Parametric Derivatives and Normal Equations for a Curve with Gradient 1

[sooyong94]

Homework Statement

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.

Homework Equations

Parametric differentiation, chain rule

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...

 

[tiny-tim]

hi sooyong94!

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...

hint: what is x + y ?

 

[sooyong94]

t ? :P

 

[HallsofIvy]

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.

 

[sooyong94]

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.

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

 

[tiny-tim]

But that's $t=\frac{\pi){4}$,

no, its -π/4

 

[sooyong94]

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

 

[tiny-tim]

??

do you mean x+y = -π/4 ?​

 

[sooyong94]

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

 

[Mark44]

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