A line segment has one endpoint, A, on the x-axis and the other endpoint, B, on the y-axis. It passes through the point (1,2). If O is the point (0,0), for what value of angle OAB is the length of the line segment AB a minimum? What is the minimum length?
let A(x,0) , B(0,y) and P(1,2) be the above points
let L = AB
L^2 = x^2 + y^2
but
(1-x)/2 = 1/(2-y)
2 - y - 2x + xy = 2
y(x-1) = 2x
y = 2x/(x-1)
L^2 = x^2 + 4x^2/(x-1)^2
2L dL/dx = 2x - 8x/(x-1)^3 , used quotient rule on last part and simplified
= 0 for a min of L
2x = 8x/(x-1))^3
(x-1)^3 = 4
x-1 = 4^(1/3)
x = 4^(1/3) + 1 = appr 2.587
y = 3.26
let Ø be the acute angle formed with the x-axis
tanØ = y/x = 3.26/2.587 = ...
Ø = 51.56°
L^2 = 2.587^2 + 3.26^2
..
..
L = appr 4.16
if the line goes through (x,0),
tanθ = 2/(x-1)
so, secθ = √(x^2-2x+5) / (x-1)
the distance AB is
d = x secθ = x√(x^2-2x+5) / (x-1)
dd/dx = (x^3-3x^2+3x-5)/[(x-1)^2 √(x^2-2x+5)]
the denominator is never zero for x>1, so dd/dx=0 when
x^3-3x^2+3x-5 = 0
x = 1+∛4
so,
tanθ = 2/∛4 and θ=51.56°
d = 1+∛4 secθ = 4.16
Or, we could use d as a function of θ
2/(x-1) = tanθ, so x = 1+2cotθ
d = x secθ = (1+2cotθ)secθ = secθ + 2cscθ
dd/dθ = secθtanθ - 2cscθcotθ
= (sin^3θ - 2cos^3θ)/(sin^2θcos^2θ)
dd/dθ = 0 when sinθ = ∛2 cosθ
θ = 51.53°
pretty close to the other answer.
dang! we got the same answer!
To find the value of angle OAB that minimizes the length of line segment AB, we need to determine the slope of the line passing through points A(0,0), B(x,y), and (1,2).
Let's start by finding the equation of the line passing through A and B. Since A is on the x-axis and B is on the y-axis, the line can be written as y = mx, where m is the slope.
Given that (1,2) lies on the line, we can substitute these coordinates into the equation to solve for m:
2 = m(1)
m = 2/1
m = 2
So, the equation of the line passing through A and B is y = 2x.
To find the minimum length of line segment AB, we need to calculate the distance between A and B.
The length of AB can be determined by finding the y-coordinate of the point B where it intersects the line y = 2x.
Substituting this equation into y = 2x, we get:
2x = 2x
x = 1
Therefore, B is located at (1,2).
To calculate the distance between A(0,0) and B(1,2), we can use the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Plugging in the coordinates, we have:
Distance = sqrt((1 - 0)^2 + (2 - 0)^2)
Distance = sqrt(1^2 + 2^2)
Distance = sqrt(1 + 4)
Distance = sqrt(5)
Therefore, the minimum length of line segment AB is sqrt(5).
Now let's determine the value of angle OAB when the length of AB is at a minimum.
The slope of the line passing through O and A is given by the equation:
slope(OA) = (y2 - y1) / (x2 - x1)
Plugging in the coordinates of O(0,0) and A(0,0), we have:
slope(OA) = (0 - 0) / (0 - 0)
slope(OA) = 0 / 0
Since the slope is undefined, we cannot determine the value of angle OAB since there is no defined line.
Therefore, there is no specific value of angle OAB when the length of line segment AB is at a minimum.