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-   -   F''(x) of 2xy - y^3 = 5 (http://www.physicsforums.com/showthread.php?t=585162)

 LearninDaMath Mar8-12 08:48 PM

f''(x) of 2xy - y^3 = 5

1. The problem statement, all variables and given/known data

Here is my work/attempt. In blue is my answer. In green is the supposed correct answer.

The first derivative is surely correct, as it matches the answer I was supposed to get. However, if my second derivative answer is incorrect, can you determine where I I am making a mistake?

http://i1094.photobucket.com/albums/...plicit-1-1.jpg

 tjackson3 Mar8-12 09:06 PM

Re: f''(x) of 2xy - y^3 = 5

Here's a hint: your mistake comes on the third to last step (the line beginning f''(x) = -2(-2y)

 LearninDaMath Mar8-12 09:55 PM

Re: f''(x) of 2xy - y^3 = 5

Quote:
 Quote by tjackson3 (Post 3805625) Here's a hint: your mistake comes on the third to last step (the line beginning f''(x) = -2(-2y)
Would it be correct to approach this step like this:

http://i1094.photobucket.com/albums/...mplicit2-1.jpg

EDIT: On the first line, please regard the 12y^2 as just 12y. The second line is written as though the 12y is what I started off with.

 pylauzier Mar8-12 10:12 PM

Re: f''(x) of 2xy - y^3 = 5

Your mistake is in the 3rd to last step, where you have one term with a 2x-3y^2 denominator. In the following step you simplified the equation by multiplying (2x-3y^2) with the (2x-3y^2)^2 term. However the two 4y terms on the numerator were not being divided by 2x-3y^2.

 LearninDaMath Mar8-12 10:50 PM

Re: f''(x) of 2xy - y^3 = 5

Would this be the correct way to do that step?

http://i1094.photobucket.com/albums/...mplicit3-1.jpg

 LearninDaMath Mar8-12 11:59 PM

Re: f''(x) of 2xy - y^3 = 5

Okay, I was able to get the answer in the right form as the provided answer. However, it is off numerically by 2. The provided answer has 18xy and my answer has 16xy. I have always known 8 + 8 to equal 16...what gives?

http://i1094.photobucket.com/albums/...eimplicit4.jpg

 Mark44 Mar9-12 09:24 AM

Re: f''(x) of 2xy - y^3 = 5

Some comments.
1) You don't take "f' " of an expression. f'(x) is the derivative of f. The operator is d/dx. f' and f'' should not appear in your work at all.
2) You wrote f(x) = 2xy - y3 = 0. This is incorrect because a) the expression 2xy - y3 involves both x and y, not just x alone, and b) the equation 2xy - y3 = 0 defines an implicit relationship between x and y.

 LearninDaMath Mar11-12 03:47 AM

Re: f''(x) of 2xy - y^3 = 5

Ah, thats right. It's not a function is it. It's an equation, but not a function. Appreciate the input.

 stripes Mar11-12 06:18 AM

Re: f''(x) of 2xy - y^3 = 5

As Mark44 said, it's an implicit relationship. Correct me if I'm wrong, but doesn't the question seem a little strange? How can

$f(x) = 2xy - y^{3} = 5$?

That being the case, wouldn't we need to write

$f(x,y) = 2xy - y^{3} = 5$, or $f(x,y) = 2xy - y^{3} - 5$

since it is a function of two variables? Then we would just need to find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. If this isn't the case, we would just have

$2xy - y^{3} = 5$,

and we would need to find $\frac{dy}{dx}$ implicitly, not f'(x).

So guys, which is it? Or am I completely off on this one?

 Mark44 Mar11-12 01:15 PM

Re: f''(x) of 2xy - y^3 = 5

Quote:
 Quote by stripes (Post 3809320) As Mark44 said, it's an implicit relationship. Correct me if I'm wrong, but doesn't the question seem a little strange? How can $f(x) = 2xy - y^{3} = 5$? That being the case, wouldn't we need to write $f(x,y) = 2xy - y^{3} = 5$, or $f(x,y) = 2xy - y^{3} - 5$ since it is a function of two variables? Then we would just need to find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. If this isn't the case, we would just have $2xy - y^{3} = 5$, and we would need to find $\frac{dy}{dx}$ implicitly, not f'(x). So guys, which is it? Or am I completely off on this one?
No, I think you are on track here. f'(x) is meaningless here, since the left side of the equation 2xy - y3 - 5 = 0 represents a function of two variables.

The tacit assumption here, I believe, is that y is an implicit function of x, and goal is to find dy/dx using implicit differentiation. Note that this should be written as dy/dx = ..., not f'(x) = ..., so as to not cause confusion between the function (of two variables) in the first equation, and the function (of one variable) that relates x and y.

After dy/dx has been found, differentiate again to find d2y/dx2.

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