1. ## Normal question

I've posted this here because I'm concerned with finding the segment through use of a derivative for a minimum. I have the gradient of the tangent and normal to the graph, but I'm not sure how to find where the line OP intersects with the existing curve.

2. I'm not sure if this is the method that you want to solve this with, although it does use derivatives.

Use the distance formula to minimize the distance. One point is (0,0) and the other is going to be some point (x,y) on the line, so we can use that fact to replace y with 10-2x. So the two points are (0,0) and (x,10-2x).

The distance formula solves for D and has a square root, but you should notice that if you minimize D^2 = D*D, then you minimize D as well and it makes the equation much easier to work with.

So: $\displaystyle D^2=(x-0)^2+(10-2x)^2$

Now you differentiate and find the critical x-value, plug that into y=10-2x and you should have your point.

3. Would you mind elaborating on this 'distance formula'? The official answers used it as well, but I'd never seen it before. That is, they also had 10-2x, which seemed completely random to me.
If it's some new concept that will be hard to learn, feel free to try a trigonometric approach

4. Originally Posted by Naur
I've posted this here because I'm concerned with finding the segment through use of a derivative for a minimum. I have the gradient of the tangent and normal to the graph, but I'm not sure how to find where the line OP intersects with the existing curve.
That is, after all, exactly what this problem asks! 2x+y= 10 is the same as y= 10- 2x so every point on that line is of the form (x, 10-2x). The distance from that point to the origin is $\displaystyle D= \sqrt{x^2+ (10-2x)^2}$ and that is what you want to simplify.

A hint worth knowing: The derivative of $\displaystyle A^2$ with respect to x is 2A dA/dx and, for A non-zero, is 0 if and only if dA/dx is. That is, instead of finding critical points of D, with that complicating square root, it is sufficient to find the critical points of $\displaystyle D^2= x^2+ (10-2x)^2$ which is much simpler.

By the way, Since you are dealing with a straight line you don't really need the derivative (which is basically a way of replacing a curve with its tangent line). The shortest distance from a point to a line is along the line perpendicular to the given line. Since y= 10- 2x has slope -2, the slope of any perpendicular line is -1/(-2)= 1/2 and the line through the origin with that slope is y= (1/2) x. Find the point of intersection of y= 10- 2x and y= (1/2)x.

Try it both ways and see if you don't get the same answer.

5. I found x = 4 for the first method, and x = 6.6667 for the second. I'm obviously doing something wrong. I'd say 6.6667 is right, as I'm quite sure I got that one right...

Got one of my two maths exams in an hour. Wish me luck! My ability is based on a lot of the help obtained on this site.