Find the differential equations of the following family of curves.
Differential Equations
Find the differential equations of the following family of curves.
1. Parabolas with axis parallel to the y – axis with distance vertex to focus fixed as a.
2. Parabolas with axis parallel to the x – axis with distance vertex to focus fixed as a.
3. All ellipses with center at the origin and axes on the coordinate axes.
4. Family of cardioids.
5. Family of 3 – leaf roses.
For number 1.
1
(1) Upwards and downwards
(1) Upwards and downwards parabolas with latus rectum equal to 4a.
$y - k = \pm 4a(x - h)^2$
$y' = \pm 8a(x - h)$
$y'' = \pm 8a$
y−k=±4a(x−h)2 is this the
In reply to (1) Upwards and downwards by Jhun Vert
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
I made a mistake in there, it
In reply to y−k=±4a(x−h)2 is this the by Bingo
I made a mistake in there, it should be (x - h)2 = ±4a(y - k). The solution for (1) should go this way:
$(x - h)^2 = \pm 4a(y - k)$
$2(x - h) = \pm 4ay'$
$2 = \pm 4ay''$
$y'' = \pm \frac{1}{2a}$
y−k=±4a(x−h)2 is this the
In reply to (1) Upwards and downwards by Jhun Vert
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
Yes you are right, please
In reply to y−k=±4a(x−h)2 is this the by Bingo
Yes you are right, please refer to my reply above
so for number 2,
so for number 2,
x-h=±4a(y-k)2
x'=±8a(y-k)
x''=±8a
am i wrong?
It is better to express your
In reply to so for number 2, by Bingo
It is better to express your answer in terms of y' rather than x'. Although x' will do and simpler.
$(y − k)^2 = \pm 4a(x − h)$
$2(y − k)y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$
$y'' = \dfrac{\mp 8a}{y'}$
$y'' \, y' = \pm 8a$
our prof. gave the same
In reply to so for number 2, by Bingo
our prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay"=0 are they the same with your answer sir ?
In the above solution, I made
In reply to our prof. gave the same by ChaCha Ingal Cortez
They are different. I made a mistake in $y - k$ of the above solution.
$(y − k)^2 = \pm 4a(x − h)$
$2(y - k) \, y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
$y - k = \dfrac{\pm 2a}{y'}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{\pm 2a}{y'} \right)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$
$y'' = \dfrac{\mp (y')^3}{2a}$
$2a \, y'' = \mp (y')^3$
$\pm (y')^3 + 2a \, y'' = 0$
Thank You so much sir.
Thank You so much sir.
sir, do you know the standard
sir, do you know the standard or general equations for the last 3 problems?
what equations should i use?
I think you are from ADZU
In reply to sir, do you know the standard by Bingo
I think you are from ADZU haha
Good evening, sir. Do you
In reply to I think you are from ADZU by Shandy
Good evening, sir. Do you have the solution for the last three problems? Thank you!
Sir do you know how to
Sir do you know how to eliminate arbitrary constant here in this equation?
(y-33)²=4a(x-h)?