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Implicit Differentiation Help (C4)

Hi :smile:

Really stuck on 6b of the Jan 09 AQA C4 paper. (I've linked mark scheme and question below). 6a is fine, following this I've rearranged to find dy/dx in terms of x and y, then equated to zero. Tried to find a substitution as similar to y=(x), with no luck.

Would anybody mind pointing me in the right direction, and explaining how to approach this Q? :smile:

http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Core%204/2009%20-%20Jan/C4AQAJan09MS.pdf - mark scheme

http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Core%204/2009%20-%20Jan/C4AQAJan09QP.pdf - question is at the bottom of page 3
Reply 1
Avatar for Zacken
Zacken
22
Original post by Serine
Hi :smile:

Really stuck on 6b of the Jan 09 AQA C4 paper. (I've linked mark scheme and question below). 6a is fine, following this I've rearranged to find dy/dx in terms of x and y, then equated to zero. Tried to find a substitution as similar to y=(x), with no luck.

Would anybody mind pointing me in the right direction, and explaining how to approach this Q? :smile:

http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Core%204/2009%20-%20Jan/C4AQAJan09MS.pdf - mark scheme

http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Core%204/2009%20-%20Jan/C4AQAJan09QP.pdf - question is at the bottom of page 3


What do you get for dy/dx? :smile:
Reply 2
Avatar for Serine
Serine
OP
1
Original post by Zacken
What do you get for dy/dx? :smile:


dy/dx=2(1-xy) / x^2 +3y^2 :smile:
Reply 3
Avatar for Zacken
Zacken
22
Original post by Serine
dy/dx=2(1-xy) / x^2 +3y^2 :smile:


Yes, that's good! What do you get when you set that equal to zero?
Reply 4
Avatar for razzor
razzor
10
Set dy/dx = 0, rearrange the equation to find y in terms of x then substitute that back into the original equation to eliminate the y.
Reply 5
Avatar for Serine
Serine
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1
Original post by Zacken
Yes, that's good! What do you get when you set that equal to zero?


0=2(1-xy) - this sounds really silly but am I correct in rejecting the denominator of the expression as a potential source of solutions? :/
Reply 6
Avatar for Zacken
Zacken
22
Original post by Serine
0=2(1-xy) - this sounds really silly but am I correct in rejecting the denominator of the expression as a potential source of solutions? :/


Definitely!! That's perfect. Good work. You can simplify and re-arrange that to 2(1−xy)=0⇒1−xy=0⇒1=xy2(1-xy) = 0 \Rightarrow 1-xy = 0 \Rightarrow 1 = xy, agreeing with me so far?

So now you have two equations that xx and yy satisfy: xy=1xy = 1 and x2y+y3=2x+1x^2y + y^3 = 2x + 1.

What if you re-arrange for yy in the first equation and plug it into the second?
Reply 7
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Serine
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ImageUploadedByStudent Room1460198080.327113.jpg


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Reply 8
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Zacken
22
Original post by Serine
...


Ta-da! Well done. :biggrin:
Reply 9
Avatar for Serine
Serine
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I think I can see it now - in essence this is just like finding a corresponding value of y for a value of x given by f'(x) at a stationary point, because this function is defined implicitly dy/dx can give the value of x or y at the stationary point in terms of the other variable. Hence when substituting this into the original expression you get an expression of the x co-rd! :smile:
Reply 10
Avatar for Serine
Serine
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Original post by Zacken
Ta-da! Well done. :biggrin:


Thank you so much!!! :h:
Reply 11
Avatar for Zacken
Zacken
22
Original post by Serine
I think I can see it now - in essence this is just like finding a corresponding value of y for a value of x given by f'(x) at a stationary point, because this function is defined implicitly dy/dx can give the value of x or y at the stationary point in terms of the other variable. Hence when substituting this into the original expression you get an expression of the x co-rd! :smile:


Basically. The question gives you two bits of information - the equation of the curve, so x,yx,y satisfies this and then asks you to find the stationary point, so you know that dydx=0\frac{\mathrm{d}y}{\mathrm{d}x} = 0 this gives you two equations and the rest is just simultaneous equations except leaving your answer as an equation in xx instead of solving for it.

P.S: The reason why you can ignore the denominator is because if you have something of the form ab=0\frac{a}{b} = 0, multiply both sides by bb to get ab×b=0×b\frac{a}{b} \times b = 0\times b , simplifying: a=0a = 0.

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