P3 Parametric Differentiation -- Quick Question Watch

This discussion is closed.
devesh254
Badges: 0
#1
Report Thread starter 14 years ago
#1
A curve, C, is given by

x = 2t + 3, y = t^3 - 4t

where t is a parameter. The point A has parameter t = -1 and the line L is the tangent to C at A. The line L also intersects the curve at B.

(a) Show that an equation for L is 2y + x = 7.
(b) Find the value of t at B.


I have done (a), it is (b) which I can't figure out! Could someone please help! Thx! :confused:
0
RichE
Badges: 15
Rep:
?
#2
Report 14 years ago
#2
(Original post by devesh254)
A curve, C, is given by

x = 2t + 3, y = t^3 - 4t

where t is a parameter. The point A has parameter t = -1 and the line L is the tangent to C at A. The line L also intersects the curve at B.

(a) Show that an equation for L is 2y + x = 7.
(b) Find the value of t at B.


I have done (a), it is (b) which I can't figure out! Could someone please help! Thx! :confused:
If you put

x = 2t + 3, y = t^3 - 4t

into 2y+x = 7 you get

(2t^3-8t) +(2t+3) = 7

which rearranges to

2t^3 - 6t - 4 = 0

We know t = -1 is a root, in fact a repeated one as the line's a tangent. The cubic factorises as

2(t+1)^2 (t-2) = 0

and we see t = 2 is the other possibility.
0
devesh254
Badges: 0
#3
Report Thread starter 14 years ago
#3
ahhh k....cheers mate
0
devesh254
Badges: 0
#4
Report Thread starter 14 years ago
#4
(Original post by RichE)
If you put

x = 2t + 3, y = t^3 - 4t

into 2y+x = 7 you get

(2t^3-8t) +(2t+3) = 7

which rearranges to

2t^3 - 6t - 4 = 0

We know t = -1 is a root, in fact a repeated one as the line's a tangent. The cubic factorises as

2(t+1)^2 (t-2) = 0

and we see t = 2 is the other possibility.

sorry mate just wanted to know, how did u factorise the cubic?

i just managed to make f(t) = 2(t^3 - 3t - 2) and then used different values of t to find the roots. i know what i did is right...but just wanted to know if there was any easy way to factorise the cubic? thx
0
RichE
Badges: 15
Rep:
?
#5
Report 14 years ago
#5
(Original post by devesh254)
sorry mate just wanted to know, how did u factorise the cubic?

i just managed to make f(t) = 2(t^3 - 3t - 2) and then used different values of t to find the roots. i know what i did is right...but just wanted to know if there was any easy way to factorise the cubic? thx
Well t=-1 had to be a root because the line passed through the curve at A. Taking out a factor of (t+1) would you leave with a quadratic to solve. Or with a bit more thought, you can spot that (t+1) has to be at least a double root as the line not only intersects at A but touches the curve at A.
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

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

Personalise

University open days

  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 16 Oct '19
  • Teesside University
    All faculties open Undergraduate
    Wed, 16 Oct '19
  • University of the Arts London
    London College of Fashion – Cordwainers Footwear and Bags & Accessories Undergraduate
    Wed, 16 Oct '19

How has the start of this academic year been for you?

Loving it - gonna be a great year (110)
17.92%
It's just nice to be back! (166)
27.04%
Not great so far... (221)
35.99%
I want to drop out! (117)
19.06%

Watched Threads

View All