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

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SpacelimbusBest ResponseYou've already chosen the best response.1
implicite differentiation, in this case y(x), such that y is a function of x.
 one year ago

SpacelimbusBest 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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I"m working on the separable equations section of my differential equations chapter.
 one year ago

SpacelimbusBest 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.
 one year ago

MathSofiyaBest 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 \[1=2ky\frac{dy}{dx}\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
the gradient would be orthogonal.
 one year ago

MathSofiyaBest 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
 one year ago

SpacelimbusBest 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.
 one year ago

MathSofiyaBest 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. \[x^2+\frac{y^2}{2}=C\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
for y=k/x I get lny=lnx+C What do you think?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
is this the integrated form?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yep. \[\int \frac1ydy=\int\frac1x dx\]
 one year ago

SpacelimbusBest 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.
 one year ago
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