# DifferentiationWatch

#1
Q3b). Find the equation of the normal to the curve with parametric equations x=1 - cos2t, y=sin2t at the point P, where t=pi/6

i got dy/dx to be: 2cos2t/2sin2t, is that the same as 1/tan2t? and i can't seem to work out the rest, don't know what i'm doing wrong....!

any help would be much appreiated!! thanx
0
13 years ago
#2
i got dy/dx to be: 2cos2t/2sin2t, is that the same as 1/tan2t?
Yes ; 2cos2t/2sin2t => cos2t/sin2t => 1/tan2t => cot2t

any help would be much appreiated!! thanx
*appreciated
13 years ago
#3
(Original post by V.P. Keys)
...=>...=>...=>...
* ... = ... = ... = ...
0
13 years ago
#4
(Original post by Wrangler)
* ... = ... = ... = ...
I'm pretty sure it is implies ...
13 years ago
#5
(Original post by V.P. Keys)
I'm pretty sure it is implies ...
I'm sure you're sure. But I'm right.

I wouldn't normally be so pedantic - but I felt you needed a taste of your own medicine!
0
13 years ago
#6
Q3b). Find the equation of the normal to the curve with parametric equations x=1 - cos2t, y=sin2t at the point P, where t=pi/6

i got dy/dx to be: 2cos2t/2sin2t, is that the same as 1/tan2t? and i can't seem to work out the rest, don't know what i'm doing wrong....!

any help would be much appreiated!! thanx
x = 1 - cos2t => dx/dt = 2sin2t
y = sin2t => dy/dt = 2cos2t

dy/dx = dy/dt * dt/dx = 2cos2t / 2sin2t = cot(2t)
At P: dy/dx = 1/[tan(Pi/3)] = (1/2) / [√3 / 2] = (√3)/3
:. At P: Gradient (Normal) = -1/[√3)/3]= -3/(√3) = -√3
At P: x = 1 - cos(Pi/3) = 1 - (1/2) = 1/2
At P: y = sin(Pi/3) = √3 / 2
=> P(1/2, √3 / 2)

:. Equation of normal to curve at P:
y - (√3 / 2) = -√3(x - 1/2)
=> y = -(√3)x + √3

--------------

(Original post by V.P. Keys)
I'm pretty sure it is implies ...
It's equals.
2 + 2 = 4, for example
0
13 years ago
#7
(Original post by Nima)
2 + 2 = 4, for example
Agreed.

2 + 2 = 4, is a computation

y = 1/sinx, is a function only known to mathematicians, it implies that y = cosecx. so;

1/sinx => cosecx

13 years ago
#8
(Original post by V.P. Keys)
y = 1/sinx, is a function only known to mathematicians, it implies that y = cosecx. so;

1/sinx => cosecx

y = sin(x) <=> y = 1/cosec(x)

There is no truth or fallacy to the statement "sin(x)", so we can't infer anything from it. However, there is an obvious logical argument to "y = sin(x)", and this is equivalent to saying "y = 1/cosec(x)". Hope this clears up your confusion.

I didn't want to stray from the thread starters topic!
0
#9
thanx

--------------

Q3a). Find the equation of the normal to the curve with parametric equations x=e^t, y=e^t + e^-t, at point P, where t=2

again, i've done all of it, but at da back of the book it says da answer is x=1, but should it not be y=2, becuaue the gradient is 0, n when u put it in
y-y1=m(x-x1) it comes out as y=2?

can any1 explain this plz....
thanx
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of the Arts London
MA Performance Design and Practice Open Day Postgraduate
Thu, 24 Jan '19
• Coventry University
Sat, 26 Jan '19
• Brunel University London
Sat, 26 Jan '19

### Poll

Join the discussion

#### Are you chained to your phone?

Yes (78)
19.35%
Yes, but I'm trying to cut back (163)
40.45%
Nope, not that interesting (162)
40.2%