Implicit diff (2 var), error or what?

[trancefishy]

i am working on a homework assignment. it's easy, or, so i think...



Given.
3x^2 - xy^3 + sin(x^3 - y) = 4

Find \frac{dy}{dx}

not a problem. i ended up with

\frac{dy}{dx} = \frac {6x - y^3 + 3x^2 cos(x^3 - y)}{3xy^2 + cos(x^3 - y)}

using implicit differention.

now, "What is the slope of the curve defined by the equation at the point (0, \pi)?"

at first, i plugged it into the derivative, ended up with \pi^3, but, upon inspection, i noticed that the point (0, \pi) does not exist in the original equation. am I to find the slope of the curve of the derivative, or is this a typo, or am i looking at this problem incorrectly? thanks.

 

[courtrigrad]

wouldn't you just use point slope form: y-y_{1} = m(x-x_{1})? Just plug in the values into the derivative and work off from the above form?

 

[trancefishy]

um, point slope form looks messy, but that's besides the point. perhaps this wasn't clear enough. the point (0, \pi) doesn't exist in the original equation. not on the graph. sin(-\pi) does not equal 4. that's the crux of my problem.

IF the point was on the original graph, no problem, plug the values into the derivative and what I get is \pi^3.

 

[Pseudo Statistic]

um, point slope form looks messy, but that's besides the point. perhaps this wasn't clear enough. the point (0, \pi) doesn't exist in the original equation. not on the graph. sin(-\pi) does not equal 4. that's the crux of my problem.

Take the limit as x-> 0 and y -> Pi or look at the graph?

 

[Jameson]

I think it is a typo. There are in infinite many tangent lines to a point... so the point would have to be on the curve in order for the derivative you found to be relavent.

 

[trancefishy]

looking at the graph is what caused me to realize the point isn't on there. it's not even close. now, as x approches 0 from the left, y approaches 2pi. the slope would also be increasing without bound. but that's for 2pi, which has no bearing on my problem.

it's becoming obvious to me that this must be a typo, though, it's so obviously incorrect i just dont' know how it got by

 

[trancefishy]

thank you, jameson