Shortest distance possible...

**realintegerz** · November 23rd 2008, 07:12 PM

I see the graphs of these, using Geogebra, and I don't know how I can prove them..

1) Point A is @ (0,2) and point B is on the graph of y=x^2 + 1, what should be the approximate location of B so that the distance from A to B is as short as possible?

2) Point A is @ (3,1) and B is on a circle with center (1,2) and radius of 4. What is the closest point B should come to point A? Approximate to nearest thousandth.

And I'm having difficulty with this problem as well, but it doesn't use the same ideology of the above two.

A window is being built such that the bottom is a rectangle and the top is a semicircle. If there are 12 meter of framing material what should be the width of the window in order to let in the most light?

Any help is appreciated...

**realintegerz** · November 24th 2008, 08:03 PM

Ok, I got help [from outside MHF] for the last 2 problems but I still don't get the 1st one...

can anyone please help?

**o_O** · November 24th 2008, 08:14 PM

Let $d$ be the distance from (0,2) to the curve.

Now if you remember your distance formula, ( $d = \sqrt{(x-x_1)^2 + (y-y_1)^2}$ ), we have this relationship: $d^2 = (x-0)^2 + (y-2)^2 \ \iff \ d^2 = {\color{blue}x^2} + (y-2)^2$

Now, we can rearrange the equation of the curve to get: $\color{blue}x^2 = y - 1$

So our relationship becomes: $d^2 = {\color{blue}(y-1)} + (y-2)^2 \ \iff \ d^2 = y^2 - 3y + 3$

Differentiate both sides and solve for when the derivative $d'$ is equal to 0 (as this will be the point where you have a minimum).

Two things to note:
(1) We are working in terms of $y$ . So whatever your answer will be, it will give you the y-coordinate in which the distance is shortest from (0,2) to the curve. Plug this in to your equation of the curve to get the x-coordinates.

(2) $d$ is a function of y. So you must implicitly differentiate.

**realintegerz** · November 24th 2008, 08:27 PM

Ok, so I solved for y as d was 0 on the left side

I got y = 3/2 + sq. root ( - 3/4), assuming I ignored the negative root, y equaled 2.36

and x is 1.16....

I looked at a graph and the shortest distance is supposed to be 0.87, and the coordinates for B are (0.61, 1.38) to (0.79, 1.62)

**realintegerz** · November 24th 2008, 09:12 PM

I got help earlier in the day about this problem, and my friend said that B is closest when it is perpendicular to the function

**o_O** · November 24th 2008, 10:02 PM

Are you sure you solved it correctly?

You should get: $y = \frac{3}{2} \ \Rightarrow \ x = \pm \frac{\sqrt{2}}{2}$ ..... i.e., $\left(\pm \frac{\sqrt{2}}{2}, \frac{3}{2}\right)$

_______________________

Yes, you can use the fact that the shortest distance from the point to a curve is when the line segment is perpendicular. This involves solving: ${\color{red}\frac{y-2}{x-0}} = {\color{blue}-\frac{1}{y'}} \ \iff \ \frac{x^2 - 1}{x} = -\frac{1}{2x}$

The red represents the slope of the line segment from (0,2) to the a point on your curve. The blue represents the negative reciprical of the tangent at a given point. Whatever x values we get from solving this will be the values such that the line joining (0,2) to the curve is perpendicular to the curve. Note that x = 0 cannot be a solution to the equation but drawing a line segment from (0,2) to (0,1) fits the description above.

Much shorter actually.

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