#1

Yep, need maths help again lol

this time its about differentiations:

the graph of y=ax^3 + bx has gradient 4 when x=-1 and gradient 31 when x=2.

find the values of a and b.

Ive done the whole exercise except this question and i need help lol.

this time its about differentiations:

the graph of y=ax^3 + bx has gradient 4 when x=-1 and gradient 31 when x=2.

find the values of a and b.

Ive done the whole exercise except this question and i need help lol.

#2

Never did Gradient, is that related to slope or derivative?

#3

derivative lol :S

#4

differentiate y

dy/dx = ax^2+b

m = ax^2 +b

when x = -1, m =4

when x = 2, m = 31

simultaneous equation

4 = a + b

31 = 4a + b

Solve 1st and 2nd equation

3a = 27

a = 9

b = 4-9

b = -5

I hope that's right

dy/dx = ax^2+b

m = ax^2 +b

when x = -1, m =4

when x = 2, m = 31

simultaneous equation

4 = a + b

31 = 4a + b

Solve 1st and 2nd equation

3a = 27

a = 9

b = 4-9

b = -5

I hope that's right

#5

differentiate y

dy/dx = ax^2+b

m = ax^2 +b

when x = -1, m =4

when x = 2, m = 31

simultaneous equation

4 = a + b

31 = 4a + b

Solve 1st and 2nd equation

3a = 27

a = 9

b = 4-9

b = -5

I hope that's right

thanks mate, except when you differentiate y it becomes 3ax^2 + b

But thank you very much! I didnt think of simultaneous equations lol

#6

m = dy/dx = 3ax^2 + b

4 = 3a + b

b = 4 - 3a

31 = 6a + b

31 = 6a + 4 - 3a

31 = 3a + 4

3a = 27

a = 9

b = 4 - 3a

b = 4 - 27

b = -23

Different to the guy above, but I'm quite sure dy/dx = 3ax^2 and not just ax^2.

EDIT: Yeah, I'm definitely right.

4 = 3a + b

b = 4 - 3a

31 = 6a + b

31 = 6a + 4 - 3a

31 = 3a + 4

3a = 27

a = 9

b = 4 - 3a

b = 4 - 27

b = -23

Different to the guy above, but I'm quite sure dy/dx = 3ax^2 and not just ax^2.

EDIT: Yeah, I'm definitely right.

*Last edited by JamieB at Nov 14, 2007,*