Hello jiboom Originally Posted by

**jiboom** thanks for the confirmation. I maybe wondered if they wanted something else but surely the increase in y is the delta y you have.

For the curve question i simplify dy/dx as

dy/dx= 5a(5cos t-cos 5t)/[-5a(sin t+ sin 5t)]

=cos 5t-cos t/[sin 5t+sin t]

= -2sin3tsin2t/[2sin3tcos2t]

=-sin2t/cos2t

so grad of normal is cos(2t)/sin(2t) hence normal is given by

y-a(5 sin t -sin 5t)= cos(2t)/sin(2t)[x-a(5cos t+cos 5t)]

sin(2t)y-cos(2t)x=a[5sin(t)sin(2t)-sin(5t)sin(2t)-cos(5t)cos(2t)-5cos(t)cos(2t)

.......................=a[-5(cos(2t)cos(t)-sin(2t)sin(t))-(cos(5t)cos(2t)+sin(5t)sin(2t)

sin(2t)y-cos(2t)x=a[-5cos(3t)-cos(3t)]=-6acos(3t)

so need slick way of finding the other points in first quadrant which normal is normal at another point.

Ah, I see you spotted the mistake you'd made in the sign, and you've corrected it. This is a very tricky question (unless I'm overlooking something obvious - which is entirely possible!).

Yes, I agree that the equation of the normal at $\displaystyle p$ is:$\displaystyle y\sin2p=x\cos2p-6a\cos3p$

Therefore the normal at $\displaystyle q$ is:$\displaystyle y\sin2q=x\cos2q-6a\cos3q$

The normal at $\displaystyle p$ is normal to the curve again at $\displaystyle q$, if these are one and the same line; i.e. if (making $\displaystyle y$ the subject and equating coefficients):$\displaystyle \frac{\cos2p}{\sin2p}=\frac{\cos2q}{\sin2q}$ (1)

and$\displaystyle \frac{\cos3p}{\sin2p}=\frac{\cos3q}{\sin2q}$ (2)

Equation (1) gives:$\displaystyle \tan2q=\tan2p$

$\displaystyle \Rightarrow 2q = 2p + n\pi, n = 1, 2, 3$ (Note: $\displaystyle n =4$ gives $\displaystyle p = q +2\pi$; i.e. $\displaystyle p$ and $\displaystyle q$ represent the same point; and $\displaystyle n > 4$ gives repeated values.)

$\displaystyle \Rightarrow q = p + n\pi/2$

Substitute into (2):$\displaystyle \frac{\cos3p}{\sin2p}= \frac{\cos (3p+3n\pi/2)}{\sin(2p +n\pi)}$

There's a bit of work to do now with the three different values of $\displaystyle n$, but I think when you simplify everything, you get:

$\displaystyle n=1:$$\displaystyle \tan3p = -1$

$\displaystyle \Rightarrow p = \pi/4, 7\pi/12, 11\pi/12, ...$

$\displaystyle n=2:$$\displaystyle \cos 3p = 0$

$\displaystyle \Rightarrow p = \pi/6, \pi/2, 5\pi/6, ...$

$\displaystyle n=3:$$\displaystyle \tan3p = 1$

$\displaystyle \Rightarrow p = \pi/12, 5\pi/12, 3\pi/4, ...$

And it's then a question of choosing the values of $\displaystyle p$ that give points in the first quadrant. I think these are:$\displaystyle p = \pi/12, \pi/6, \pi/4, 5\pi/12$

I attach an Excel spreadsheet that shows the curve (with $\displaystyle a = 1$)and a normal (in fact the one where $\displaystyle p = \pi/6$), which you can (if you have Excel) play around with to confirm that these values seem to be correct.

Grandad