Thread: Find Point where tangent is Horizontal

1. Find Point where tangent is Horizontal

Find a point on the curve x3 + y3 = 5xy
other than the origin at which the tangent line is horizontal.

Here's what i got so far...

dy/xy = (5y-3x^2)/(3y^2-5x)

solved for y... y=3x^2/5

I have to plug y back into the initial equation right? I keep getting a wrong value .. I'm getting (250/27)^1/3. Can someone tell me what I did wrong and how i can get the right answer?!

2. Re: Find Point where tangent is Horizontal

Looks like you got to the right place.

Plugging $y=\dfrac{3x^2}{5}$ in and solving for $x$ you get 1 real solution

$x = \dfrac{5}{3}2^{1/3},~y = \dfrac{5}{3} 2^{2/3}$

In other words you have the right answer.

3. Re: Find Point where tangent is Horizontal

Thanks! That's good to know, but i was wondering how you solved for x algebraically? I couldn't for the life of me get x to simplify to your answer :/

4. Re: Find Point where tangent is Horizontal

Originally Posted by lc99
Find a point on the curve x3 + y3 = 5xy
other than the origin at which the tangent line is horizontal.

Here's what i got so far...

dy/xy = (5y-3x^2)/(3y^2-5x)

solved for y... y=3x^2/5

I have to plug y back into the initial equation right? I keep getting a wrong value .. I'm getting (250/27)^1/3. Can someone tell me what I did wrong and how i can get the right answer?!
Also note that the derivative of y does not actually exist at the origin. The graph crosses itself at the origin with two different slopes.

-Dan

5. Re: Find Point where tangent is Horizontal

Originally Posted by romsek
Looks like you got to the right place.

Plugging $y=\dfrac{3x^2}{5}$ in and solving for $x$ you get 1 real solution

$x = \dfrac{5}{3}2^{1/3},~y = \dfrac{5}{3} 2^{2/3}$

In other words you have the right answer.
How do you simplify the (250/27) ^1/3 into that answer is what i mean? Ahh nvmd. I figured that it was like simplifying radicals except that it's cubed root! THanks anyways!

6. Re: Find Point where tangent is Horizontal

Originally Posted by lc99
How do you simplify the (250/27) ^1/3 into that answer is what i mean?
$250=5^3\cdot 2$ so that $\sqrt[3]{\left(\dfrac{250}{27}\right)}=\dfrac{5\sqrt[3]{2}}{3}$