4
$\begingroup$

I have no idea how to do this at all. "Find the point on the line $y=2x+3$ that is closest to the origin" I am just not very smart or creative so I have no idea how to do this. I graphed it and I think I can assume that it will be closest to the origin near (1.5, 0) through (0, 3). Those are the x and y intercepts that I found (possibly correctly) and I think the formula for finding it close to the origin would be $x+y < 1.5$ since 1.5 is the closest point I have right now. Idealy I want x+y to equal 0 but it won't since that is probably not an intercept.

From here I have no idea what to do, I have tried a couple things but nothing even makes sense. I have never done a problem like this.

  • 0
    I don't understand how I would do that.2011-10-21

3 Answers 3

5

There are purely geometric ways to do the problem, but since it arose in a calculus course, let's use standard max/min procedures.

Think of a "general" (unspecified) point $P$ on the line $y=2x+3$. Suppose that the $x$-coordinate of $P$ is $x$. Then the $y$-coordinate of $P$ is $2x+3$. So $P$ is the point $(x,2x+3)$.

What is the distance from $(x,2x+3)$ to the origin? By what I hope is a familiar formula, this distance is $\sqrt{(x-0)^2 +(2x+3-0)^2}.$ This is because in general the distance between $(a,b)$ and $(c,d)$ is $\sqrt{(a-c)^2+(b-d)^2}$. Apply this formula, using $a=x$, $b=2x+3$, and $c=0$, $d=0$.

Alternately, you can find the distance from $(x,2x+3)$ to the origin by drawing a picture and using the Pythagorean Theorem.

We want to choose $x$ so as to minimize the distance from $P$ to the origin, so we want to minimize $\sqrt{x^2+(2x+3)^2}$.

We can now let $f(x)=\sqrt{x^2+(2x+3)^2}$ and find the $x$ that makes $f(x)$ smallest by the usual derivative process.

But there is a little trick that simplifies things. If $\sqrt{x^2+(2x+3)^2}$ is as small as possible, then so is $x^2+(2x+3)^2$, and vice-versa. So we try to minimize the square of the distance. Let $g(x)=x^2+(2x+3)^2.$ Can you now find the value of $x$ that makes $g(x)$ as small as possible?

Added: After some simplification, f'(x)= \frac{5x+6}{\left(x^2+(2x+3)^2\right)^{1/2}}. If you take the suggested alternate route, you will find that g'(x)=10x+12.

  • 0
    Let $f(x)=\sqrt{x^2+(2x+3)^2}$. Differentiate, find where $f'(x)=0$. Or else let $g(x)=x^2+(2x+3)^2$, find $g'(x)$, find where $g'(x)=0$. You can certainly do these things.2011-10-21
1

Let's observe picture bellow.If we define line $q$ as $q \perp p$ , and $(0,0) \in q$ we may write:

$k_pk_q=-1 \Rightarrow k_q=\frac{-1}{2}$

Since $(0,0) \in q\Rightarrow 0=\frac{-1}{2}0+n_q\Rightarrow n_q=0$

So,in order to find point $A$ that is closest to the origin you have to solve following system:

\begin{cases}

y=2x+3 \

y=\frac{-1}{2}x

\end{cases}

enter image description here

  • 0
    it means that $q$ is perpendicular to $p$ , $y=k_px+n_p$ is explicit form of the line $p$2011-10-21
1

Let's go through my checklist from the previous optimization problem. To that checklist let me add one other thing:

When you start working on the problem, you don't need to have already visualized the entire path-to-a-solution. Just try to make sure you know what your current step is, and what your next step is likely to be. Often, as you walk along the path towards a solution you start seeing a little further ahead and then you can take the next step; and from there, you can see a bit more and figure out what the next step needs to be, and so on.

  1. Read the problem carefully.

    Find the point on the line $y=2x+3$ that is closest to the origin.

  2. Identify the quantity we want to optimize.

    We want to optimize the distance from a point $P$ on the line, to the origin. Call this distance $D$. We want to find the minimum.

  3. Draw a picture; label any quantities that will play a role in computing the distance.

    You have a line $L$ of slope $2$ and intercept $3$. Place a point $P$ on the line, with coordinates $(a,b)$. Draw the distance from the point to the origin:

                y             |   /              |  / L: y = 2x+3             | /             |/             /            /|                 / |   P=(a,b)*  |         /   |  ------/----+-------------  x       /     |       /      | 
  4. Identify the constraints we have, in terms of the labels in the graph.

    The constraints are that the point $P=(a,b)$ has to be on the line $y=2x+3$. That means that if you plug in $(a,b)$ into the equation, you have to get an equality. That is, you must have $b=2a+3$.

  5. Write down a function for the quantity we want to optimize.

    The distance from $P$ to the origin is: $D(P) = \sqrt{(a-0)^2 + (b-0)^2} = \sqrt{a^2+b^2}.$ This is an application of the general "distance formula": the distance between $(a,b)$ and $(x,y)$ is $\sqrt{(a-x)^2 + (b-y)^2}.$

  6. Use the constraints from 4 to change the function from 5 into a function of a single variable. Identify the domain of the function.

    Since $b=2a+3$, then just plug in this into the formula for $D$: $\begin{align*} D &= \sqrt{a^2 + b^2}\\ &= \sqrt{a^2 + (2a+3)^2}\\ &= \sqrt{a^2 + (4a^2 + 12a + 9)}\\ &= \sqrt{5a^2 + 12a + 9}. \end{align*}$ So we have a function of a single variable: $D(a) = \sqrt{5a^2+12a+9}$. What does this function do? You give it the $x$-coordinate of a point on the line $L$ (this coordinate is $a$), and it tells you the distance from the point on the line with $x$-coordinate $a$ to the origin.

    What is the domain? In principle, all real numbers. Every point on the line. But in practice you don't actually need to consider all the point. Clearly, once you get far enough from the origin the distance will only grow, so you only need to consider some finite closed interval that is big enough to include all the points that "really matter". From the picture, the only points that matter are the points on the line that are on the second quadrant, so it is enough to consider those values of $a$ that are between $-\frac{3}{2}$ (the $x$-intercept) and $0$. So we can restrict ourselves to $[-\frac{3}{2},0]$. (I'll note what to do if you don't realize this after we are done).

  7. Optimize the function "in the usual way".

    We want to find the absolute minimum of the function $D(a) = \sqrt{5a^2+12a+9}$ on $[-\frac{3}{2},0]$. The function is defined everywhere, and is continuous, so we need to find the critical points in the interval, then evaluate at the endpoints and the critical point. The smallest value we get is what we want.

    To find the critical points, we take the derivative: $\begin{align*} D(a) &= \left(5a^2 + 12a + 9\right)^{1/2}\\ D'(a) &= \frac{1}{2}\left(5a^2 + 12a + 9\right)^{-1/2}\left(5a^2 + 12a + 9\right)'\\ &= \frac{1}{2}\left(\frac{1}{\sqrt{5a^2+12a+9}}\right)\left(10a + 12\right)\\ &= \frac{10a+12}{2\sqrt{5a^2+12a+9}}\\ &= \frac{2(5a+6)}{2\sqrt{5a^2+12a+9}}. \end{align*}$ The critical points are where the derivative is undefined and where the derivative is zero. It could be undefined where the denominator is $0$ or the square root is not defined. But looking back at our construction of $D$, we notice that $\sqrt{5a^2 + 12a + 9} = \sqrt{a^2 + (2a+3)^2}.$ This is sum of squares, so the thing inside the radical is never negative; and it cannot be equal to $0$, because you would need both squares to be zero, which requires $a=0$ and $2a+3=0$, and both cannot happen at the same time. So the derivative is always defined.

    So the only critical points are where the derivative is $0$, which are the points where the numerator is $0$. $\begin{align*} 5a + 6 &= 0\\ 5a &= -6\\ a &= -\frac{6}{5}. \end{align*}$ So the only critical point is $a=-\frac{6}{5}$.

    We now evaluate at the endpoints of the interval, and at the critical points. The smallest value will be the minimum we want: $\begin{align*} D\left(-\frac{3}{2}\right) &= \sqrt{5\left(-\frac{3}{2}\right)^2 + 12\left(-\frac{3}{2}\right) + 9}\\ &= \sqrt{5\left(\frac{9}{4}\right) - \frac{12(3)}{2} + 9}\\ &= \sqrt{\frac{45}{4} - \frac{36}{2} + 9}\\ &= \sqrt{\frac{45}{4} - \frac{72}{4} + \frac{36}{4}}\\ &= \sqrt{\frac{81-72}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}. \end{align*}$ (Alternatively, if you are willing to think about it, the point $(-\frac{3}{2},0)$ is clearly $\frac{3}{2}$ units away from the origin).

    $\begin{align*} D(0) &= \sqrt{5(0)^2 + 12(0) + 9}\\ &= \sqrt{9} = 3. \end{align*}$ (Again, you could just note that $(0,3)$ is $3$ units away from the origin). $\begin{align*} D\left(-\frac{6}{5}\right) &= \sqrt{5\left(-\frac{6}{5}\right)^2 + 12\left(-\frac{6}{5}\right) + 9}\\ &= \sqrt{5\left(\frac{36}{25}\right) - \frac{12(6)}{5} + 9}\\ &= \sqrt{\frac{36(5)}{25} - \frac{72}{5} + 9}\\ &= \sqrt{\frac{36}{25} - \frac{72}{5} + \frac{45}{5}}\\ &= \sqrt{\frac{81-72}{5}}\\ &= \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} \end{align*}$

    Now we compare the three values we got: $\frac{3}{2}$, $3$, and $\frac{3}{\sqrt{5}}$. The smallest value is the latter, because it's the one with largest denominator.

    So the minimum is $\frac{3}{\sqrt{5}}$, which is achieved when $a=-\frac{6}{5}$.

    This value of $a$ corresponds to the point with $y$ coordinate $b=2\left(-\frac{6}{5}\right)+3 = -\frac{12}{5}+3 = \frac{3}{5}$.

    So the point closest to the origin will be $(-\frac{6}{5},\frac{3}{5})$, which is $\frac{3}{\sqrt{5}}$ away from the origin.

  8. Check everything makes sense.

    The point we found is on the line, and is on the second quadrant, as expected.

  9. Answer the question. A word problem should have a word answer.

    The point on the line $y=2x+3$ that is closest to the origin is $(-\frac{6}{5},\frac{3}{5})$.

And we're done.


Comment 1. What happens if you don't realize that you can restrict yourself to the second quadrant?

Then you will consider the function $D(a)$ with domain all real numbers. You are not generally guaranteed that a function on the entire real line will have an absolute minimum. However, here, $D(a) = \sqrt{5a^2 +12a + 9},$ notice that $\lim_{a\to\infty} D(a) = \infty = \lim_{a\to -\infty}D(a),$ and this guarantees that $D(a)$ has to have an absolute minimum (it has no absolute maximum), and that the absolute minimum is at a critical point. Since the function has one and only one critical point, that critical point will be where the minimum is.


Comment 2. There is a clever trick here to avoid having to deal with the square root in the distance formula: because the distance is always positive, minimizing the distance is the same as minimizing the square of the distance. This because of $r$ and $s$ are nonnegative real numbers, then $r\lt s$ is true if and only if $r^2\lt s^2$ is true. So instead of using the function $D(a)$, we can use the function $d(a) =\left(D(a)\right)^2 = 5a^2 + 12a + 9,$ and look for the absolute minimum of $d(a)$. In the end, we will get the same answer (that the minimum is achieved when $a=-\frac{6}{5}$), but the work will be simpler because there won't be any square roots to deal with. However, you want to remember that you aren't actually minimizing the distance, but the square of the distance. So if the question were to find the smallest distance (as opposed to finding the closest point), you would need to plug the closest point into $D$, rather than into $d$, to find the smallest distance.

  • 0
    @Jo$r$dan: Remember point 1 of the checklist? "Read the problem. *Carefully*." Should I add "Remember it"?2011-10-22