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 12-26-2008 #1 (permalink) is cute + gore   Join Date: Oct 2008 Location: Canada Posts: 681 Blog Entries: 4 iTrader: (1) 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 ~_~. __________________ If only
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 12-26-2008 #2 (permalink) killing threda since 2007   Join Date: Aug 2008 Posts: 800 Blog Entries: 5 iTrader: (2) IGN: fLick3r/MrsT/Cod Class: Bow/Saw/Cannon Level: 111/58/58 Guild: Moogles 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. __________________ if the post above this is insightful/useful, rep me plz --------------------------- fLick3r(main)/Cod(92%PvP)/Secretary(PvE)
 12-26-2008 #3 (permalink) is cute + gore   Join Date: Oct 2008 Location: Canada Posts: 681 Blog Entries: 4 iTrader: (1) 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... __________________ If only
 12-26-2008 #4 (permalink) killing threda since 2007   Join Date: Aug 2008 Posts: 800 Blog Entries: 5 iTrader: (2) IGN: fLick3r/MrsT/Cod Class: Bow/Saw/Cannon Level: 111/58/58 Guild: Moogles 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? __________________ if the post above this is insightful/useful, rep me plz --------------------------- fLick3r(main)/Cod(92%PvP)/Secretary(PvE)

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