💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# 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.)

## a.(a) Using $P(15,250)$, we construct the following table:$$\begin{array}{|c|c|c|}\hline t & Q & \text { slope }=m_{P Q} \\\hline 5 & (5,694) & \frac{624-250}{5-15}=-\frac{444}{10}=-44.4 \\10 & (10,444) & \frac{464-250}{10-15}=-\frac{194}{5}=-38.8 \\20 & (20,111) & \frac{111-250}{20-15}=-\frac{139}{5}=-27.8 \\25 & (25,28) & \frac{28-250}{25-15}=-\frac{222}{10}=-22.2 \\30 & (30,0) & \frac{0-250}{30-15}=-\frac{250}{15}=-16 . \overline{6} \\\hline\end{array}$$b.$m_{a v g}=-33.3$c.$m \approx-33.3$

Limits

Derivatives

### Discussion

You must be signed in to discuss.
po

Prince O.

March 17, 2019

thanks

WS

Wilson S.

December 8, 2020

Lectures

Join Bootcamp

### Video Transcript

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. 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. I'm gonna move over, we'll have five, 10, 15, 20 25 and 30. 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. We're right at two, sorry, at 15 were right up to 50 At 20 were just over 100 at 25. We're at about A quarter of the way through the box and at 30 were empty. And keep in mind that when we started this tank was full with 1000 gallons. 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. And what we want to do is we want to find the slope from P to Q. And Q is going to change, So I'm going to call this first point Q one And then Q two and Q three. We're gonna call this Q4 And this Q five. So what we're trying to find is the slope of a C. Can't line. So the first slope we're finding is the line from P to Q one. So we'll say the slope from P to Q one will be the change in Y over the change in X. So the change in y will be 250 minus 6 94. Mhm Over the change in X, which would be 15 -5. In doing so we get negative 444 over 10. Or simplified would be negative 222 over five. Then we want to find the slope from P to Q two so slope from P 2.q. 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. Then we want to find the slope from P to Q three. So again change in y over change in X. So we'll have to 50 -111 And then 15 -20 in the denominator That will yield a slope of negative 139 over five. Then we want the slope from P to Q four. Change in Y will be 250 -28. Change in X will be 15 -25. You will get 222 over negative 10 which does simplify into negative 111 over five. And then the slope from P to Q five which is going to be changing why 250 zero over change in x 15 -30 Which is 250 over -15, which does simplify to negative 50/3. Now, second part of this is we want to average two of these and come up with the slope of the tangent line at P. So now the tangent line at P is going to be this line right here. All right, so it's bumping up against P. It's only hitting the curve at one point. So if we were to take the slope of this C can't line and the slope of this C can't line and average them together, we should be getting close to the slope of that blue line that we drew in. So I'm going to say slope of the tangent line is equal to the average. So I'm going to average the slope from P two. Cute too with the slope from P to Q three. So when I average that means I'm going to add them And since there are two of them were gonna divide by two. And in doing so you're going to get negative 333 over 10. And the final part of this is to use your graph to estimate the slope of that blue line. So if I was going to use my graph to estimate that, so I'm going to say I'm going from this point right here to approximately this point right here, which means I've got to rise 100, 300, 400, 500 and I've got to run this way. I'm going back 5 10 15. So my slope of this blue line, the slope of this tangent line would be my rise over my run, Which if I simplify 500 over negative 15, if I divide both of those by five, I'm getting about 100 over a negative three, which is about negative 33 & A Third. And what we had here was negative 33.3. So my picture and its slope comes close to or is comparable to the slope when I've averaged two of the secret lines together

WAHS

Limits

Derivatives

Lectures

Join Bootcamp