given the function

xy^2 – x^3y=6

dy/dx= (3x^2y –y^2)/(2xy-x^3)

find the x coordinate of each point on the curve where the tangent line is vertical

I would solve for "y" first, then take the derivative, but...

Asuming that your arithmatic is correct, finding the derivative is exactly the right first step.

Now, knowing that the derivative is the slope of the line tangent to the curve,

and that the slope = rise / run

the slope is HORIZONTAL when rise = 0

and VERTICAL when...

... run=0. OR, in your case, when (2xy-x^3) = 0

Since division by zero is not allowed (indeterminate), the derivative will have indeterminate "holes" in it. Find the holes, solve your problem!

Fianchetto:)

yes I understood all that to begin with but don't I need another equation to solve this I need 2 equations set equal to zero I'm having trouble understanding where that 2nd equation would come from

because in a previous problem just like this on a test several months ago the function was x^2 -xy+y^2=9 and the dervivtive was dy/dx= 2x-y/x-2y

and the equations used to solve it for the tangent line being vertical were x-2y=0 y^2 -3=0 I'm having trouble seeing where this 2nd equation comes from

Buried in your denominator are two equations - let x=0 for one set of "holes", then let (2xy-x^3)=0 for the others.

how do u get 2 equations out of the denominator I understand the rest except that

I guess if you explain to me how the later problem was done the one on the test I can figure out the one I'm working on now

The "TEST" equation had holes for both x and y being = 0. this one has two holes for x and one for y. Is that helpful?

not really because I still can't figure out how they got 2 equations out of the single denominator that would help me a lot

like the process they went through to get the holes on the test

In the denominator solve for x. let x=0.
Then repeat for y.
Anytime (2xy-x^3)=0, the equation is indeterminate and the slope of the tangent line is vertical

I don't see how on the test they could just get y^2 -3=0 I don't see where they got that from that is the only part I'm really having trouble with if you clear up mathematically where that came from then I will be good

(2xy-x^3)
I think is your key. Anytime this value is = 0 the whole thing falls apart. And that is exactly what you are looking for (in this case). The obvious occasion where this occurs is x=0. (2*0*y-0^3)=0 ==>bad news in a denominator, ==> good news for you!

that still doesn't tell me where y^2 - 3=0 came from in the first place you said there were 2 equations in the denominator how did they find the second one the one is just the denominator itself but it makes no sense to me where this other equation came from meaning the y^2-3=0
I'm sorry I'm being a pest but that is my problem here where that equation came from

I don't see away to get y^2-3=0 out of x-2y=0 if there is a way please tell me how

OK. An equation in two variables, in the denominator of a derivative. Derivative = slope of line tangent to curve. Find vertical derivatives. Vertical derivative = indeterminate slope(division by zero). Zeroes in the denominator of the derivative produce indeterminates. Therefore we shall proceed to hunt and kill any instances where the denominator is zero. We shall accomplish this by extracting the denominator, solving it for each of its variables, setting the equation equal to "0" then documenting the results.

Thank you

x-2y=0

2y=0

y=0





x-2y=0

x=2y

1/2x=y


(2xy-x^3)
I think is your key. Anytime this value is = 0 the whole thing falls apart. And that is exactly what you are looking for (in this case). The obvious occasion where this occurs is x=0. (2*0*y-0^3)=0 ==>bad news in a denominator, ==> good news for you!

In a previous post, I thought I did this, but looking back find that actually I did not.

If a product = 0, then one or both of the factors = 0 :

if ab=0 then either a = 0, or b = 0, or a & b are both =0

so, looking back at the denominator:

(2xy-x^3)=0

factor out an x:

x(2y-x^2)=0

is true when either:

x=0 or (2y-x^2)=0 or x=0 AND (2y-x^2)=0


I think that there may be a mistake for the second equation from the test - could you have made a mistake copying it? Looking at it again, I think that the two binomials you cited appear to be factors of the denominator. Then set each =0 as above.

Do not get a vertical tangent and a vertical asymptote mixed up.

A graph has a vertical line if f is continuous at x_{0} and \lim_{x\rightarrow{x_{0}}}|f'(x)|={\infty}

You should be able to find out about vertical tangents in a calc book.

Anyway, you could solve your function for y, getting:

y=\frac{\sqrt{x^{5}+24}+x^{\frac{5}{2}}}{2\sqrt{x} } \;\ and \;\ \frac{-(\sqrt{x^{5}+24}-x^{\frac{5}{2}})}{2\sqrt{x}}

Setting these equal and solving for x, we get x=-3^{\frac{1}{5}}\cdot{2^{\frac{3}{5}}}\approx{-1.888}

f'(x)=x+\frac{5x^{\frac{7}{2}}}{4\sqrt{x^{5}+24}}-\frac{\sqrt{x^{5}+24}}{4x^{\frac{3}{2}}}

I haven't delved into it. Check this point out. It may not be continuous there.

Looking at the graph it looks like it may be a point for a vertical tangent line.

Just a thought.

The tangent line is vertical when the denominator of dy/dx = (3x**2 y ? y**2)/(2xy -x**3) is zero. i.e. when 0 = 2xy - x**3 = x(2y-x**2). This happens when x=0 or when y = x**2/2. In this case 6 = xy**2 ? x**3y = x**5/4 ? x**5/2 = x**5(1/4 ?1/2), so when
x**5 = 6/(1/4 ?/2), or x = (6/(1/4 ? 1/2))**(1/5)

I see that in the original posting "?" is "-". In that case the vertical tangent will occur at x = (6/(-1/4))**(1/5) = (-24)(**1/5) = -(24)**(1/5)

OK, I have the same problem, what is the anwer, and could you take it through step by step?

That post I made in answer to the question was a year ago. Step by step is in the post. If you have any problemms with this post let me know and I'll try to clarify it.

xy^2 – x^3y=6

Differentiate w r t y (x` = dx/dy)


y^2 x` + 2x y - 3x^2 x` - x^3 = 0

x` = 0

hence 2xy - x^3 = 0

x = 0 or x^2 = 2y

There is no ppint on the curve with x = 0

You have to solve the equation of the curve along with x^2 = 2y

substitute y = x^2/2 in the eqn of the curve
and get the roots