# Thread: Polynomial and derivative problem

1. ## Polynomial and derivative problem

Hello everyone,
this is the exact question

what I have so far is that:
P'(x)= 3ax^3 + 2bx^2 + c

Then from what I understand P'(0)=0
so then:
P'(0)=0=3a0^3 + 2b0^2 + c
so c=0

2. Originally Posted by momentum
Hello everyone,
this is the exact question

what I have so far is that:
P'(x)= 3ax^3 + 2bx^2 + c

Then from what I understand P'(0)=0
so then:
P'(0)=0=3a0^3 + 2b0^2 + c
so c=0

P'(x) = 0 implies that the rate of change of speed in the y direction is constant, not that the rate of change of speed in the x direction is constant, so this isn't correct.

Condition 1 requires that P(0) = 0 and P(L) = h. The diagram appears to also require that P'(0) = 0 and P'(L) = 0. So we have:
P(0) = d = 0
P'(0) = c = 0

So P(x) = ax^3 + bx^2

P(L) = aL^3 + bL^2 = h
P'(L) = 3aL^2 + 2bL = 0

From the P'(L) condition we get that b = -(3L/2)a

So:
P(L) = aL^3 - (3L/2)a*L^2 = h

or
a = -2h/L^3

b = -(3L/2)*(-2h/L^3) = 3h/L^2

Thus
P(x) = -(2h/L^3)x^3 + (3h/L^2)x^2
has the required properties.

-Dan

3. thank you very much Dan

I have another question
how would you use conditions 2 and 3 to prove that
6hv^2/L^2 < (or = ) k

so then v= h/L
and
P(x)= -2vx^3/L^2 + 3vx^2/L
since v and L are constants
P'(x)= -6vx^3/L^2 + 6vx^2/L

but I don't understand how to find the vertical acceleration....the whole condition #3 is really confusing to me