SOLUTION: Help me solve the problem The rate of change of a quantity u with respect to t is du/dt = pt + c, p and c being constants. find u in terms of t given that the quantity has a max

Algebra ->  Circles -> SOLUTION: Help me solve the problem The rate of change of a quantity u with respect to t is du/dt = pt + c, p and c being constants. find u in terms of t given that the quantity has a max      Log On


   

Question 1056385: Help me solve the problem
The rate of change of a quantity u with respect to t is du/dt = pt + c, p and c being constants. find u in terms of t given that the quantity has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
du%2Fdt=pt%2Bc
u=%28pt%5E2%29%2F2%2Bct%2Bd where d is a constant.
.
.
.
In vertex form,
u=a%28t-1%29%5E2%2B5.5
u=a%28t%5E2-2t%2B1%29%2B5.5
u=at%5E2-2at%2B%28a%2B5.5%29
Comparing terms,
a=p%2F2
c=-2a=-p
d=a%2B5.5
.
.
.
Using the rate of change,
-3=p%282%29%2Bc
2p%2Bc=-3
2%28-c%29%2Bc=-3
-2c%2Bc=-3
-c=-3
c=3
So then,
p=-3
a=-3%2F2
and
d=-3%2F2%2B11%2F2
d=8%2F2
d=4
So,
u=-%283%2F2%29%28t-1%29%5E2%2B5.5