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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0implicite differentiation, in this case y(x), such that y is a function of x.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But 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
 4 years ago
Best ResponseYou've already chosen the best response.0I"m working on the separable equations section of my differential equations chapter.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0the gradient would be orthogonal.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I haven't learn anything about gradients yet. This is only chapter 9 of stewart's calculus

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Another 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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 4 years ago
Best ResponseYou've already chosen the best response.0for y=k/x I get lny=lnx+C What do you think?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is this the integrated form?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep. \[\int \frac1ydy=\int\frac1x dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \large \int \frac{1}{y}dy =  \int \frac{1}{x}dx \] So you can distribute the minus sign and you don't need to carry it inside your logarithmn.
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