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 $\displaystyle P(0) = 30$
. . to find: .$\displaystyle P(1),\;P(2),\;P(3),\;P(4),\;P(5)$
On $\displaystyle [0,2],\;P'(t) \,=\,-1$
Integrate: .$\displaystyle P(t) \:=\:-x +C_1$
Since $\displaystyle P(0) = 30\!:\;\;-0 + C_1 \:=\:30\quad\Rightarrow\quad C_1 \:=\:30$
Hence, on $\displaystyle [0,2],\;P(t) \:=\:-x + 30$
. Therefore: .$\displaystyle {\color{blue}P(1) =29,\;\;P(2) =28}$
On $\displaystyle [2,4],\;P'(t) \:=\:x - 3$
Integrate: .$\displaystyle P(t) \:=\:\frac{1}{2}x^2-3x+C_2$
Since $\displaystyle P(2) =28,\;\;\frac{1}{2}(2^2) - 3(2) + C_2 \:=\:28\quad\Rightarrow\quad C_2 = 32$
Hence, on $\displaystyle [2,4],\;P(t) \:=\:\frac{1}{2}x^2 - 3x + 32$
. Therefore: .$\displaystyle {\color{blue}P(3) = \frac{55}{2},\;\;P(4) = 24}$
On $\displaystyle [4,5],\;P'(t) \:=\:1$
Integrate: .$\displaystyle P(t) \:=\:x + C_3$
Since $\displaystyle P(4) = 24,\;\;4 + C_3 \:=\:24\quad\Rightarrow\quad C_3 = 20$
Hence, on $\displaystyle [4,5], \;P(t) \:=\:x + 20$
. Therefore: .$\displaystyle {\color{blue}P(5) = 25}$
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...