# Thread: Any method for calculating differentiations?

1. about a month ago we were set some homework (so no worries it's overdue now 8) )

the question was

if y=ax^3 + bx^2 + cx

when dy/dx=15 and x=2

what is a, b and c

now I still havn't figured it out (forgot to be honest ) BUT I'm not looking for a simple answer to the problem

the only way I can think of to solve this is using trial and improvement, which is a problem because even assuming a,b,c are not over 5 (which could be wrong) there are still alot of posibilities

So is there some kind of method or formula to use to work this out? or should I just continue guessing?

2.

3. unfortunately the monks in nepal didn't teach me the secret method of determining unknown constants in differential equations, but i believe you will need 3 conditions to determine your 3 unknwon constants.

at the moment you only have one condition y' = 15 @ x=2

are you sure that was the whole problem?

4. sadly yes, although it's possible my teach screwed up again and missed a chunk

I was toying with the idea of a=-3 b=-2 and c=7.5 which would essentially give the solution y=0 + 0 + (7.5x2) but that I think defeats the purpose

5. i think you really have an infinite number of solutions to the problem as it stands.
someone can correct me if im wrong, but any values which satisfy the equation

15 = 12a + 4b + c

are solutions to the problem

6. Originally Posted by Booms
about a month ago we were set some homework (so no worries it's overdue now 8) )

the question was

if y=ax^3 + bx^2 + cx

when dy/dx=15 and x=2

what is a, b and c

now I still havn't figured it out (forgot to be honest ) BUT I'm not looking for a simple answer to the problem

the only way I can think of to solve this is using trial and improvement, which is a problem because even assuming a,b,c are not over 5 (which could be wrong) there are still alot of posibilities

So is there some kind of method or formula to use to work this out? or should I just continue guessing?
If then and if then so there are a lot of ways to make one is a=b=0 and c=15. Another is a=5/4 and b=c=0.

7. didnt i just say that?

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