# parametric word problem

• Jul 29th 2010, 08:36 PM
Veronica1999
parametric word problem
1. A bug is moving along the line 3x + 4y = 12 with constant speed 5 units per second.
The bug crosses the x-axis when t=0 seconds. It crosses the y-axis later. When? t=1
Where is the bug when t=2? (-4 , 6) when t= -1? ( 8, -3)
when t = 1.5 ? (-2 ,18/4)
What does a negative t value mean?(Headbang)

I could not answer the last problem.

2. Find the parametric equations to describe the line
3x + 4y = 12. ( t , -3/4t +3 ) Use your equations to find the coordinates for the point that is three-fifths of the way from (4,0) to (0,3). (8/5 , 9/5)
By calculating some distances verify that you have the correct point.

• Jul 30th 2010, 02:03 AM
HallsofIvy
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Originally Posted by Veronica1999
1. A bug is moving along the line 3x + 4y = 12 with constant speed 5 units per second.
The bug crosses the x-axis when t=0 seconds. It crosses the y-axis later. When? t=1

When y= 0, 3x= 12 so x= 4. The bug crosses the x-axis at (4, 0). When x= 0, 4y= 12 so y= 3. The bug crosses the y-axis at (0, 3). The distance from (4, 0) to (0, 3) is [tex]\sqrt{4^2+ 3^2}= 5. Since the bug's speed is 5 units per second, it requires 1 second to cross the from the x-axis to the y-axis. Your answer is correct.

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Where is the bug when t=2? (-4 , 6) when t= -1? ( 8, -3)
when t = 1.5 ? (-2 ,18/4)
Since, in each second, the bug moves to the left 4 units and up 3 units, for t= 2, just subtract 4 from the x value when t= 1 and add 3 to the y value when t= 1: (0-4, 3+ 3)= (-4, 6) as you say. When t= -1, since we are "going backwards", do the opposite. Add 4 to the x value when t= 0 and subtract 3 from the y value when t= 0: (4+ 4, 0- 3)= (8, -3), again as you say. For t= 1.5, the simplest thing to do is to "average" the positions at t= 1 and t= 2: ((0- 4)/2, (3+ 6)/2)= (-2, 9/2) which is the same as your (-2, 18/4) though I would advise you to reduce the fraction.

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What does a negative t value mean?(Headbang)
It means the time before t= 0 of course. That is, any time before the bug was at (4, 0).

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I could not answer the last problem.

2. Find the parametric equations to describe the line
3x + 4y = 12. ( t , -3/4t +3 )

The change in x value going from (4,0) to (0, 3) is -4. If we set x= 4- 4t, when t= 0, x= 4 and when t= 1, x= 4- 4= 0 as we want. The change in y value going from (4, 0) to (0, 3) is +3. If we set y= 3t+ 0= 3t, when t=0, y= 0 and when t= 1, y= 3. In general, the straight line from (a, b) to (c, d) can be represented by the parametric equations x= a+ (c-a)t and y= b+ (d-b)t. Parametric equations for the line are x= 4- 4t and y= 3t.

Those are not the equation given because there exist an infinite number of sets of parametric equations to describe any path. If you take their t (which I will call t' to distinguish it from my t) to be my 4- 4t, then, of course, x= 4- 4t= t' and, since -4t= t'- 4 or t= 1- (1/4)t', y= 3t= 3(1- (1/4)t')= 3- (3/4)t' as given. One difference is that their "t" is NOT the time while in my equations it is- taking t= 0 in their equation gives x= 0, y= 3 which is where the but was at time 1 second.

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Use your equations to find the coordinates for the point that is three-fifths of the way from (4,0) to (0,3). (8/5 , 9/5)
Using my equations (in which t does correspond to time) the bug will be at (4, 0) when t= 0 and at (0, 3) when t= 1. It will be 3/5 of the way between when t= 3/5. Put t= 3/5 in my equations.

Using the equations you give, since x= t, the bug is at (4, 0) when t= 4 and at (0, 3) when t= 0. "3/5" of the way from 4 to 0 is 1- 3/5= 2/5 of the way from 0 to 4: (2/5)(4)= 8/5. Set t= 8/5 in the parametric equations.

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By calculating some distances verify that you have the correct point.
After you have found the point that is 3/5 of the way, calculate the distance from it to (4, 0). It should be 3/5 of the distance from (4, 0) to (0, 3) (which, recall, is 5).

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