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# URGENT maths helppp!!! watch

1. The gradient of a curve at any point is directly proportional to the x-coordinate of that point. The curve passes through point A with coordinates 3,2. The gradient of the curve at A is 12. Find the equation of the curve
2. Y = 4x +2 maybe
3. if the gradient is proportional to the x coordinate you know the dy/dx = kx where k is a constant value
then sub in the values you know (gradient at a certain x coordinate) to find k then you can integrate to find the original equation (then sub back in the coordinate you know to find c)
4. (Original post by slaybae)
The gradient of a curve at any point is directly proportional to the x-coordinate of that point. The curve passes through point A with coordinates 3,2. The gradient of the curve at A is 12. Find the equation of the curve
Solve for k (using given values).

Then using the equation for the gradient, and the point you are given, you can find the equation of the curve.
5. Is this GCSE?
6. (Original post by Y11_Maths)
Is this GCSE?
No - AS, as it requires polynomial calculus.
7. Substitute y & x into the formula. 2=3x12 + c (y=mx+c)
Now find c, and then you have formulated the equation.
8. (Original post by Pokefan123!)
Substitute y & x into the formula. 2=3x12 + c (y=mx+c)
Now find c, and then you have formulated the equation.
this assumes that it is a linear equation - if the gradient is proportional to the x coordinate, the gradient won't be constant so unless i'm being really thick, i think you need to use calculus
9. (Original post by mupsman2312)
No - AS, as it requires polynomial calculus.
Phew ok thanks
10. (Original post by AliG_)
Y = 4x +2 maybe
not really curvy enough

11. gradient is proportional to kx
therefore 12=3k
k=4
dy/dx=4x
intergrate 4x = 2x^2 +c
y = 2x^2+c
sub in y and x
2 = 2(3)^2+c
c = -16
final equation is y = 2x^2-16

someone tell me if i made a stupid mistake(i often do)

EDIT: just checked on calculator and it looks right
12. (Original post by jb791)
therefore 12=3k
k=4
dy/dx=4x
intergrate 4x = 2x^2 +c
y = 2x^2+c
sub in y and x
2 = 2(3)^2+c
c = -16
final equation is y = 2x^2-16

someone tell me if i made a stupid mistake(i often do)

EDIT: just checked on calculator and it looks right
That's what I got
13. @thekidwhogames

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