A tank holds 1000 gallons of water which drains from the...

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 $V$ of water remaining in the tank (in gallons) after $t$ minutes. $$ \begin{array}{|c|c|c|c|c|c|c|} \hline t(\mathrm{~min}) & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline V(\mathrm{gal}) & 694 & 444 & 250 & 111 & 28 & 0 \\ \hline \end{array} $$ (a) If $P$ is the point (15,250) on the graph of $V$, find the slopes of the secant lines $P Q$ when $Q$ is the point on the graph with $t=5,10,20,25,$ and 30 (b) Estimate the slope of the tangent line at $P$ by averaging the slopes of two secant lines. (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 after 15 minutes.)

Transcript

00:01 So you have 1000 gallon water tank that is leaking or draining water and you have a chart talking about time in minutes and volume and gallons and after five minutes of draining, there are still 694 gallons in the tank after 10 minutes there are 444 gallons after 15 minutes there are 250 gallons after 20 minutes there's 111 gallons remaining in the tank after 25 minutes there's 28 gallons in the tank and after 30 minutes there's no gallons in the tank.

00:52 So my suggestion is that we start by graphing this and i'm going to let each of these marks on my grid paper represent 100 and i'm going to have each one down the bottom here.

01:19 I'm gonna move over, we'll have five, 10, 15, 20 25 and 30.

01:34 And if we were to plot those points at five, we're just slightly under 700 at 10, we're slightly under 4 50 at 11.

01:52 We're right at two, sorry, at 15 were right up to 50 at 20 were just over 100 at 25.

02:03 We're at about a quarter of the way through the box and at 30 were empty.

02:11 And keep in mind that when we started this tank was full with 1000 gallons.

02:19 So we've got a curve, we've got a decreasing curve going through these points and your goal is we're going to focus on point p, which is the 15- 50.

02:38 And what we want to do is we want to find the slope from p to q.

02:45 And q is going to change, so i'm going to call this first point q one and then q two and q three.

02:54 We're gonna call this q4 and this q five.

02:59 So what we're trying to find is the slope of a c.

03:02 Can't line.

03:03 So the first slope we're finding is the line from p to q one.

03:11 So we'll say the slope from p to q one will be the change in y over the change in x.

03:21 So the change in y will be 250 minus 6 94.

03:29 Mhm over the change in x, which would be 15 -5.

03:37 In doing so we get negative 444 over 10.

03:42 Or simplified would be negative 222 over five.

03:48 Then we want to find the slope from p to q two so slope from p 2.q.

03:57 The second time again, we're going to do change in why it's 2 15 minus 4 44 over change in x, which is 15 minus 10 and we will get a result of -194/5.

04:18 Then we want to find the slope from p to q three...

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