Question

A pool is in the shape of a rectangular prism. On Monday, water was pumped out of the pool at a constant rate, starting at 12:00pm. At 12:15 pm, the water in the pool was 45 inches deep. At 12:35 pm, the water in the pool was 41 inches deep.

Part A
How many inches does the depth of the water decrease each minute?

Part B
Write an equation that represents y, the depth of the water (in inches) after x minutes.

Answer 1

the equation is y = (-2/15 in/min)x + 43 in

Explanation:

We'll represent elapsed time with the letter t. Then 12 p.m. corresponds to t = 0 sec. Between 12 p.m. and 12:15 p.m., the change in time was 15 min and the change in water depth was (45 - 41) in., or 4 in.

Thus, the water depth changes by 4 in (a decrease) over 30 min.

Part A: The water depth rate of change is -4 in/30 min, or -2 in/15 min, or

(-2/15 in)/min (the depth is decreasing).

Part B: We need to derive a linear function that describes the water depth at any given time x (or t). Two points on the graph of this line are

(15 min, 45 in) and (35 min, 41 in): the 'run' is 30 min and the 'rise' is -4 in.

As in Part A, the slope of this line is (-2/15 in)/min.

The desired equation is then y = mx + b, or 45 in = (-2 in/15 min)(15 min) + b, or 45 in = 2 in + b, or 43 in = b. Then, in its most general form,

the equation is y = (-2/15 in/min)x + 43 in