P3 Parametric Differentiation -- Quick QuestionWatch

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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! 0
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! 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
Thread starter 14 years ago
#3
ahhh k....cheers mate 0
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
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.
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