Thread: derivative graph

You are given the following graph.

Use the figure above and the fact that P = 30 when t = 0 to find the values below.
P(1) =
P(2) =
P(3) =
P(4) =
P(5) =

dP/dt = -1_____0 <= t <=2
x-3____2<= t <= 4
1____4 <= t <= 5

start from integrating the function in the first interval:

P = -t+C_____0<=t<=2

now we know that P(0) = 30 thus C = 30 ----> P = -t+30_____0<=t<=2
no because the derivative is continues we can use P(2) = 28 as the initial condition for the function in the next interval
P = (t^2)/6 - 3t + C____2<= t <=4, P(2) = 28----> P(t) = (t^2)/6 - 3t +100/3

continue in the same way and solve for the last interval...

Hello, mathaction!

This is a long and messy one . . .

You are given the following graph.

Use the figure above and the fact that
. . to find: .

On

Integrate: .

Since

Hence, on

. Therefore: .



On

Integrate: .
Since
Hence, on
. Therefore: .



On

Integrate: .

Since

Hence, on

. Therefore: .

thank you for the help, however i think you miss calculated p(4) and p(5)...for p(4), the equation...1/2x^2-3x+32...plug in 4 for x

16/2 - 12 +32= 8+20=28 p(4)=28

P(t)=x-C3...4-C3=28=24
P(t)=x+24...p(5)=5+24 = 29

p(5)=29


thanks for the other ones tho...