Find a vector function that represents the curve of intersection of following two surfaces: the cone $z=\sqrt{x^2+y^2}$ and the plane $z=1+y$.
This is a question right out of Stewart's calculus text. Since the equation of a cone in parametric form is
$$x = t \cos(t),\space y = t\sin(t), \space z = t,$$
I believed the answer to this question should be
$$\vec{r}(t)= (t \cos t)\vec{i} + (t \sin t)\vec{j} + (1 + t \sin t)\vec{k},$$
but it's not the correct answer.
I would appreciate if someone can help me find the right answer.
$\endgroup$2 Answers
$\begingroup$One easy way to solve this is to use Cartesian coordinates to get the relations between the variables, then to translate that to a vector function.
In this case, you have two expressions for $z$. Equate them and we get
$$\sqrt{x^2+y^2}=1+y$$
$$x^2+y^2=1+2y+y^2$$
$$y=\frac 12x^2-\frac 12$$
We need to be careful of not introducing extraneous solutions in our squaring the equation, but none seem to be in evidence here.
We can set $x$ to the parameter and use the second of your equations to get
$$\vec{r}(t)=(t)\vec{i} + \left(\frac 12t^2-\frac 12\right)\vec{j} + \left(\frac 12t^2+\frac 12\right)\vec{k}$$
There are, of course, infinitely many other possibilities.
As a quick geometric check, we see that the problem asks for the intersection of a cone with a plane parallel to an "edge" of the cone. The ancient Greeks knew that was nothing or a parabola. The point $(1,0,1)$ is in the intersection, so the correct answer here is not nothing. Our answer gives a "tilted parabola" so it fits.
$\endgroup$ 0 $\begingroup$If one considers the intersection as living inside of $R^3$ then the "solution" of $ y= f(x)$ would imply that you are leaving z free to vary which would be an 'extrusion' in the z-direction, which is obviously wrong.
In fact, checking $y=\frac{1}{2}x^2-\frac{1}{2}$ at x = 1 yields y = 0 and you don't get the z = 1 without the equation z=1+y.
The "solution" y = f(x) is the projection of the intersection into the x-y plane.
Now, if you believe that the intersection is 'unbounded', like a parabola, then the parametric choices of x=t, y=t or z=t are reasonable choices for an unbounded part of the intersection.
However, in order for the parameterization to work, the other variables would need to be functions of the remaining variable.
A look at your attempt with $r(t)=tsint \hat{i} + tcost \hat{j} + (1+tcost)\hat{k}$ is that you satisfied the equation z=1+y but did not satisfy the equation $z=\sqrt{x^2+y^2}$
It is easy to see that if you choose $y=t$ then the $z= 1+y$ makes $z(t)=1+t$ using these 'in' $z=\sqrt{x^2+y^2}$ it becomes clear that x can be solved or $x(t)=\sqrt{(1+t)^2-t^2}$. But the question about the ranges of x values arises.
If you choose, $z=t$, then $z=1+y$ yields $y(t)= t-1$ and then the first equation, again, yields x(t). With the exact same concerns.
However, the choice of $x=t$ is not clear how you get $y(t)$ and $z(t)$ easily. Retracing Mr. Daulton's answer you can see how the relationship for y and x is found upon substitution and from $x(t)$ you get $y(t)$ then finally and again, using the first equation, and you arrive at $z(t)$
Please notice, that nowhere did an unbounded parametric solution begin with the original parametric form of the cone you gave. Why? Because that parametrization is best thought of as the result of intersecting the plane $z=c$ with the cone!
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