anonymous
  • anonymous
how is the differentiation of x=ky^2 equal to \[1=2ky\frac{dy}{dx}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
implicite differentiation, in this case y(x), such that y is a function of x.
anonymous
  • anonymous
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.
anonymous
  • anonymous
I"m working on the separable equations section of my differential equations chapter.

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anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[1=2ky\frac{dy}{dx}\]
anonymous
  • anonymous
the gradient would be orthogonal.
anonymous
  • anonymous
I haven't learn anything about gradients yet. This is only chapter 9 of stewart's calculus
anonymous
  • anonymous
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.
anonymous
  • anonymous
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. \[x^2+\frac{y^2}{2}=C\]
anonymous
  • anonymous
for y=k/x I get ln|y|=ln|-x|+C What do you think?
anonymous
  • anonymous
is this the integrated form?
anonymous
  • anonymous
yep. \[\int \frac1ydy=\int-\frac1x dx\]
anonymous
  • anonymous
\[ \large \int \frac{1}{y}dy = - \int \frac{1}{x}dx \] So you can distribute the minus sign and you don't need to carry it inside your logarithmn.
anonymous
  • anonymous
ok ln|y|=-ln|x|+C
anonymous
  • anonymous
perfect.
anonymous
  • anonymous
Thank you!

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