# Implicit diff (2 var), error or what?

1. Feb 28, 2005

### 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.

2. Feb 28, 2005

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?

3. Feb 28, 2005

### 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 $$\pi^3$$.

Last edited: Feb 28, 2005
4. Feb 28, 2005

### Pseudo Statistic

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

5. Feb 28, 2005

### 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.

6. Feb 28, 2005

### 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

7. Feb 28, 2005

### trancefishy

thank you, jameson