# Thread: horizontal tangent on a curve

1. ## horizontal tangent on a curve

The curve C in the xy-plane is determined by the equation:
3x5 + 30xy + 12y3

(a) Find an expression in x and y for dy/dx.

(b) At what points different from (0,0) does C have a horizontal tangent?

I was able to get part (a) but don't know how to get part (b). Any help is greatly appreciated.

2. Originally Posted by jarrod573
The curve C in the xy-plane is determined by the equation:
3x5 + 30xy + 12y3

(a) Find an expression in x and y for dy/dx.

(b) At what points different from (0,0) does C have a horizontal tangent?

I was able to get part (a) but don't know how to get part (b). Any help is greatly appreciated.
Put the value of dy/dx you got in part a as 0(zero) and all the values which satisfy the equality is you your answer

dy/dx=0

3. Originally Posted by jarrod573
The curve C in the xy-plane is determined by the equation:
3x5 + 30xy + 12y3

(a) Find an expression in x and y for dy/dx.

(b) At what points different from (0,0) does C have a horizontal tangent?

I was able to get part (a) but don't know how to get part (b). Any help is greatly appreciated.
a) I've got:

$\dfrac{dy}{dx}=-\dfrac{5(x^4 + 2·y)}{2·(5·x + 8·y^2 )}$

b) $\dfrac{dy}{dx} = 0 ~\implies~x^4+2y=0~\implies~x= \pm \sqrt[4]{-2y}$ which will yield real results if y < 0.

Plug in this term instead of x into the original equation to calculate y. I've got: $y = -\sqrt[7]{2^5}~\vee~y = 0$

Re-substitute this value into the term of x. I've got:

$x = -\sqrt[7]{2^3}$

The tangent is drawn in blue.