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abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Find \[\frac{ ∂y }{ ∂x }\] remember two curves are orthogonal if the slopes of their tangent lines at their intersection points are negative reciprocals.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0I guess I just dont understand how to get that. I am looking at the example and in the example they use the problem y^2 = Cx^3 and they rewrite it as y' = 3y/2x I don't understand how to get to that point from the example or from my problem I listed above.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Well, they found the p. derivative, but it must be in terms of x.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0how is the derivative of that equal to what they gave.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix can you help?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm I haven't seen this type of problem before. I think I understand what they did in the example, lemme see if I can explain it.

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0I am okay once I find y'.... but I don't understand how to get that in the example or in the problem I have.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large y^2=Cx^3 \qquad \rightarrow \qquad \color{royalblue}{C=\frac{y^2}{x^3}}\]Keep that blue piece in mind, we'll need it later. Differentiating the black equation with respect to x,\[\large 2yy'=3Cx^2\] Solving for y',\[\large y'=\frac{3x^2}{2y}\color{royalblue}{C} \quad = \quad \frac{3x^2}{2y}\color{royalblue}{\frac{y^2}{x^3}}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh you got this figured out already? :D hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large x=Cy^4 \qquad \rightarrow \qquad \color{royalblue}{C=\frac{x}{y^4}}\]Derivative of black part,\[\large 1=4Cy^3y'\qquad \rightarrow \qquad y'=\frac{1}{\color{royalblue}{C}4y^3}\] \[\large y'=\frac{1}{4y^3}\color{royalblue}{\frac{y^4}{x}}\] \[\large y'=\frac{y}{4x}\] Hmm what do we need to do from here :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Or is this related to the example? Maybe I misunderstood lol

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0got it... thank you... are you asking me how to solve the problem now? or if what we did solved the problem?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I'm just not familiar with these types of problems. Wasn't sure if that was what you were looking for :D lol Grr why didn't we cover this in Diff EQ _

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0ohhh, well once we have y', we are supposed to replace y' with 1/y'... since they are orthogonal... we then solve for y' again... so we would get y' = 4x/y then we have to separate the variables... y y' = 4x and integrate y^2/2 = 2x^2 + C then move the x's over y^2/2 + 2x^2 = C and we are done... just have to graph it now

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0I just didnt get how to find the y' first, but I do now... thank you.
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