Instantaneous and average rate of change calculus (level of water as pool is drained over 4 hrs)

Can someone please help me go through the steps together. I need a little help and would appreciate someone helping. There is no answer sheet so I would like some help with the solutions to these problems.
 
a) the graph tells you that the water is draining slower as more time passes
b) I am not sure how to start b and how to solve it.
 
Hello. Part (a) could be done a couple ways. If you've already learned that slopes along a curved graph represent rates at which the quantity is changing, then think about slopes along the given graph, over time. What's happening to the (absolute) value of these slopes, as time moves forward? (You don't need actual slope values, to do this. Think graphically about slopes of lines tangent to the curve, in general, as t increases.)

If you haven't yet learned the relationship between slope and rate, then you could estimate some volumes on the graph, to calculate average rate of change, both near the beginning (t=0) and near the end (t=240), and then compare those rates.

For example, I estimate the following (t, V) values from the graph:

(0, 2150)
(15, 1900)

Using these values, can you say what happened to the volume during the first 15 min? That change in volume is the average rate of change over the first 15 min.

(225, 550)
(240, 400)

From these values, can you find the volume's average rate of change over the last 15 min shown in the graph?

Compare those two average rates of change. What's happening to the rate, as time moves forward?


For parts (b)i and (b)ii, find the average rate of change the same way as above (using estimated coordinates from the graph). They want these averages calculated for the first and last hours (not the first and last 15 min, as in my examples).

Show us what you get.

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Last edited:
davidchoi01 said:
a) the graph tells you that the water is draining slower as more time passes
b) I am not sure how to start b and how to solve it.
Click to expand...
I had lost my Internet connection, while typing my first reply, so I wasn't alerted to your part (a) answer above.

Yes, that's correct. The volume's rate of change is decreasing over time (i.e., the absolute value of tangent-line slopes is getting smaller.)

Some physics: When the pool is full, all that water generates a large weight force, which pushes out water near the bottom at a high rate. As the water volume decreases, there is less weight force pushing downward, so water near the bottom is pushed out the drain at slower and slower rates.

Average rate-of-change calculations involve only numbers at the beginning and end points of the interval in question. See my examples for average rates over the first and last 15 min (post #5).

Please show all of your steps, if you need more help. Cheers

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