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.

-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

And for the second curve notice that

Thus for the first curve we reach our point at    and for the second curve we reach our point at

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

    and at      we have that  

Similarly

The the cross product between the two vectors would be

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

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)

d

step-by-step explanation: