I am confused on number 28-- the problem with the particle in the xy-plane. Could somebody please explain this a little further? This link has a copy of the problem, and the work to the solution: http://www.pascack.k12.nj.us/cms/lib5/NJ…
I would just like a better/different explanation of the problem. Thank you!
I would just like a better/different explanation of the problem. Thank you!
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Hi,
This word processor is not suited at all for discussing this problem. I’m not sure where you were needing a little more clarification but I suspect it might have been the part below the line. Accordingly, I’ll skip down to the part after the horizontal line to begin my comments. So, the components of the velocity, dx/dt and dy/dt are a right angles to each other and we can obtain their resultant speed by the Pythagorean theorem, whose basic form is, of course, c^2 = a^2 + b^2. And we are given the magnitude of that speed as 2√10. So, we have the following:
√((dx/dt)^2 + (dy/dt)^2) =2√10.
Now, we substitute the equivalent 3(dx/dt)^2 for dy/dt to get the following:
√((dx/dt)^2 + (3dy/dt)^2) =2√10 which simplifies to
√((dx/dt)^2 + 9(dy/dt)^2) =2√10.
Now, squaring both sides we have this:
(dx/dt)^2 +9 (dx/dt)^2) =4*10.
10(dx/dt)^2 = 40 (Adding like terms.)
(dx/dt)^2 = 4 (Divide by 10.)
Take the square root of both sides and we get this:
dx/dt = 2 (We’re using the positive root only.)
Now, since dy/dt = 3dx/dt and dx/dt = 2 by substitution we can write
dy/dt=3dx/dt
=3(2)
=6
Not sure if this is what you wanted, but I hope it helps a little.
formeng
This word processor is not suited at all for discussing this problem. I’m not sure where you were needing a little more clarification but I suspect it might have been the part below the line. Accordingly, I’ll skip down to the part after the horizontal line to begin my comments. So, the components of the velocity, dx/dt and dy/dt are a right angles to each other and we can obtain their resultant speed by the Pythagorean theorem, whose basic form is, of course, c^2 = a^2 + b^2. And we are given the magnitude of that speed as 2√10. So, we have the following:
√((dx/dt)^2 + (dy/dt)^2) =2√10.
Now, we substitute the equivalent 3(dx/dt)^2 for dy/dt to get the following:
√((dx/dt)^2 + (3dy/dt)^2) =2√10 which simplifies to
√((dx/dt)^2 + 9(dy/dt)^2) =2√10.
Now, squaring both sides we have this:
(dx/dt)^2 +9 (dx/dt)^2) =4*10.
10(dx/dt)^2 = 40 (Adding like terms.)
(dx/dt)^2 = 4 (Divide by 10.)
Take the square root of both sides and we get this:
dx/dt = 2 (We’re using the positive root only.)
Now, since dy/dt = 3dx/dt and dx/dt = 2 by substitution we can write
dy/dt=3dx/dt
=3(2)
=6
Not sure if this is what you wanted, but I hope it helps a little.
formeng
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If the speed along the curve is 2sqrt(10),
the square of the speed along the curve is 40,
and this must be the sum of (dx/dt)^2 + (dy/dt)^2.
The usefulness of knowing that dy/dx = 2x - 1
is that dy/dx = dy/dt divided by dx/dt.
At the point (2,2), dy/dx = 3,
and certainly the choice dy/dt = 6 "works,"
since it would give dx/dt = 2 and
you can see that 6^2 + 2^2 = 40.
This was my fastest way of getting it...
the square of the speed along the curve is 40,
and this must be the sum of (dx/dt)^2 + (dy/dt)^2.
The usefulness of knowing that dy/dx = 2x - 1
is that dy/dx = dy/dt divided by dx/dt.
At the point (2,2), dy/dx = 3,
and certainly the choice dy/dt = 6 "works,"
since it would give dx/dt = 2 and
you can see that 6^2 + 2^2 = 40.
This was my fastest way of getting it...
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y=x^2-x
dy/dt = dy/dx * dx/dt = (2x-1)dx/dt
If v = particle velocity then
v^2 = (dy/dt)^2 + (dx/dt)^2 = 40*10 = 40
At (2,2)
dy/dt = (4-1)dx/dt = 3dx/dt
dx/dt = (1/3)(dy/dt)
(dy/dt)^2+(1/9)(dy/dt)^2 = 40
(10/9)(dy/dt)^2 = 40
(dy/dt)^2 = 360/10 = 36
dy/dt = 6
Answer D
dy/dt = dy/dx * dx/dt = (2x-1)dx/dt
If v = particle velocity then
v^2 = (dy/dt)^2 + (dx/dt)^2 = 40*10 = 40
At (2,2)
dy/dt = (4-1)dx/dt = 3dx/dt
dx/dt = (1/3)(dy/dt)
(dy/dt)^2+(1/9)(dy/dt)^2 = 40
(10/9)(dy/dt)^2 = 40
(dy/dt)^2 = 360/10 = 36
dy/dt = 6
Answer D