Locus question. (2 Viewers)

[shaon0]

Show that the tangent at the point P(x1,y1) on the general degree 2 curve:
ax^2 + by^2 + 2cxy + 2dx + 2ey + f = 0
is
ax.x1 by.y1 + c(x1.y + x.y1) + d (x + x1) + e(y +y1) + f = 0

 

[kony]

differentiating,

2ax + 2byy' + 2c(y+xy') + 2d + 2ey' = 0

2byy' + 2cxy' + 2ey' = -2(ax+cy+d)

y' = -(ax+cy+d)/(by+cx+e)

for m at P, we just sub x=x1 and y=y1.

then, use point gradient formula.

y-y1=m(x-x1).

at some point, you'll need to use the fact that ax1²+by1²+2cx1y1+2dx1+2ey1+f=0.

it's just like finding the tangent for a parabola.

 

[tommykins]

回复: Locus question.

Show that the tangent at the point P(x1,y1) on the general degree 2 curve:
ax^2 + by^2 + 2cxy + 2dx + 2ey + f = 0
is
ax.x1 by.y1 + c(x1.y + x.y1) + d (x + x1) + e(y +y1) + f = 0

Couldn't do this question without implicit differentiation which is in Extension 2...but if you did implicit differentiate, kony has given you the solutions.

 

[shaon0]

differentiating,

2ax + 2byy' + 2c(y+xy') + 2d + 2ey' = 0

2byy' + 2cxy' + 2ey' = -2(ax+cy+d)

y' = -(ax+cy+d)/(by+cx+e)

for m at P, we just sub x=x1 and y=y1.

then, use point gradient formula.

y-y1=m(x-x1).

at some point, you'll need to use the fact that ax1²+by1²+2cx1y1+2dx1+2ey1+f=0.

it's just like finding the tangent for a parabola.

hey thanks for the answer.

 

[lolokay]

is it supposed to be [ax.x1 + by.y1]?

 

[shaon0]

is it supposed to be [ax.x1 + by.y1]?

yea the 1 is meant to be sub-script.

 

[lolokay]

I mean the + bit, which you left out (just checking that I solved it correctly)

 

[shaon0]

I mean the + bit, which you left out (just checking that I solved it correctly)

oh yea sorry.

 

[Aerath]

Re: 回复: Locus question.

Couldn't do this question without implicit differentiation which is in Extension 2...but if you did implicit differentiate, kony has given you the solutions.

Isn't implicit differentiation just:
x^2 + y^2 = 4
2x + 2yy' = 0

In which case we learnt that during 2U Differentiation. =\
*confused*

 

[tommykins]

Re: 回复: Locus question.

Isn't implicit differentiation just:
x^2 + y^2 = 4
2x + 2yy' = 0

In which case we learnt that during 2U Differentiation. =\
*confused*

Really? No-one in my 2u/3u cohort knows it (well I'm assuming they don't )

 

[lyounamu]

Re: 回复: Locus question.

Isn't implicit differentiation just:
x^2 + y^2 = 4
2x + 2yy' = 0

In which case we learnt that during 2U Differentiation. =\
*confused*

That's awesome!

I didn't know about that either. lol

 

[tommykins]

回复: Re: 回复: Locus question.

That's awesome!

I didn't know about that either. lol

Quick lesson for you then -

Implicit differentiation is when you have nasty graphs with x^2 , y^2 and xy's.

For example, x^2 + xy = 4

The normal differentiation scheme would be make y the subject, then dy/dx.

Implicit differentiation is the differntiation of each term, where y turns into dy/dx and x^n becomes nx^(n-1)

The above example can be done like this -
u = x v = y
u' = 1 v' = dy/dx

2x + (xdy/dx + y) = 0 (as 4' = 0)

then 2x + xdy/dx + y = 0
xdy/dx = -y - 2x
dy/dx = -(y+2x)/x

Try it out here -
Find the differential of -

x^3 + x^2y^4 + xy + y^3 + 9 = 0

 

[Aerath]

Re: 回复: Re: 回复: Locus question.

Yeah, teacher just said, basically, everything you dash something (when it's with respect to x), everytime you see something that isn't x, you have to dag along a dy/dx or whatever. But standard differentiation rules still apply (product rule, quotient etc).

I sure hope he's right. =\

Really? No-one in my 2u/3u cohort knows it (well I'm assuming they don't )

Hahahaha.

 

[lyounamu]

Re: 回复: Re: 回复: Locus question.

Quick lesson for you then -

Implicit differentiation is when you have nasty graphs with x^2 , y^2 and xy's.

For example, x^2 + xy = 4

The normal differentiation scheme would be make y the subject, then dy/dx.

Implicit differentiation is the differntiation of each term, where y turns into dy/dx and x^n becomes nx^(n-1)

The above example can be done like this -
u = x v = y
u' = 1 v' = dy/dx

2x + (xdy/dx + y) = 0 (as 4' = 0)

then 2x + xdy/dx + y = 0
xdy/dx = -y - 2x
dy/dx = -(y+2x)/x

Try it out here -
Find the differential of -

x^3 + x^2y^4 + xy + y^3 + 9 = 0

how do you do that again?

 

[tommykins]

回复: Re: 回复: Re: 回复: Locus question.

how do you do that again?

Product rule.

say xy = 1

y = 1/x -> dy/dx = -1/x^2

but using implicit

u = x v = y
u' = 1 v' = dy/dx

y + xdy/dx = 0

xdy/dx = -y
dy/dx = -y/x -> but y = 1/x

hence dy/dx = -(1/x)/x = -1/x^2

 

[lyounamu]

Re: 回复: Re: 回复: Re: 回复: Locus question.

Product rule.

say xy = 1

y = 1/x -> dy/dx = -1/x^2

but using implicit

u = x v = y
u' = 1 v' = dy/dx

y + xdy/dx = 0

xdy/dx = -y
dy/dx = -y/x -> but y = 1/x

hence dy/dx = -(1/x)/x = -1/x^2

Is that 4 Unit maths? I don't think we have to learn this.

 

[tommykins]

回复: Re: 回复: Re: 回复: Re: 回复: Locus question.

Is that 4 Unit maths? I don't think we have to learn this.

I don't know if it is, Aerath says he's learnt it but noone in my 2u/3u cohort knows it.

It's not that hard anyways, learn it now in case they ask you to differentiate a circle or something

 

[lyounamu]

Re: 回复: Re: 回复: Re: 回复: Re: 回复: Locus question.

I don't know if it is, Aerath says he's learnt it but noone in my 2u/3u cohort knows it.

It's not that hard anyways, learn it now in case they ask you to differentiate a circle or something

Can you put up the full solution to the question you asked me? I really want to see how you achieve the answer. I am a bit confused.

 

[Aerath]

Re: 回复: Re: 回复: Re: 回复: Re: 回复: Locus question.

I don't know if it is, Aerath says he's learnt it but noone in my 2u/3u cohort knows it.

It's not that hard anyways, learn it now in case they ask you to differentiate a circle or something

Hmmm, I think everyone in my 2U/3U class knows it. But I'm going to guess that it was just something that my teacher taught that was outside the syllabus. =\

 

[tommykins]

Re: 回复: Re: 回复: Re: 回复: Re: 回复: Locus question.

Can you put up the full solution to the question you asked me? I really want to see how you achieve the answer. I am a bit confused.

Original question -

x^3 + x^2y^4 + xy + y^3 + 9 = 0

Sorry for the messy working out.