Given that near (1,1,1) the curve of intersection of the surfaces x^4 + y^2 + z^6 - 3xyz = 0 and xy + yz + zx - 3z^8 = 0 has the parametric equations x = f(t), y = g(t), z = t with f, g differentiable. (a) What are the values of the derivatives f'(1), g'(1)? (b) What is the tangent line to the curve of intersection at (1,1,1) given that z = 1 + s? (Find what x and y are equal to.) Answers: f'(1) = 4g'(1) = 7x = 1 + 4sy = 1 + 7s I took the gradients of each surface (the first one being f(x,y,z) and the second one being g(x,y,z) respectively). I tried to set the respective components equal to each other in an attempt to solve for something but I ended nowhere useful. I also tried to think of z = t = 1 and in order to try and find x and y. Then, I was like "Oh, wait, I have a point (1,1,1), let me just shove it into the gradients." all in order to get the correct answers and then to try and make sense of what I did but I didn't succeed .Any help in solving this problem would be greatly appreciated! Thanks in advance! P.S. Why does my nicely formatted text get automatically bundled up in this forum?