A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume of water remaining in the tank (in gallons) after t minutes.
a.
If P is the point (15,250) on the grap V, find the slopes of the secant lines PQ when Q is the point on the graph with t=5,10,20,25, and 30
t=5 ->Q(5, 694)
the slope of the secant lines PQ passing through P (15,250) and Q(5, 694) is
m[pq]=(694-250)/(5-15)=444/-10=-44.4
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t=10 ->Q(10, 444)
the slope of the secant lines PQ passing through P (15,250) and Q(10, 444) is
m[pq]=(444-250)/(10-15)=194/-5=-38.8
the slope of the secant lines PQ passing through P(15, 250) and Q(20, 111) is
m[pq]= (111- 250)/(20-15) =-139/5=-27. 8
b. the slope of the tangent at P
avarage of m[pq]=-38.8 and m[pq]=-27. 8, which is m[tangent]=(-38.8 + -27. 8)/2=-33. 3
We can thus estimate that the slope of the tangent line at the point P(15, 250) is -33. 3.
However, we certainly don’t know that this is the exact value of the slope of the tangent line and, furthermore, there is no way to find the exact value because we are only given a discrete set of data points. In order to find the exact value of the slope of the tangent line at P(15, 250), we would need to be given the values of V at all values of t in some (at least small) time interval
containing t = 15.
c. Use a graph of the function to estimate the slope of the tangent line at P.
(This slope represents the rate at which the water is flowing from the tank at the instant t =15 min. In other words, it is the instantaneous flow rate at t =15 min.)
Once again (just as in part b), our estimate is bound to be rather crude because it relies on guessing from incomplete information.
Using the hand-drawn guess at a tangent line shown in the picture below, we arrive at
m =(170 -315)/(17 -12. 5) =
-32. 2
This is remarkably close to the answer that we got in part b!
View attachment 612
