Is it possible to find the implicit parallel curve of $f(x) = x^2$? If so what would it be and how does one find it?
75978
I have a simple parabola in the form $y = a + bx^2$. I would like to find the formula for a curve which is parallel to this curve by distance $c$. By parallel I mean that there is an equal distance along a line perpendicular to the tangent to my curve at all points.  I've established that the curve isn't in the form $y = a + c + dx^2$, whilst I can make this satisfy for $x=0$ and $x$ equal to one other number is isn't valid across the range.  Any help much appreciated.  Rob
31188
Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ without any further restriction. If $f(f(x))=x^2-x+1$, how can one find $f(0)$?   Thanks in advance.
64533
I'm trying to find the minimal distance between the surfaces described by $z=x^2+y^2$ and $x+y-2z=8$. I would imagine there are several approaches including the use of Lagrange multipliers. I attempted to find the spot when their normal vectors are parallel ( since i believe if these vectors are not parallel there is a direction that will bring the distance lower). So if $U(x,y,z)=x^2+y^2$ and $V(x,y,z)=x+y-2z$ are functions describing these surfaces then their normal vectors are $(2x,2y,-1)$ and $(1,1-2)$ respectively. Then I want their cross product to be zero; this is at $(1/4,1/4,z)$. But when i solve for $z$ and I cannot simultaneously satisfy both equations. What gives?! Am I along the right line of thinking?
89472
While an explicit function $y(x)$'s arc length $s$ is easily obtained as  $$s = \int \sqrt{1+|y'(x)|^2}\,dx,$$   is there any formula for implicit functions given by $f(x,y) = 0$? One can use the implicit differentiation $y'(x) = -\frac{\partial_y f}{\partial_x f}$ to obtain  $$s = \int\sqrt{1 + |\partial_y f / \partial_x f|^2}\,dx,$$  but that still requires (locally) solving for $y(x)$. Is there any formulation that does not require this, e.g. another implicit equation involving $s$?    Thoughts so far:  One could rewrite $s$ as  $$s = \int |\nabla f|\, |\partial_x f|dx,$$  or symmetrize to  $$s = \int |\nabla f|\, \underbrace{(|\partial_x f|dx + |\partial_y f|dy)}_{(*)}/2$$  where $(*)$ might be strongly related to $|df|$ I guess (though it's not identical due to the $|\cdot|$), but then?
82046
In multivariable calculus, given a function like $F(x,y,z) =0$, the implicit function $z=f(x,y)$ exists if and only if $\frac{\partial F}{\partial z} \not = 0$. And the implicit function is given by $\frac{dz}{dy}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$.   The theorem can be proved by mathematical reasoning. But I want to know whether there is some intuitive understanding of the theorem, say geometry intuition in the three dimensional space, etc.
88104
Find the singular points and the corresponding tangent lines of $X^4+Y^4-X^2Y^2$.    I get that the singular points are those of the form $(a,a)$ and $(a,-a)$. The corresponding tangent lines are $X=-a, Y=-a$ and $X=-a, Y=a$ respectively. Is this correct? Thanks.
50981
Find all $f(x)$ satisfying $f(f(x)) = x^2 - 2$.   Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further.   Possibly Helpful Links: Information on similar problems can be found  and .  Source:  It is about to be closed for containing too many problems in one question. I'm posting each problem separately.
91168
What would the graph of $(-1)^{x}$ look like?   I know that the value of the function alternates between $1$ and $-1$ when it is defined so I think it would just be points spread over the lines $x=1$ and $x=-1$.   Is this correct? Also, will there be any definite pattern of the points.
36077
Prove that for any piecewise smooth curve it is possible to find the parametrisation $\phi$ that is consistent with its length, ie.  length of a curve segment between $\phi(a)$ and $\phi(b)$ is equal to $|b-a|$.  It seems intuitive, but I don't have any idea how can I show this in the formal way.
2508
Find all (if that's possible) functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(f(x))=\sin x.$$  This is a problem a friend gave me. He said he saw it on a book a long time ago, but he couldn't remember what book. I literally have no idea how to solve this so I can't post my attempt here.
4007
$f(x) = \frac{x}{x^2 + 1}$.  How do you find $f^{-1}$?  Is try and solve for y and I get the following:  $y = \frac{x}{x^2 + 1}$  $y(x^2 + 1) = x$  $yx^2 + y = x$  $y = -yx^2 + x$  $y = x(-yx + 1)$  It seems whatever I do I can't isolate the x variable, I just go in circles...
15145
Given two vectors $u,v\in\mathbb{R}^3$, find the parametric equations for the plane through the origin parallel to $u$ and $v$.  I'm having a difficult time understanding this question. How can a plane through the origin in $\mathbb{R}^3$ be parallel to two vectors, both of which start from the origin? Surely they all intersect at the origin and thus not parallel?  More concretely, $u=(1,2,3)$ and $v=(-2,1,1)$. The cartesian equation of the plane is $x+7y-5z=0$, which is essentially the solution. But I want to know why.
21201
We all know some curves can be described by $y=f(x)$ and some surfaces can be described by $z=f(x,y)$ However, there exists curves and surfaces which cannot be described by those, such as a circle and a sphere. Therefore, we introduce parameterized vector equations, which can describe them. For example, circle: $\vec r(t)=r\cos(t)\hat i+r\sin(t)\hat j$ sphere: $\vec r(u,v)=\rho\cos(u)\sin(v)\hat i+\rho\sin(u)\sin(v)\hat j+\rho\cos(v)\hat k$ However, all curves described by $y=f(x)$ and  $z=f(x,y)$ can be parameterized. For curves, $$\vec r(t)=t\hat i+f(t)\hat j$$ For surfaces, $$\vec r(u,v)=u\hat i+v\hat j+f(u,v)\hat k$$ Therefore, I think this suggests the set of all parameterized surfaces (or curves) is the super-set of the set of all surfaces (or curves) described by $z=f(x,y)$ (or $y=f(x)$). Is that correct? Now, here comes the the real challenge. A curve can also be described by an implicit function $f(x,y)=0$ and a surface can also be described by an implicit function $f(x,y,z)=0$ I have 3 questions regarding this.   Can all surfaces (curves) described by an implicit function be parameterized? (If yes, then what is the general way?) Can all surfaces (curves) described by parametric vector equations be represented using implicit function? (If yes, then what is the general way?) Compare the set of all parameterized surfaces (curves) and the set of all surfaces (curves) represented by implicit function. (which is which super-set?)   Sorry for the use of nontechnical terms. I use them because I don't know the technical ones. I have only started learning vector calculus last year in university.  EDIT: I think my question is not too clear, so I will give an example of writing the surface $f(x,y,z)=0$ into $\vec r(u,v)$ We want to parameterize a sphere. $$x^2+y^2+z^2-\rho^2=0$$ Let $x=\rho\cos(u)\sin(v)$, $y=\rho\sin(u)\sin(v)$, $z=\rho\cos(v)$, $$\rho^2\cos^2(u)\sin^2(v)+\rho^2\sin^2(u)\sin^2(v)+\rho^2\cos^2(v)-\rho^2$$ $$=\rho^2\sin^2(v)(\cos^2(u)+\sin^2(u))+\rho^2\cos^2(v)-\rho^2=\rho^2-\rho^2=0$$ I want to know if there is a general way of finding $x=x(u,v)$, $y=y(u,v)$ and $z=z(u,v)$ for any given $f(x,y,z)=0$
111869
The problem is to show that a polynomial $f(x) \in F[x]$ (F is a field) has no repeated roots if and only if f(x) and f'(x) (the derivative of f(x)) are relatively prime. I've managed to prove one direction of this equivalence (if there are no repeated roots, f(x) and f'(x) are relatively prime), but the other one gives me headaches... can anyone help out?
89922
If we have an implicit surface $g(x, y, z)=...$, the gradient is a normal vector. But why? Do I have to start with tangent vectors and then show the gradient is perpendicular? Not sure how to approach this.
46840
1.find a for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ such that for every m real, f has real roots 2.find m such that for every a real, f has real roots  My ideea is to demonstrate that $\delta=4(m-a)^2 - 4(3am-2) = 4(m^2-5am+a^2+2) &gt;0$ But I have no ideea how to do this.
94740
I'm trying to solve this question in the classical  (page 23):        A regular parametrized curve $\alpha$ has the property that all its tangent lines pass through a fixed point. Prove that the trace of $\alpha$ is a (segment of a) a straight line.      My attempt  Following the statement of the question, we have $\alpha(t)+\lambda(s)\alpha'(s)=const$.  Taking the derivative of both sides we have $\alpha'(s)+\lambda'(s)\alpha'(s)+\lambda(s)\alpha''(s)=0$ which is equal to $(1+\lambda'(s))\alpha'(s)+\lambda(s)\alpha''(s)=0$.  Since $\alpha'(s)$ and $\alpha''(s)$ are linearly independent, we have $\lambda'(s)=-1$ and $\lambda(s)=0$ for every $s$ which I found strange, since the derivative of the zero function is zero.  I need a clarification at this point and a hand to finish my attempt of solution.