# Horizontal/Vertical Tangents

1. Sep 15, 2008

### mzmad

Give the (x,y) coordinate of each point where a horizontal/vertical tangent for the curve: x=(3t)/(1+t^3), y=(3t^2)/(1+t^3). from -inf to inf.

Can someone please guide me through this?! I'm not sure where to begin.

2. Sep 15, 2008

### edziura

Chain rule?

$$\frac{dy}{dx}$$ = $$\frac{dy}{dt}$$ $$\frac{dt}{dx}$$

3. Sep 16, 2008

### Feldoh

Horizontal tangents => dy/dx = 0

Vertical tangents => dy/dx = undef.

4. Sep 16, 2008

### mzmad

Horizontal Tangent: dy/dt= (3t^2)/(1+t^3)= -3t(t^3-2)/(t^3+1)^2=0. How do I solve?

5. Sep 16, 2008

### Feldoh

Assuming you did the derivative correctly, setting the numerator equal to zero will give you the horizontal tangents as long as the denominator is a non-zero.

Setting the denominator of dy/dx equal to zero will give you an undefined slope, meaning the tangent line is vertical.

Last edited: Sep 16, 2008
6. Sep 16, 2008

### mzmad

horizontal tan dy/dt=o=3t(t^3-2)/(1+t^3)^2, t=0, -2^1/3, 2^1/3.?

7. Sep 16, 2008

### Feldoh

Looks right.

8. Sep 16, 2008

### mzmad

and so my coordinated would be (x,y)=(0,0), (-2^1/3,0), and (2^1/3,0)??

9. Sep 16, 2008

### Feldoh

You solved for the parameter value t for when the derivative is 0. However x and y are both functions of t, so you still need to find x and y

10. Sep 16, 2008

### mzmad

ahhh, I see. (0,0)(3*2^1/3)/(1+(2^1/3)^3),0) and -3*2^1/3/(1-(2^1/3)^3),0) ???

11. Sep 16, 2008

### edziura

"horizontal tan dy/dt..."

The coordinates of this curve are defined in terms of a parameter t.

The tangent to the curve is not dy/dt, it is dy/dx. To find dy/dx, there are two possibilities.

You solve one of the given equations for either x or y (depending on which equation you choose), subtitute into the other equation and get y in terms of x, then differentiate to get dy/dx. It would be messy to do so in this case.

Or, use the chain rule.

$$\frac{dy}{dx}$$ = $$\frac{dy}{dt}$$ $$\frac{dx}{dt}^{-1}$$

= $$\frac{-3t^{4} + 6t}{(1 + t^{3})^{2}}$$ $$\frac{(1 + t^{3})^{2}}{-6t^{3} + 3t}$$ =...

Simplify and set = 0; solve for t; sub into the original equations to get (x,y).

12. Sep 16, 2008

### Feldoh

Yes, but edziura, as you said:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

So you can individually solve dy/dt and dx/dt for the points of vertical and horizontal tangency.

13. Sep 16, 2008

### edziura

I agree.

14. Sep 16, 2008

### mzmad

Now, I'm a bit confussed. What did I do wrong? I did dy/dt and dx/dt separately.

15. Sep 17, 2008

### edziura

"horizontal tan dy/dt=o=3t(t^3-2)/(1+t^3)^2, t=0, -2^1/3, 2^1/3."

t = 0 and t = 2^1/3 are correct, but not t = -2^1/3; you are finding an odd root, not an even one.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?