Let S be the locus of all points
(x, y) in the first quadrant such that
(x/t)+(y/(1− t))= 1 for some t with 0<t<1. Find the area of S.
S is a collection of line segments with x- and y-intercepts at
(0,1-t) and (t,0)
Divide the interval (0,1) into n intervals of width 1/n. Then the two lines at t=k/n and t=(k+1)/n intersect at
y = (n-k)(n-k-1)/n^2
So, adding the (k+1)st line adds a thin triangle of base 1/n and height (n-k)(n-k-1)/n^2. It has area
(n-k)(n-k-1)/(2n^3)
Thus, the sum of the n triangles is
n-1
∑(n-k)(n-k-1)/(2n^3) = (1/6)(1 - 1/n^2)
k=0
as n->∞ the sum -> 1/6
So, the area of S is 1/6
*whew*
any other ideas?
To find the area of the locus S, we first need to understand the equation representing S and then use integral calculus to calculate the area.
The equation (x/t) + (y/(1 - t)) = 1 represents a line passing through the point (0, 1) and (t/(1 - t), 0) for some t with 0 < t < 1.
Let's analyze the equation further. By multiplying through by t(1 - t), we can simplify the equation to:
x(1 - t) + yt = t(1 - t)
Rearranging, we get:
x(1 - t) = t(1 - t) - yt
Dividing both sides by (1 - t), we obtain:
x = t - yt/(1 - t)
Since we are considering the first quadrant, both x and y are positive. Therefore, the equation represents a line segment within the first quadrant.
To calculate the area of S, we need to compute the integral of the line segment from x = 0 to x = t/(1 - t).
Now, let's evaluate the integral.
∫[0, t/(1 - t)] (y dx)
First, let's express y in terms of x using the given equation:
(x/t) + (y/(1 - t)) = 1
Simplifying, we have:
y = (1 - t) - (x/t)
Substituting this expression for y in the integral, we get:
∫[0,t/(1 - t)] [(1 - t) - (x/t)] dx
Expanding the integral:
∫[0,t/(1 - t)] (1 - t) dx - ∫[0,t/(1 - t)] (x/t) dx
Integrating the first term:
[(1 - t) x] evaluated from 0 to t/(1 - t)
Simplifying:
[(1 - t) (t/(1 - t))] - [(1 - t) (0)]
= t - t^2/(1 - t)
Integrating the second term:
[x/t] evaluated from 0 to t/(1 - t)
Simplifying:
[t/(1 - t)] - [0]
= t/(1 - t)
To find the area of S, we subtract the two integrals:
Area S = (t - t^2/(1 - t)) - (t/(1 - t))
Simplifying:
Area S = (t - t^2 - t)/(1 - t)
Area S = (-t^2)/(1 - t)
Finally, we need to find the definite integral of Area S with respect to t, from t = 0 to t = 1:
∫[0,1] (-t^2)/(1 - t) dt
Integrating:
[(1/3) t^3 + (1/2) t^2] evaluated from 0 to 1
[(1/3) (1)^3 + (1/2) (1)^2] - [(1/3) (0)^3 + (1/2) (0)^2]
= (1/3) + (1/2)
= 5/6
Therefore, the area of S is 5/6 square units.