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I'm studying calculus on my own with the MIT OCW book. Question 61 on page 113 asks: "What point on $y= -x^2$ is closest to what point on $y = 5 -2x$?" I have been looking all over the internet for an example but can only find examples of this when we are finding a point on a curve closest to a fixed point and not another curve. Since both points can move, I tried using the distance formula with and setting $y = 5 - 2a$ as a substitute but after an hour fiddling with it I can't figure out how to do this. Can anyone please help?

https://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/Calculus.pdf

Bernard
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maybedave
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  • You can start by fixing some point $(x_0,y_0)$ on one curve and compute the minimal distance $d(x_0,y_0)$. Then you get a formula for that and you are left to minimize $d(x_0, y_0)$ over all points $(x_0,y_0)$ yon that curve. – Severin Schraven Nov 05 '21 at 12:15
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    The problem gives a hint: "At the nearest points, the graphs have the same slope." Did you consider that? Did you follow the other instructions in the problem, which you can do before you find the closest points: "Sketch the graphs"? – David K Nov 05 '21 at 12:15
  • Yes, I did graph it and can see approximately where the shortest distance would be by eye (around (1, -1) but I don't understand how to setup the algebra to do this because the input for the line and the input of the curve would be two different variables. – maybedave Nov 05 '21 at 12:17
  • I set y = 5 - 2x to y = 5 - 2a. Then my distance formula is sqrt((x-a)^2 - (5 - 2a - x^2)^2) But when I expand this all out I get a function of f(x) = sqrt(5a^2 - 2ax - 20a + x^4 -8x^2 +4ax^2 + 25). Then I find the derivative which is (2x^3 + (4a - 8)x -a)/sqrt(x^4 - 4ax^2 - 8x^2 -2ax + 5a^2 - 20a + 25). Then I find the zero of the derivative to find the stationary points and am left with 2x^3 -8x + a(4x -1) = 0 and I don't know how to finish this because I have two inputs 'a' and 'x' – maybedave Nov 05 '21 at 12:21
  • It's good that you have made so much of an attempt. To make the question better, please edit all of this work into the question itself (not the comments) and use MathJax to format it so people can read it: https://math.stackexchange.com/help/notation – David K Nov 05 '21 at 12:27
  • BTW the problem is on page 105 according to the page numbering of the book, which comes out on the 113th page of the PDF. – David K Nov 05 '21 at 12:29
  • Notice the comment of @DavidK, which has received 2 upvotes, indicating that other mathSE reviewers regard the comment as valuable. The graph of the line has constant slope $(-2)$, while the function $y = f(x) = -x^2,$ has slope [i.e. first derivative $f'(x)$] of $-2x$. So, the first question that I would ask is: at what point on the curve $f(x) = -x^2$ will the first derivative of $-2x$ equal the value $(-2)$? ...see next comment – user2661923 Nov 05 '21 at 12:33
  • Minimizing the square of the distance is the same and it makes life much easier – Claude Leibovici Nov 05 '21 at 12:37
  • Having identified the corresponding point $(x_0,y_0)$ on the curve $y = f(x) = -x^2$, the next question is: when you construct the line perpendicular to the tangent of the curve of $f(x)$ at $(x_0, y_0)$ what will the slope of this perpendicular line be. Answer: if a given line has slope $M$, the perpendicular line has slope $\displaystyle \frac{-1}{M}.$ Next question: You have a (perpendicular) line with a specific slope, and one of the endpoints of the line is the point on $y = -x^2$ at $(x_0, y_0)$. ...see next comment – user2661923 Nov 05 '21 at 12:37
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    Therefore, you should be able to identify the point $(x_1, y_1)$ on the line $y = 5 - 2x$ such that the perpendicular from $(x_0, y_0)$ passes through the point $(x_1, y_1)$. Then, having identified the two point $(x_0, y_0)$ and $(x_1, y_1)$, all that remains is to compute the distance between the two points. ...see next comment – user2661923 Nov 05 '21 at 12:40
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    As other comments have already suggested, the approach outlined by my comments (as originally suggested by the comment of David K) is not the only approach to the problem. You can also let $(x,y)$ be any point on the curve $y = f(x) = -x^2$, and compute the minimum distance between this point and the line $y = 5 - 2x$. This gives you a distance function in terms of any point $(x,y)$ on the curve $y = f(x) = -x^2.$ ...see next comment – user2661923 Nov 05 '21 at 12:44
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    Then, you can take the first and second derivative of this distance function (or as the comment of @ClaudeLeibovici suggested) the square of the distance function. Having done this, you can use the first and second derivatives of the distance function to determine the minimum value possible from the distance function (or the square of the distance function). You end up determining the minimum distance possible without even calculating the (2nd) endpoint of the corresponding line segment. – user2661923 Nov 05 '21 at 12:46
  • I noticed that you already left a comment following the posted question, where you put analysis in the comment. This is bad, per mathSE protocol. I suggest that you take another look at my comments, and everybody else's comments, and then edit your question directly, showing every effort that you have made, in as clear a presentation as possible, and include any questions that you need answered in order to conquer the problem. Then, your posting will become high enough quality as to permit any mathSE reviewer to post a complete answer, rather than helpful comments. – user2661923 Nov 05 '21 at 12:52

2 Answers2

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Take the parametric point on the parabola as $(t,-t^2)$ it perpendicular distance from $y=5-2x$ is $D(t)=|(-t^2-5+2t)|/\sqrt{5}$, next $dD/dt=0$ leads to $t=1$. So the required point is $(1,-1).$

Z Ahmed
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Consider the squared distance between the points $(u,-u^2)$ and $(v, 5-2v)$, which is

$$(u-v)^2+(u^2+5-2v)^2.$$

To find the minimum, which is a stationary point, we cancel the gradient components, giving the system

$$\begin{cases}2(u-v)+4u(u^2+5-2v)=0,\\2(u-v)-4(u^2+5-2v)=0.\end{cases}$$

If $u^2+5-2v\ne0$, we have $u=-1$ and $v=\dfrac{13}3$. If $-u^2+5-2v\ne0$, that implies $u=v$ and $u=v=1\pm2i$.

Bobby Laspy
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