13
$\begingroup$

I could use some help solving the following problem. I have many more like this but I figured if I learn how to do one then I can figure out the rest on my own. Thanks in advance!

A curve described by the equation $y=\sqrt{16x^2+5x+16}$ on a Cartesian plane. What is the shortest distance between coordinate $(2,0)$ and this line?

$\endgroup$
3
  • $\begingroup$ Can you elaborate on what it means for a line to follow a curve? Is the question simply to find the shortest distance between $(2,0)$ and that curve? $\endgroup$
    – Kaj Hansen
    Commented May 3, 2017 at 22:28
  • $\begingroup$ the equation of the line. ill edit to clear that up. thx $\endgroup$ Commented May 3, 2017 at 22:29
  • $\begingroup$ yes, the solution would be to find the shortest distance between the point (2,0) and the curve $\endgroup$ Commented May 3, 2017 at 22:32

3 Answers 3

10
$\begingroup$

Start by finding the distance from some point on the curve to $(2,0)$ in terms of $x$. Using the distance formula, we get $$D=\sqrt{(x-2)^2+(\sqrt{16x^2+5x+16}-0)^2}$$ $$D=\sqrt{x^2-4x+4+16x^2+5x+16}$$ $$D=\sqrt{17x^2+x+20}$$ This will end up being a messy derivative. However, since the distance $D$ will never be negative, we can minimize $D^2$ instead of $D$ and still get the same answer. So now we get $$D^2=17x^2+x+20$$ $$\frac{dD^2}{dx}=34x+1$$ Now we set this equal to $0$ and solve for $x$: $$34x+1=0$$ $$x=-\frac{1}{34}$$ So the distance is minimized at $x=-\frac{1}{34}$, and to find the minimum distance, simply evaluate $D$ when $x=-\frac{1}{34}$.

$\endgroup$
2
  • 1
    $\begingroup$ Let's suppose that D can be negative, then can we still simplify using $D^2$? $\endgroup$
    – Piano Land
    Commented Apr 11, 2018 at 6:47
  • $\begingroup$ @PianoLand No, but there is something wrong with your D formula if you can't do that. $\endgroup$
    – Xwtek
    Commented May 27, 2021 at 2:46
4
$\begingroup$

Since distance is positive and the square root function is increasing, it suffices to find the smallest value the squared distance between $(x,y)$ on the curve and the point $(2,0)$ can take. This is $$ L(x) = (x-2)^2 + (y-0)^2 = (x-2)^2+y^2 = x^2-4x+4 + 16x^2+5x+16 = 17x^2+x+20. $$ A minimum can only occur if $L'(x)=0$. So $$ L'(x) = 34x+1, $$ so there is a turning point at $x=-1/34$. Moreover, the derivative is negative on the left and positive on the right, so the point is a minimum. Hence the minimum distance is $$ \sqrt{L(-1/34)} = \sqrt{\frac{1359}{68}} \approx 4.47. $$

$\endgroup$
4
  • 1
    $\begingroup$ Why not a negative $x$? $\endgroup$ Commented May 3, 2017 at 22:50
  • $\begingroup$ Brain fart. Thanks for pointing that out... $\endgroup$
    – Chappers
    Commented May 3, 2017 at 22:56
  • $\begingroup$ Truly an anomaly from someone with a 33.5k reputation! $\endgroup$ Commented May 3, 2017 at 22:57
  • $\begingroup$ Let us hope so! $\endgroup$
    – Chappers
    Commented May 3, 2017 at 23:06
1
$\begingroup$

Hint 1: take a point $(x, y)$ on the curve, calculate its distance from $(2, 0)$. The fact that the point is on the curve allows you to express that distance in terms of $x$ alone. Then find the minimum (but check the second hint first).

Hint 2: instead of minimizing the distance, minimize the square of the distance.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .