Let A be the point (0,1) and let B be the point (2,4) . Find the length of the s

Question

Let A be the point (0,1) and let B be the point (2,4) . Find the length of the shortest path that connects points A and B if the path must touch the x - axis. In other words, the path goes from point A to somewhere on the x - axis, and then to B . (This is the "line of sight" path from A to B if the x -axis is a mirror.) See below for a sketch of such a path (not to scale). https://www.mathclass.org/WebPages/Homes/afdee97c-a004-478a-bc1f-0aafdd1b6bb2/hws/set16_S11/mirror.png The minimum length equals?

Explanation / Answer

first, we will draw a line from (0,1) to some point (x,0) on the x axis. And we will draw a line from that point to the final point, (2,4).

Now write the lengths of the two lines using the pythagorean theorem:

length of first segment = sqrt( x^2 + 1^2) = sqrt (x^2 + 1)

length of second segment = sqrt[ (2-x)^2 + 4^2 ] = sqrt ( 20 - 2x + x^2)

Now... call total length L so that

L = sqrt (x^2 + 1) + sqrt(20 - 2x + x^2)

we want to find the value of x that makes L the shortest ! So take the derivative, dL/dx , and set equal to zero

dL/dx = (1/2) (x^2+1)^-1/2 * 2x + (1/2) (20 - 2x + x^2)^-1/2 * (2x - 2)

when you set equal to zero, multply both sides by (1/2) and mutiply both sides by the sq root terms

0 = (20 - 2x + x^2)^1/2 * 2x +   (x^2+1)^1/2 * (2x-2)

or

(1 - x) (x^2+1)^1/2 = (20 - 2x + x^2)^1/2 * x

square both sides

(1-x)^2 (x^2 + 1) = (20-2x+x^2) * x^2

(1 - 2x + x^2) (x^2 + 1) = 20x^2 - 2x^3 + x^4

x^2 - 2x^3 + x^4 + 1 - 2x + x^2 = 20x^2 - 2x^3 + x^4

0 = 18^2 + 2x - 1   

Use quadratic formula. x = 0.1866

And the length of the line is

L = sqrt (x^2 + 1) + sqrt(20 - 2x + x^2) =

= sqrt(.1866^2 + 1) + sqrt(20-2*.1866 + .1866^2) =

= 5.451

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