Question

Find the differential equation of the family of curves and of the orthogonal trajectories.

y = c - 2x

Needing a little help on this one...

Thanks

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- April 10th 2007, 09:29 PMcoolmath123Family of Curves
Question

Find the differential equation of the family of curves and of the orthogonal trajectories.

y = c - 2x

Needing a little help on this one...

Thanks - April 11th 2007, 06:39 PMPupka
Differential equation:

first, separate C:

c = y + 2x

Differentiate:

0 = y' + 2

y' = -2

Slope must be negative reciprocal. therefore m = 1/2;

dy/dx = 1/2

S2dy = Sdx (S - integration sign)

2y = x + C

C = 2y - x; - April 11th 2007, 07:06 PMcoolmath123
Thanks,

The rest of my question states

Find the differential equation of the family of curves and of the

orthogonal trajectories. Then solve the differential equation to find the

formula of the orthogonal trajectories.

I have to find these things below and I dont have to graph it.

Given the family

y = c - 2x

Differential equation of the family?

y' = 2

Differential equation of the orthogonal trajectories?

y' = -1 / 2

Orthogonal trajectories?

y = -x / 2 + c

This our my answers above

I think it is correct but I want to make sure - April 11th 2007, 07:54 PMPupka
Note, y' is -2, not 2. Therefore the slope of the family is negative reciprocal 1/2.

Differential equation: y' = -2

Slope (differential equation) of the orthogonal family: y' = 1/2

Orthogonal family: y = (x+C)/2 - April 11th 2007, 09:16 PMcoolmath123
ok, the only reason I had positive 2 is because the given family is Y = c - 2x.

When you showed me the steps you had

Differential equation:

first, separate C:

c = y + 2x

Does that change anything in the answer I had since it is y = c - 2x instead of plus???

- April 12th 2007, 09:11 AMPupka
Sorry, but I don't quite undrstand what you mean. :confused: The equation is y = c - 2x, how could you have positive 2 in there? When you differentiate you get y' = -2, and that is the slope of that family. If you graph both families that might help. The orthogonal family must have a negative reciprocal slope because by deinition its curves intersect the original family at right angles. Sorry, if that doesn't help, ask again.

- April 12th 2007, 11:56 AMcoolmath123
No you are correct. I was going over my work today and I saw were the error was. Thanks for your help:)