# Thread: Value of function from table of it's rates of change

1. ## Value of function from table of it's rates of change

This is from an early portion of Integral Calculus. So the Fundamental Theorem is about all that has been covered thus far.

If the value of a function p = 2 at time = 0 and the function is changing at a rate of -1 at t= 0, -1 at t=1, -1 at t=2 then at 0 at t=3 and 1 at t=4 and 1 at t=5 , Give the values of this function at t = 1 through t = 5

I assumed that I would just take the value of p at 0 and add to it or subtract from it by the rate of change at each t value.

In doing so, I would get p(1) = 2 + -1 = 1 However, a solution I saw showed the following:

p(0) + integral from 0 to 1 of p'(t) dt which the way I would use this method would yield an answer of 2 + (-1 - (-1)) which is equal to two. However, the solution shows:
2 + (-1 + -1) / (1/1) and therefore the answer is 0

I am trying to figure out my error in reasoning. Thanks for any help, Frostking

2. The simplest thing to do is to treat this as a series of constant slope segments. We know that, at t= 0, p= 2 and its rate of change is -1. Okay, p= 2- t is a line with slope -1 such that p(0)= 2. At t= 1, p(1)= 2- 1= 1. Now we know that p(1)= 1 and its rate of change is -1. A line through (1, 1) with slope -1 is p= 1+ (-1)(t- 1)= 2- t again. (Since the slope at t= 0 and t= 1 are the same, we can use the same line for both.)

p(2)= 2- 2= 0 and we now have rate -1 again. We could find the line with slope -1 through (2, 0) but since that is again the same slope as at 0 and 1, we know the line is still p(t)= 2- t.

p(3)= 2- 3= -1 and now the slope is 0. That's a horizontal line. p(t)= -1 for all t between 3 and 4.

p(4)= -1 and now the slope is 1: p(t)=-1+ (1)(t- 4)= -1+ t- 4= t- 5.

p(5)= 5- 5= 0.