A point P lies on the center of a ladder. As the ladder slips, what type of curve does P trace?
I know that it traces a circle, but I don't know how to show this.
I know that it traces a circle, but I don't know how to show this.
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Assume that the midpoint traces out a circle. Then we must find the equation of the circle.
Assume that the ladder is of length 2. Let the ladder touch the y-axis at (0, h) and the x-axis at (l, 0) at any time t. Then the equation of the line that represents the ladder is y = (-h/l)x + h (at any time t).
If we take the case where l = h, then the midpoint is (1/√2, 1/√2). Now, we have three points of the circle, (0, 1), (1, 0). (1/√2, 1/√2), and from this, it can be determined that the equation of the circle is x^2 + y^2 = 1.
Then x^2 + y^2 = 1 and y = (-h/l)x + h must intersect at x = l/2, y = h/2.
x^2 + y^2 = (l/2)^2 + (h/2)^2 = (l^2 + h^2)/4 = 4/4 = 1
and that point goes through the line (you can check).
So we have verified that the ladder's midpoint does indeed trace a circle.
Assume that the ladder is of length 2. Let the ladder touch the y-axis at (0, h) and the x-axis at (l, 0) at any time t. Then the equation of the line that represents the ladder is y = (-h/l)x + h (at any time t).
If we take the case where l = h, then the midpoint is (1/√2, 1/√2). Now, we have three points of the circle, (0, 1), (1, 0). (1/√2, 1/√2), and from this, it can be determined that the equation of the circle is x^2 + y^2 = 1.
Then x^2 + y^2 = 1 and y = (-h/l)x + h must intersect at x = l/2, y = h/2.
x^2 + y^2 = (l/2)^2 + (h/2)^2 = (l^2 + h^2)/4 = 4/4 = 1
and that point goes through the line (you can check).
So we have verified that the ladder's midpoint does indeed trace a circle.
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Take a ladder that intersects the ground at (a, 0) and (0, b). Assume that the ladder falls at a rate of -x ft/s. The ladder is b - xt feet of the ground after t seconds. Using the Pythagorean Theorem, the base of the ladder is at (√(a^2 + 2xbt - x^2*t^2), 0) at time t. (cont'd)
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Now, the top of the ladder is located at (0, b - xt) while the base is at (√(a^2 + 2xbt - x^2*t^2), 0), so the center of the ladder is at (√(a^2 + 2xbt - x^2*t^2)/2, (b - xt)/2). Taking the center to be at the intersection of the ground the wall, (cont'd)
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the distance between (0, 0), the center of the circle, and (√(a^2 + 2xbt - x^2*t^2)/2, (b - xt)/2) is √(a^2 + b^2)/2, a constant. Thus, the distance between the center of the ladder and the intersection between the ground the wall is constant, which, by definition, traces a circle. Q.E.D.
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since the base of the ladder does not move, the point P is a constant distance from that point.the definition of a circle is that the points are equidistant from the center.