# Differentiating Q - C1Watch

This discussion is closed.
#1
Curve C has the equation

y= x^3 - 7x^2 + 15x + 3 x>(or equal to) 0

The point P, on curve C, has xcoord 1 and the point Q is the min. turning poitn of C.

a) find dy/dx.

I did this and I got 3x^2 -14x + 15

b) Find the coord of Q

Now how would I do that??

Thank
0
14 years ago
#2
Well you'd start with saying that the minimum point occurs when dy/dx=0. Are you familiar with that?
0
14 years ago
#3
(Original post by koldtoast)
Curve C has the equation

y= x^3 - 7x^2 + 15x + 3 x>(or equal to) 0

The point P, on curve C, has xcoord 1 and the point Q is the min. turning poitn of C.

a) find dy/dx.

I did this and I got 3x^2 -14x + 15

b) Find the coord of Q

Now how would I do that??

Thank
Turning points are where the gradient is zero.

so 3x^2 -14x + 15=0
so (3x-5)(x-3)=0, therefore x=5/3 or x=3

Min turning points occur when the second derivative is >0

so d2y/dx2=6x-14
when x=5/3, d2y/dx2=-4
and when x=3, d2y/dx2=4

so the minimum turning point, is when x=3, i.e at the point (3,12)
the point found by substituting back in x=3 to the intiial curve equation.
0
14 years ago
#4
y= x^3 - 7x^2 + 15x + 3 x>(or equal to) 0

dy/dx = 3x^2 -14x + 15

dy/dx = 0 at min turning point since, by definition gradient at turning point = 0

so: at dy/dx = 0,
3x^2 -14x + 15=0
(3x+1)(x-5)=0
3x+1=0 or x-5=0
3x=-1 x=5
x=-1/3

X cannot = -1/3 because question states x>(or equal to) 0

subst x=5 into y= x^3 - 7x^2 + 15x + 3
y= (5)^3 - 7(5)^2 + 15(5) + 3
y=125-175+65+3=18

therefore min is at x=5,y=18 i.e Q(5,18)

This can (not neccesary for this question) be veryfied as shown below,

At min turning point, second derivative is greater that 0

d2y/dx2 = 6x-14
at x=6 , d2y/dx2=6(5)-14=16 i.e >0 therefore min
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#5
(Original post by Gemini)
Well you'd start with saying that the minimum point occurs when dy/dx=0. Are you familiar with that?
No, I don't think i've ever heard of that in C1, am I meant to know that?
0
14 years ago
#6
Its something that your teachers really should have taught you...

Can't think of a way to explain it clearly, hopefully someone else will help *hinthint*
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#7
and also, show that PQ is parallel to the x axis.

gemini, i've looked through it and I think I'll be able to grasp the concept fairly well enough to know how to use it in an exam, I think it sort of understand it

Thanks!
0
14 years ago
#8
(Original post by koldtoast)
and also, show that PQ is parallel to the x axis.

gemini, i've looked through it and I think I'll be able to grasp the concept fairly well enough to know how to use it in an exam, I think it sort of understand it

Thanks!
If you want you can, instead of using the d2y/dx2, you can just work out the gradient just before and after...

So for a min point, gradient before should be negative and gradient just after should be positive so it's like \_/

Which board are you doing? Is it the same board you got the question paper from? I'm doing Edexcel and this stuff isn't till C2!
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#9
I do Edexcel as well, I got it from the Pure Maths June 04 paper, our teacher went through them and crossed out the ones we wouldn't be able to do since the C1 is so much more different than the old P1, maybe she thought we would be able to tackle this Q!
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