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Consider z=f(x,y)=x^3 + 2xy + y

Find all points on the surface where the tangnent plane is parallel to the plane 5x+3y-z=0

So I took the gradient of f (x,y,z) and got (2x^2 i , 2x+1 j , -1 k) And this vector must be colinear to (5, 3, -1)

So (2x^2 , 2x+1 , -1 ) = (5, 3, -1) I get: (x, y, z) = (1, 3/2, -1)

My problem is I was supposed to get multiple points (from subsequent questions), not one, so I must have done something wrong but I can't find it.

Jacob
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1 Answers1

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The plane is the graph of the function $g(x, y) = 5x + 3y$. The gradient of this function is $(5, 3)$. The gradient of $f(x, y)$ is $(3x^2 + 2y, 2x + 1)$. Now set one equal to the other, and solve.

Arthur
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  • $(3x^2+2y, 2x+1)$, isn't it? – Brian Tung May 01 '15 at 18:58
  • @BrianTung You're right. – Arthur May 01 '15 at 19:01
  • Thanks. This is actually what I tried myself above, but without success. From equating I get the point (1,1,1) (not (1, 3/2, -1) as I wrote above, which had a calculation error). The next question states "for each of the points you have found give an equation to the tangent plane at that point". So there must be more points I am not finding. Or is the language used just not right? – Jacob May 02 '15 at 16:34
  • @Jacob You're right, it does seem like there is only one solution. I assume you've copied the problem correctly as well. However, the point is $(1, 1, 4)$, not $(1, 1, 1)$, since $f(1, 1) = 4$. Although, when they say "for each of the points", technically it can still be just one point. – Arthur May 02 '15 at 22:53