. Find an equation of the tangent plane at P.

Answer 1

Answer:

-30(x-4)  + 29 (y-1)  -12(z-3)  =  0

Step-by-step explanation:

You can find the perpendicular vector to both curves using the cross product  between them, and that will give you all the information you need to find the equation of the tangent plane to the surface S.

Let's find the parameter of each curve for the specific point.

For the first curve  notice that

$r( 0 ) = (4,1,3)$

And for the second curve notice that

$r2(1) = (4,1,3)$

Thus for the first curve we reach our point at $t=0,$    and for the second curve we reach our point at $u = 1 .$

Then we compute the tangent vector to our curves, and it would be given by the derivative so

$r1'(t) = (2,-2t,-5+2t)$     and at  $t = 0$    we have that  

$r1'(0) = (2,0,-5)$

Similarly

$r2'(1)=(3+2(1) , 6(1)^2 ,2) = (5,6,2)$

The the cross product between the two vectors would be

$(5,6,2) \times (2,0,-5) = (-30,29,-12)$

Since (-30,29,-12) would be the perpendicular vector to the tangent plane and (4,1,3) would be a point of the tangent  then we would have that the equation of the tangent plane is given by

-30(x-4)  + 29 (y-1)  -12(z-3)  =  0