Rate of Separation of Two Moving Points

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turophile

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The problem:

A point moves along the parabola y = x[sup:36463p5o]2[/sup:36463p5o] with its x-coordinate increasing at 2 units per second. Another point moves along the line y = x in such a way as to keep the y-coordinates of the two points the same. How fast are the points separating when they cross at (1, 1)?

My work so far:

Let f[sub:36463p5o]1[/sub:36463p5o](x[sub:36463p5o]1[/sub:36463p5o]) = y[sub:36463p5o]1[/sub:36463p5o] = x[sub:36463p5o]1[/sub:36463p5o][sup:36463p5o]2[/sup:36463p5o] and let f[sub:36463p5o]2[/sub:36463p5o](x[sub:36463p5o]2[/sub:36463p5o]) = y[sub:36463p5o]2[/sub:36463p5o] = x[sub:36463p5o]2[/sub:36463p5o]. Let a be the distance between the two points. Since the y-value of the two points is always the same, x ? 0 and a = | x[sub:36463p5o]1[/sub:36463p5o] – x[sub:36463p5o]2[/sub:36463p5o] | = | y[sub:36463p5o]1[/sub:36463p5o][sup:36463p5o]1/2[/sup:36463p5o] – y[sub:36463p5o]2[/sub:36463p5o] | . We want to find da/dt.
We are given d(x[sub:36463p5o]1[/sub:36463p5o])/dt = 2, x[sub:36463p5o]1[/sub:36463p5o] = x[sub:36463p5o]2[/sub:36463p5o] = 1, and y[sub:36463p5o]1[/sub:36463p5o] = y[sub:36463p5o]2[/sub:36463p5o] = 1.

I'm not sure that I have this set up correctly nor what the next step should be.
 
turophile said:
The problem:

A point moves along the parabola y = x[sup:2i5ebln6]2[/sup:2i5ebln6] with its x-coordinate increasing at 2 units per second. Another point moves along the line y = x in such a way as to keep the y-coordinates of the two points the same. How fast are the points separating when they cross at (1, 1)?

My work so far:

Let f[sub:2i5ebln6]1[/sub:2i5ebln6](x[sub:2i5ebln6]1[/sub:2i5ebln6]) = y[sub:2i5ebln6]1[/sub:2i5ebln6] = x[sub:2i5ebln6]1[/sub:2i5ebln6][sup:2i5ebln6]2[/sup:2i5ebln6] and let f[sub:2i5ebln6]2[/sub:2i5ebln6](x[sub:2i5ebln6]2[/sub:2i5ebln6]) = y[sub:2i5ebln6]2[/sub:2i5ebln6] = x[sub:2i5ebln6]2[/sub:2i5ebln6]. Let a be the distance between the two points. Since the y-value of the two points is always the same, x ? 0 and a = | x[sub:2i5ebln6]1[/sub:2i5ebln6] – x[sub:2i5ebln6]2[/sub:2i5ebln6] | = | y[sub:2i5ebln6]1[/sub:2i5ebln6][sup:2i5ebln6]1/2[/sup:2i5ebln6] – y[sub:2i5ebln6]2[/sub:2i5ebln6] | . We want to find da/dt.
We are given d(x[sub:2i5ebln6]1[/sub:2i5ebln6])/dt = 2, x[sub:2i5ebln6]1[/sub:2i5ebln6] = x[sub:2i5ebln6]2[/sub:2i5ebln6] = 1, and y[sub:2i5ebln6]1[/sub:2i5ebln6] = y[sub:2i5ebln6]2[/sub:2i5ebln6] = 1.

I'm not sure that I have this set up correctly nor what the next step should be.
Click to expand...

This is an interesting problem. Let's say

y[sub:2i5ebln6]1[/sub:2i5ebln6] = x[sub:2i5ebln6]1[/sub:2i5ebln6][sup:2i5ebln6]2[/sup:2i5ebln6]

(d/dt)(y[sub:2i5ebln6]1[/sub:2i5ebln6]) = 2x[sub:2i5ebln6]1[/sub:2i5ebln6] * 2 = 4x[sub:2i5ebln6]1[/sub:2i5ebln6]

and

y[sub:2i5ebln6]2[/sub:2i5ebln6] = x[sub:2i5ebln6]2[/sub:2i5ebln6]

Another point moves along the line y = x in such a way as to keep the y-coordinates of the two points the same.

d/dt[y[sub:2i5ebln6]1[/sub:2i5ebln6] - y[sub:2i5ebln6]2[/sub:2i5ebln6]] = 0 ? d/dt[y[sub:2i5ebln6]1[/sub:2i5ebln6]] = d/dt[y[sub:2i5ebln6]2[/sub:2i5ebln6]]

y[sub:2i5ebln6]2[/sub:2i5ebln6] = x[sub:2i5ebln6]2[/sub:2i5ebln6]

d/dt[x[sub:2i5ebln6]2[/sub:2i5ebln6]] = d/dt[y[sub:2i5ebln6]2[/sub:2i5ebln6]] = d/dt[y[sub:2i5ebln6]1[/sub:2i5ebln6]]

Let the distance between two particel be 's' and since the y-coordinates of the two points the same ?

ds/dt = d/dt [x[sub:2i5ebln6]2[/sub:2i5ebln6] - x[sub:2i5ebln6]1[/sub:2i5ebln6]] = d/dt [x[sub:2i5ebln6]2[/sub:2i5ebln6]] - d/dt [x[sub:2i5ebln6]1[/sub:2i5ebln6]] = d/dt[y[sub:2i5ebln6]1[/sub:2i5ebln6]] - 2 = 4x[sub:2i5ebln6]1[/sub:2i5ebln6] - 2 = 4 - 2 = 2

.
 
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