So I was given the following table along with the following information:
"The height of water in a storage tank during a $15$-hour period is given by a twice-differentiable function $h$, where $h(t)$ is measured in feet and $t$ is measured in hours since midnight for $0<t<15$. The graph of $h$ is concave up on the interval $0<t<15$. Selected values of the derivative of $h$, $h'(t)$ are given in the table above. At time $t=2$, the height of the water is $7$ feet."
After looking at the question that came with this prompt:
"Use the tangent line approximation for $h$ at time $t=2$ to estimate $h(2.8),$ the height of the water at time $t=2.8$."
I'm a bit lost, I realize that the values given for $h'(t)$ are respectively the rate at which the water is accumulating in the tank, but I'm confused on how I'd go about estimating the total accumulation of water at the given time. Any help would be appreciated!
Best Answer
Since you are given h' at only 6 values you can only approximate h and there may be many equally valid approximations.
If the height at some time, t0, were h0 and h' were the constant, a, between t0 and t1, then the height a t1 would be h(t1)= h0+ a(t1- t0).
You can, for example, take h(2)= 7 and then, taking h' to be 0.3 for all t from 2 to 4, h(4)= 7+ 0.3(4- 2)= 7.6 feet. Then h(5)= 7.6+ 0.6(1)= 8 feet, etc..
Some people might think it better to use the average of values at two consecutive points.
h'(2)= 0.3 and h'(4)= 0.6 so the average is (0.3+ 0.6)/2= 0.45. That gives h(4)= 7+ 0.45(2)= 7.9 feet, (0.6+ 0.7)/2= 0.65 so h(5)= 7.9+ 0.65(1)= 8.55 feet, etc.