suppose $(x_0,y_0)$ is a point on some curve $g(t):R->R_2$ that is Closest to a point $(x_1,y_1)$ NOT on the curve .

show that the line through $(x_0,y_0)$ and $(x_1,y_1)$ is orthogonal to the tangent line to the curve at the point $(x_0,y_0)$

really though question i have tried everything

let $g(t) = (x(t) , y(t))$

$g'(t) = (x'(t),y'(t))$

so we can say that the tangent line $(x_0,y_0) + s(x',y')$ and the line that passes through the points : $(x_0,y_0)+k(x_1-x_0,y_1-y_0)$

and we need to prove that those lines orthogonals , i tried dot product equals to 0 but i always find that i miss something to prove.

i am sure that this missing clue is that we need to use the fact that this 2 points are the closest to each other and that no other point ON THE CURVE is closer.

  • 1
    $\begingroup$ What is $R_2$ here? Is it $\mathbf{R}^2$? $\endgroup$ – Partha Nov 13 '18 at 10:40
  • $\begingroup$ true its R2 like you wrote a vector with 2 coordinates $\endgroup$ – Razi Awad Nov 13 '18 at 10:45
  • 1
    $\begingroup$ It's a minimization problem. $(x_0, y_0)$ is the point that minimizes $(x-x_1)^2+(y-y_1)^2$ for $(x,y)$ on the curve. $\endgroup$ – saulspatz Nov 13 '18 at 10:48
  • $\begingroup$ the problem given as a calculas question $\endgroup$ – Razi Awad Nov 13 '18 at 10:50
  • $\begingroup$ wait what ! i had an idea now lets say that the point (x1,y1) is a center of a circule with radius squrt((x−x1)2+(y−y1)2) that is the minimum distance given by the question . so the question become trivial right ? because the radius is always orthogonal to the tangent of a circule $\endgroup$ – Razi Awad Nov 13 '18 at 10:54

Your attempted solution is incorrect. Firstly, the closest-point condition is not sufficient to imply that the curve is differentiable there. So the question actually must have that extra condition. Secondly, you wrongly assumed that the curve is a graph of $(x,y)$ where $y$ is some function of $x$. That was not given. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

Let me use variables $x,y$ instead of $x_0,y_0$, varying with parameter $t$. We can also assume that $x_1 = y_1 = 0$.

Let $r = x^2+y^2$. Consider any point on the curve (specified by parameter $t$ and coordinates $x,y$). As $Δt \to 0$ we have $\lfrac{Δr}{Δt} = \lfrac{((x+Δx)^2+(y+Δy)^2) - (x^2+y^2)}{Δt}$ $= 2(x\lfrac{Δx}{Δt}+y\lfrac{Δy}{Δt}) + (Δx\lfrac{Δx}{Δt}+Δy\lfrac{Δy}{Δt})$. If the curve is differentiable there, then we also have $Δx,Δy \approx 0$ and $\lfrac{Δx}{Δt} \approx \lfrac{dx}{dt}$ and $\lfrac{Δy}{Δt} \approx \lfrac{dy}{dt}$, and so $\lfrac{Δr}{Δt} \approx 2(x\lfrac{dx}{dt}+y\lfrac{dy}{dt})$, and hence $\lfrac{dr}{dt} = 2(x\lfrac{dx}{dt}+y\lfrac{dy}{dt})$. If moreover $r$ is minimum there, then $\lfrac{dr}{dt} = 0$, and so $x\lfrac{dx}{dt}+y\lfrac{dy}{dt} = 0$, equivalently $(x,y)·(\lfrac{dx}{dt},\lfrac{dy}{dt}) = 0$, and hence $(x,y)$ and $(\lfrac{dx}{dt},\lfrac{dy}{dt})$ are orthogonal.

Note that $(x,y)$ and $(\lfrac{dx}{dt},\lfrac{dy}{dt})$ are orthogonal if and only if they are perpendicular or one of them is zero. So there is actually no need for the question to state that $(x_1,y_1)$ is not on the curve. If you really want to get perpendicularity, even that condition is insufficient; you also have to add an extra condition that the curve's derivative at that point is nonzero. Otherwise it is possible that $(\lfrac{dx}{dt},\lfrac{dy}{dt})$ is zero (the point tracing the curve stops momentarily) but there is no (geometric) tangent there; for example $(\cos(t)+\sin(t),\sin(2t))$ has zero derivative when $t = \lfrac14π$.

  • 1
    $\begingroup$ Of course, you can take the shortcut once you understand the derivation, namely $\lfrac{dr}{dt} = \lfrac{d(x^2)}{dx} \lfrac{dx}{dt} + \lfrac{d(y^2)}{dy} \lfrac{dy}{dt} = 2x \lfrac{dx}{dt} + 2y \lfrac{dy}{dt}$ and continue from there. $\endgroup$ – user21820 Nov 13 '18 at 17:22
  • $\begingroup$ but you used the fact that the curve is differentiable in your soultion i think that it is neseccary that the curve is differntiable in order to solve the question $\endgroup$ – Razi Awad Nov 13 '18 at 17:26
  • $\begingroup$ @RaziAwad: Isn't that exactly what I said? I said "the question actually must have that extra condition.", whereas you did not say that, neither in your question nor in your answer! Also, the curve only needs to be differentiable at that single point, as shown by my proof. $\endgroup$ – user21820 Nov 13 '18 at 17:27
  • $\begingroup$ absolutely i totaly agree that in a formal solution i should have said that the curve is diffrentiable in that point . but that doesn't mean my answer is incorrect it is partially correct . $\endgroup$ – Razi Awad Nov 13 '18 at 17:32
  • 1
    $\begingroup$ @RaziAwad: If you differentiate with respect to $t$, you would be doing no different from what I did in my post, so yes. The key point to remember is that the derivative of a general curve is always with respect to the free parameter, and not to any coordinate. Also, it is almost always easier to minimize the squared distance than the actual distance. $\endgroup$ – user21820 Nov 14 '18 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.