# Parametric derivatives

1. Jan 21, 2014

### sooyong94

1. The problem statement, all variables and given/known data
The parametric equations of a curve are
$x=\frac{1}{2}(sint cost+ t), y=\frac{1}{2} t-\frac{1}{4} sin2t$,
$-\pi/2<t\leq0$. P is a point on the curve such that the gradient at P is 1. Find the equation of the normal at P. Hence, determine if the normal at P meets the curve again.

2. Relevant equations
Parametric differentiation, chain rule

3. The attempt at a solution

I have found $\frac{dy}{dt}$ and $\frac{dt}{dx}$. ..Then $\frac{dy}{dx}=\frac{1-cos 2t}{1+cos2t}$

I have also found out that when $\frac{dy}{dx}=1$, $t=\frac{-\pi}{4}$. Then the equation of the normal is $y=-x-\frac{\pi}{4}$. Now I don't know how to determine if the normal at P meets the curve again...

2. Jan 21, 2014

### tiny-tim

hi sooyong94!
hint: what is x + y ?

3. Jan 21, 2014

### sooyong94

t ? :P

4. Jan 21, 2014

### HallsofIvy

Staff Emeritus
Yes, using the fact that sin(2t)= 2sin(t)cos(t), adding the parametric equations for x and y gives x+ y= t.
But $y= -x- \frac{\pi}{4}$ tells you that $x+ y= \frac{\pi}{4}$. Wherever the curve and line intersect, they must both be true.

5. Jan 21, 2014

### sooyong94

But that's $t=\frac{\pi){4}$, which is not within the domain of t. :/

6. Jan 21, 2014

### tiny-tim

no, its -π/4

7. Jan 21, 2014

### sooyong94

:hmm: Wasn't there should another coordinate that intersects with the curve other than x=-pi/4?

8. Jan 21, 2014

### tiny-tim

??

do you mean x+y = -π/4 ?​

9. Jan 21, 2014

### sooyong94

I mean the normal line cuts the curve at other points... is it possible for that?

10. Jan 21, 2014

### Staff: Mentor

Questions about derivatives belong in the Calculus & Beyond section...