The surface with equation $z = x^{3}+xy^{2}$ intersects the plane with equation $2x−2y = 1$ in a curve. What is the slope of that curve when $x = 1$ and $y = -\frac{1}{2}$?

I'm a little confused about how to go about this question, so, could somebody please help me get started, particularly about how to find the intersection in this case. I presume that the intersection equation must have an $x$ and $y$ remaining as we would have to substitute the given values but I'm not sure how to do it. Could someone help me get going? Details on how to find the intersection (curve) and a hint on how to proceed would help greatly! Thanks!


1 Answer 1


To find the intersection you could substitute your second equation into the first one, since $y=\frac{2x-1}{2}$ we get:

$z=x^3+xy^2 \Rightarrow z=x^3+x\left (\frac{2x-1}{2} \right )^2$

We now have $y$ and $z$ expressed as functions of $x$ which allows us to give a parametrisation for the curve (that is the intersection between the two surfaces):

$$\vec r(t)=(x(t),y(t),z(t))=\left (t,\frac{2t-1}{2},t^3+t\left (\frac{2t-1}{2} \right )^2 \right)$$

Now find the tangent vector to the curve at your specified point.

Remark: I noticed that the $x=1,y=-1/2$ does not verify the plane equation. So this means that you either have a typo ($x=1,y=1/2$ works) or you have two points: first at $x=1$ and then at $y=-1/2$. If it is the latter case then you need to find what value of $x$ gives you $y=-1/2$ from the plane equation and then substitute this into $\vec r\ '(t)$ (with $x=t$, the parameter).

EDIT: If you haven't covered parametrised curves yet there's another (perhaps even simpler) way of doing this.

Let $f(x,y,z)=x^3+xy^2-z$ and $g(x,y,z)=1-2x+2y$, then your surfaces are the level surface $f(x,y,z)=0$ and $g(x,y,z)=0$.

Since since $f$ class $\mathscr{C}'$ then the gradient of $f$ at point $P_0\in \left \{P:f(P)=0 \right \}$, $\nabla f(P_0)$, is orthogonal to the level surface at the point $P_0$.

Similarly the gradient of $g$ is orthogonal to the surface level of $g$.

Now the intersection is a curve which is contained in both surfaces, hence it is orthogonal to both $\nabla f(P)$ and $\nabla g(P)$. Therefore the curve is parallel to the vector product $\nabla f(P) \times \nabla g(P)$. We have therefore obtained a vector which is parallel to the curve at the desired point.

  • $\begingroup$ Thanks so much for your response! Yes, the fact that the x and y values do not satisfy the plane equation threw me off a little bit. I think it must be two separate points. However, we haven't convered parametrisation yet. So, if these are two separate points can I substitute the plane equation into the surface to obtain an equation in only x, then differentiate and substitute x=1 and then substitute the plane equation again, this time to get an equation in only y, differentiate and substitute y=-1/2? Would this give a correct answer to the question? $\endgroup$ Mar 2, 2015 at 0:36
  • $\begingroup$ Check my edit for an alternative method with doesn't use parametrisations. If there are two separate points you'll get two separate slopes for each point by applying the method in my answer twice. I'm not sure I understand what you mean by "then substitute again". $\endgroup$
    – Reveillark
    Mar 2, 2015 at 1:14
  • $\begingroup$ I mean first write the plane equation in terms of y and substitute into the surface, differentiate and then substitute x=1. Then, we write the plane equation in terms of x and substitute into the surface, differentiate and substitute y=-1/2? $\endgroup$ Mar 2, 2015 at 1:18
  • $\begingroup$ That would work assuming $z$ is a function of $x$ and $y$. By substituting you reduce it to a function of 1 variable. It ultimately boils down to what you interpret by slope: either a number when $z$ is a function of $x$ or $y$ (post substitution) or as a vector tangent to the curve as I've done in my answer. If you think about it, these two concepts are very related, in fact the calculations involved are the same as parametrising (I could have also chosen y as my parameter). If my answer suits you, feel free to accept it so others can see it has been resolved. $\endgroup$
    – Reveillark
    Mar 2, 2015 at 1:41

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