Question

how is the differentiation of x=ky^2 equal to
\[1=2ky\frac{dy}{dx}\]

Answer 1

implicite differentiation, in this case y(x), such that y is a function of x.

Answer 2

But that's only one guess in this case, the mechanical way I remember for implicit differentiation is derivate as if you were deriving something in terms of x and then just multiply it by dy/dx, chain rule.

Answer 3

I"m working on the separable equations section of my differential equations chapter.

Answer 4

I will just add this, maybe it helps you, this is the chain rule for multivariable calculus.

The above equation you can write like that
\[z=f(x,y)=x-ky^2\]

So the multivariable chain rule says

\[ dz = f_x dx + f_ydy \]
\[ \frac{dz}{dx}= f_x+f_y\frac{dy}{dx}\]
This is far from a proof, but you can read some application out of it. Implicit differentiation doesn't selectively deal with partial derivatives though.

Answer 5

The original question reads. Find the orthogonal trajectories of the family of curves x=ky^2, where k is an arbitrary constant.
And the first thing they did was differentiate x=ky^2 to get
\[1=2ky\frac{dy}{dx}\]

Answer 6

the gradient would be orthogonal.

Answer 7

I haven't learn anything about gradients yet. This is only chapter 9 of stewart's calculus

Answer 8

Another relationship for orthogonal functions is

\[ m_n \cdot m_y = -1 \] where \(m_n\) is normal to \(m_y\) but I don't see why they apply this sort of differentiation here.

Answer 9

Oh I think I see what they've done.
They rearranged the equation to a separable equation, did the integral.
Then stated:
THe orthogonal trajectories are the family of ellipses given by the following equation.
\[x^2+\frac{y^2}{2}=C\]

Answer 10

for y=k/x
I get
ln|y|=ln|-x|+C
What do you think?

Answer 11

is this the integrated form?

Answer 12

yep.
\[\int \frac1ydy=\int-\frac1x dx\]

Answer 13

\[ \large \int \frac{1}{y}dy = - \int \frac{1}{x}dx \]

So you can distribute the minus sign and you don't need to carry it inside your logarithmn.

Answer 14

ok
ln|y|=-ln|x|+C

Answer 15

perfect.

Answer 16

Thank you!