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12-26-2008   #1 (permalink)
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Default Calculus Help

Given that for each real value of t, there is a point with coordinates (x, y), where x = 2t - 5 and y = t - 1, let m be the line formed by the set of these (x, y) points. Determine the area bounded by line m, the x-axis and the two lines x = - 2 and x = - 1

[Sol] When x = - 2, t = 3/2
........When x = - 1, t = 2

As shown on the graph..blah blah (I'm not gonna draw it, too complicated)

S = ∫ydx (-2 to -1)
...= ∫y*dx/dt*dt (3/2 to 2)
...= ∫2ydt (3/2 to 2)

I don't really get it. What's the point of switching from x to t? In fact I don't even exactly get the question ~_~.
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12-26-2008   #2 (permalink)
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so the area is basically a trapezoid of sorts; another way of putting it would be depicting the problem as a function f(x) = ax + b, and asking what's the integral from -2 to -1 of f(x) dx. or, replacing y with f(x) ∫y dx (-2 to -1)
This is the first step in your solution.

The problem here is that we have parametric coordinates. Y is defined in terms of t, so ∫ydx is unsolvable as it is (we don't know what y with respect to x is). So in order for the integral ydx to be solvable, we have to either substitute dx for t, or substitute y in terms of t. In the above situation, they chose to substitute dx for t. Here's the math they left out:

x=2t-5
dx/dt=2
dx=2dt

So...
∫ydx (x = -2 to -1)
turn into (just plug in 2dt for dx)
∫y*dx/dt*dt = ∫y*2dt (as x = -2 to -1).
We now have an integral with respect to t, but with bounds with respect to x.
To get to the second step, we need to figure out the bounds on t as x goes -2 to -1.
We do this like this:
x=2t-5
when x = -2
3=2t, t = 3/2
when x = -1
4=2t, t = 2
So the integral needs to be rewritten as follows:
∫y*2*dt (as t goes from 3/2 to 2)
which equals this.
∫2ydt (as t goes 3/2 to 2)

So the final solution would be:
∫2ydt (as t goes 3/2 to 2) =
∫2*(t-1) dt (blah)=
... simple calculus from there, i think.
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12-26-2008   #3 (permalink)
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oh...oooh oh omg i'm dumb. totally missed that concept. well thx for clearing it up. i'm just suppose to sub y with the terms of t >_> dumb...
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12-26-2008   #4 (permalink)
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np ^^ took me a while to figure out what it was asking as well. i was like... area...? why not just 1/2 (base1 + base2)*height?
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