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# Maths is this correct? watch

1. The gradient of the curve is given by dy/dx= (x^2+3)^2 divided by x^2, x is not 0
(A) Show that day/dx = x^2+6+9x^-2
My answer is : proven
The point. (3,20) lies on C
(B) find an equation for the curve C in the form y=f(x)
I have got:
X^2 divided by 2/3 +3/2(x^2)
Y=(x^2+6+9x^-2) +c
20= 3(x^2+6+9x^-2) +c
20= 3x^2 +18 +27x^-2 +c
3x^2+27x^-2-2+c
2. (Original post by Musicanor)
The gradient of the curve is given by dy/dx= (x^2+3)^2 divided by x^2, x is not 0
(A) Show that day/dx = x^2+6+9x^-2
My answer is : proven
The point. (3,20) lies on C
(B) find an equation for the curve C in the form y=f(x)
I have got:
X^2 divided by 2/3 +3/2(x^2)
Y=(x^2+6+9x^-2) +c
20= 3(x^2+6+9x^-2) +c
20= 3x^2 +18 +27x^-2 +c
3x^2+27x^-2-2+c
(Original post by Musicanor)
The gradient of the curve is given by dy/dx= (x^2+3)^2 divided by x^2, x is not 0
(A) Show that day/dx = x^2+6+9x^-2
My answer is : proven
The point. (3,20) lies on C
(B) find an equation for the curve C in the form y=f(x)
I have got:
X^2 divided by 2/3 +3/2(x^2)
Y=(x^2+6+9x^-2) +c
20= 3(x^2+6+9x^-2) +c
20= 3x^2 +18 +27x^-2 +c
3x^2+27x^-2-2+c
It helps to use words in the exam and in answers - to me it's not clear where your first line in part b) comes from, or what it's referring to.
3. (Original post by Musicanor)
The gradient of the curve is given by dy/dx= (x^2+3)^2 divided by x^2, x is not 0
(A) Show that day/dx = x^2+6+9x^-2
My answer is : proven
The point. (3,20) lies on C
(B) find an equation for the curve C in the form y=f(x)
I have got:
X^2 divided by 2/3 +3/2(x^2)
Y=(x^2+6+9x^-2) +c
20= 3(x^2+6+9x^-2) +c
20= 3x^2 +18 +27x^-2 +c
3x^2+27x^-2-2+c
So for part b, integrate dy/dx = x^2 + 6 + 9x^-2
this should give y = (1/3)x^3 + 6x - 9x^-1 + c
substitute x = 3 and y = 20 to work out c
20 = (1/3)3^3 +6(3) - 9(3^-1) + c
20 = 9 + 18 - 3 + c
c = -4
so equation of the curve is y = (1/3)x^3 + 6x - 9x^-1 - 4
4. [QUOTE=Anonymous62;76001834]So for part b, integrate dy/dx = x^2 + 6 + 9x^-2
this should give y = (1/3)x^3 + 6x - 9x^-1 + c
substitute x = 3 and y = 20 to work out c
20 = (1/3)3^3 +6(3) - 9(3^-1) + c
20 = 9 + 18 - 3 + c
c = -4
so equation of the curve is y = (1/3)x^3 + 6x - 9x^-1 - 4[/QUOTEyay Thank you

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