# alevel maths help please

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please could someone help me with this question

Last edited by inara_30; 1 month ago

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please could someone help me with this question

**inara_30**)please could someone help me with this question

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#3

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i'm confused with where to differentiate because i though tany differentiated is sec2y ?

**inara_30**)i'm confused with where to differentiate because i though tany differentiated is sec2y ?

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Only if you differentiate with respect to y. Youre differentiating (implicitly) with respect to x.

**mqb2766**)Only if you differentiate with respect to y. Youre differentiating (implicitly) with respect to x.

x^2sec^2ydy/dx + 2xtany = 0

so dy/dx = -2xtany / x^2sec^2y

i know that 1 + tan^2y = sec^2y so i substituted that leaving with:

dy/dx = -2xtany / x^2 + x^2tan^2y

the question already says x^2tan^2y = 9 so i substituted that leaving me with:

dy/dx = -2xtany / x^2 + 9

and now i'm stuck ....

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#5

(Original post by

ah ok so when differentiating implicitly i got:

x^2sec^2ydy/dx + 2xtany = 0

so dy/dx = -2xtany / x^2sec^2y

i know that 1 + tan^2y = sec^2y so i substituted that leaving with:

dy/dx = -2xtany / x^2 + x^2tan^2y

the question already says

dy/dx = -2xtany / x^2 + 9

and now i'm stuck ....

**inara_30**)ah ok so when differentiating implicitly i got:

x^2sec^2ydy/dx + 2xtany = 0

so dy/dx = -2xtany / x^2sec^2y

i know that 1 + tan^2y = sec^2y so i substituted that leaving with:

dy/dx = -2xtany / x^2 + x^2tan^2y

the question already says

**x^2tan^2y = 9**so i substituted that leaving me with:dy/dx = -2xtany / x^2 + 9

and now i'm stuck ....

Last edited by mqb2766; 1 month ago

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(Original post by

Bold is not quite what the question says, but get that corrected and you should be ok. You should substitute for tan(y) on the numerator and denominator.

**mqb2766**)Bold is not quite what the question says, but get that corrected and you should be ok. You should substitute for tan(y) on the numerator and denominator.

ok so if i rewrite that

dy/dx = -2xtany / x^2 + tany(x^2tany)

now substitute x^2tany = 9 into that leaving with:

dy/dx = -2xtany / x^2 + 9tany

i then rearranged the question so that tany = 9/x^2 and substituted that so:

dy/dx = (-2x x 9/x^2) / (x^2 + 9(9/x^2))

which i simplify into to :

dy/dx = -18x / x^4 + 81

is that right now ?

thanks for your help

but for part b i'm really confused how do you know which number to select either side of the value given in this question to prove it's an inflection point ...

i know that an inflection point is when there's a change in sign but no matter what numbers i use it always comes out as a positive...

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#7

(Original post by

ah silly me

ok so if i rewrite that

dy/dx = -2xtany / x^2 + tany(x^2tany)

now substitute x^2tany = 9 into that leaving with:

dy/dx = -2xtany / x^2 + 9tany

i then rearranged the question so that tany = 9/x^2 and substituted that so:

dy/dx = (-2x x 9/x^2) / (x^2 + 9(9/x^2))

which i simplify into to :

dy/dx = -18x / x^4 + 81

is that right now ?

thanks for your help

but for part b i'm really confused how do you know which number to select either side of the value given in this question to prove it's an inflection point ...

i know that an inflection point is when there's a change in sign but no matter what numbers i use it always comes out as a positive...

**inara_30**)ah silly me

ok so if i rewrite that

dy/dx = -2xtany / x^2 + tany(x^2tany)

now substitute x^2tany = 9 into that leaving with:

dy/dx = -2xtany / x^2 + 9tany

i then rearranged the question so that tany = 9/x^2 and substituted that so:

dy/dx = (-2x x 9/x^2) / (x^2 + 9(9/x^2))

which i simplify into to :

dy/dx = -18x / x^4 + 81

is that right now ?

thanks for your help

but for part b i'm really confused how do you know which number to select either side of the value given in this question to prove it's an inflection point ...

i know that an inflection point is when there's a change in sign but no matter what numbers i use it always comes out as a positive...

tan(y) = 9/x^2

It would simplify the derivation a bit, but what you have done is fine. Sometimes its worth spending a minute or two seeing if a transformation may help.

For b) inflection is when the second derivative changes sign. So not exactly sure what you've done?

Last edited by mqb2766; 1 month ago

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(Original post by

For a) as noted in another thread, if you did

tan(y) = 9/x^2

It would simplify the derivation a bit, but what you have done is fine. Sometimes its worth spending a minute or two seeing if a transformation may help.

For b) inflection is when the second derivative changes sign. So not exactly sure what you've done?

**mqb2766**)For a) as noted in another thread, if you did

tan(y) = 9/x^2

It would simplify the derivation a bit, but what you have done is fine. Sometimes its worth spending a minute or two seeing if a transformation may help.

For b) inflection is when the second derivative changes sign. So not exactly sure what you've done?

54(x^4 - 27) / (x^4 + 81)^2

as there is x^4 - 27 on the numerator how do i prove that it's an inflection point do i just show:

x^4 - 27 = 0

x^4 = 27

therefore x = 4

*root*27 and so it's a point of inflection ??

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#9

(Original post by

yes i realised that after i posted but my d^2y/dx^2 has come out as:

54(x^4 - 27) / (x^4 + 81)^2

as there is x^4 - 27 on the numerator how do i prove that it's an inflection point do i just show:

x^4 - 27 = 0

x^4 = 27

therefore x = 4

**inara_30**)yes i realised that after i posted but my d^2y/dx^2 has come out as:

54(x^4 - 27) / (x^4 + 81)^2

as there is x^4 - 27 on the numerator how do i prove that it's an inflection point do i just show:

x^4 - 27 = 0

x^4 = 27

therefore x = 4

*root*27 and so it's a point of inflection ??y = x^4

which has a zero second derivative at 0 but no sign change so no point of inflection.

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(Original post by

Pretty much. If the numerator is zero (and the denominator isn't) then the expression is zero. Strictly speaking you should show the sign changes at that point, otherwise you could have something like

y = x^4

which has a zero second derivative at 0 but no sign change so no point of inflection.

**mqb2766**)Pretty much. If the numerator is zero (and the denominator isn't) then the expression is zero. Strictly speaking you should show the sign changes at that point, otherwise you could have something like

y = x^4

which has a zero second derivative at 0 but no sign change so no point of inflection.

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#11

(Original post by

ok how will i be able to show the sign changes from this second derivative - i'm confused about that

**inara_30**)ok how will i be able to show the sign changes from this second derivative - i'm confused about that

The denominator is positive.

What is the sign of the numerator when x is greater than or less than this value. So ...

Last edited by mqb2766; 1 month ago

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(Original post by

Thought it would be fairly easy?

The denominator is positive.

What is the sign of the numerator when x is greater than or less than this value. So ...

**mqb2766**)Thought it would be fairly easy?

The denominator is positive.

What is the sign of the numerator when x is greater than or less than this value. So ...

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#13

(Original post by

the sign of the numerator will be positive so ah ok there will be a point of inflection as it would be negative as well

**inara_30**)the sign of the numerator will be positive so ah ok there will be a point of inflection as it would be negative as well

Last edited by mqb2766; 1 month ago

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(Original post by

I think what you mean is the (numerator) second derivative is negative when x^4<27 and positive when x^4>27?

**mqb2766**)I think what you mean is the (numerator) second derivative is negative when x^4<27 and positive when x^4>27?

i'm assuming you're in year 13 as well so good luck with exams if you have any !

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#16

(Original post by

mqb2766 hello sorry to bother you but you're the only one that seems to be active in this maths forum that's why i'm, asking you questions

for this question please check if i have got the write answer :

a). A = 3, B = 4, C=-2

b). attached working for part b

thanks

**inara_30**)mqb2766 hello sorry to bother you but you're the only one that seems to be active in this maths forum that's why i'm, asking you questions

for this question please check if i have got the write answer :

a). A = 3, B = 4, C=-2

b). attached working for part b

thanks

+ 2/(2x-1)

to keep things a bit simpler, but your working is fine.

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(Original post by

Both look good. Id have written the C term as

+ 2/(2x-1)

to keep things a bit simpler, but your working is fine.

**mqb2766**)Both look good. Id have written the C term as

+ 2/(2x-1)

to keep things a bit simpler, but your working is fine.

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#18

(Original post by

yay thank you !!

**inara_30**)yay thank you !!

https://www.wolframalpha.com/input/?i=partial+fractions+%28-6x%5E2--11x--1%29%2F%28%28x-3%29%281-2x%29%29

Means I can be lazy ...

Last edited by mqb2766; 1 month ago

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(Original post by

Glad your happy and

https://www.wolframalpha.com/input/?...9%281-2x%29%29

Means I can be lazy ...

**mqb2766**)Glad your happy and

https://www.wolframalpha.com/input/?...9%281-2x%29%29

Means I can be lazy ...

i didn't know there will be one to solve partial fractions

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#20

(Original post by

oh wow thanks for this website

i didn't know there will be one to solve partial fractions

**inara_30**)oh wow thanks for this website

i didn't know there will be one to solve partial fractions

Give me 5 to sort it out.

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