# Thread: nature of turning points

1. ## nature of turning points

dy/dx = (cos t - sin t)/(sin t - cos t)

The second derivative is 0.

I am trying to find the nature of turning points but the second derivative can't be used and a nature table can't be made because dy/dx always equals -1 for any value of t.

2. ## Re: nature of turning points

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

When $\frac{dy}{dx}=0$, you have turning points. What can you deduce?

3. ## Re: nature of turning points

The turning points are where 0=cos t - sin t. I already have the points. How do I find their nature?

4. ## Re: nature of turning points

Originally Posted by Stuck Man
The turning points are where 0=cos t - sin t. I already have the points. How do I find their nature?
You clearly didn't read Quacky's Post.

Notice that \displaystyle \begin{align*} \frac{\cos{t} - \sin{t}}{\sin{t} - \cos{t}} &= \frac{-\left(\sin{t} - \cos{t}\right)}{\sin{t} - \cos{t}} \\ &= -1 \end{align*}

So the derivative is \displaystyle \begin{align*} -1 \end{align*}. You only have stationary points when the derivative is \displaystyle \begin{align*} 0 \end{align*}. What does this tell you about the function?

5. ## Re: nature of turning points

The curve is a spiral and it does have tuning points between 0<=t<=2pi. It is x=e^tsint, y=e^tcost.

6. ## Re: nature of turning points

The bottom line of dy/dx should have a plus symbol. I have finished this question now, thanks.

7. ## Re: nature of turning points

Originally Posted by Stuck Man
The bottom line of dy/dx should have a plus symbol. I have finished this question now, thanks.
This is why we always advise everyone to post the ENTIRE question, so that our correct guidance isn't completely useless.