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

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Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
implicite differentiation, in this case y(x), such that y is a function of x.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I"m working on the separable equations section of my differential equations chapter.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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 \[z=f(x,y)=xky^2\] 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.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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 \[1=2ky\frac{dy}{dx}\]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
the gradient would be orthogonal.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I haven't learn anything about gradients yet. This is only chapter 9 of stewart's calculus
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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. \[x^2+\frac{y^2}{2}=C\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
for y=k/x I get lny=lnx+C What do you think?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
is this the integrated form?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yep. \[\int \frac1ydy=\int\frac1x dx\]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
\[ \large \int \frac{1}{y}dy =  \int \frac{1}{x}dx \] So you can distribute the minus sign and you don't need to carry it inside your logarithmn.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
ok lny=lnx+C
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
perfect.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Thank you!
 2 years ago
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