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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.$m \approx-33.3$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 1

The Tangent and Velocity Problems

Limits

Derivatives

Prince O.

March 17, 2019

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Wilson S.

December 8, 2020

Missouri State University

Harvey Mudd College

Boston College

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This is problem number one for Stuart. Eighth edition sectional two point one Ah Tank holds a thousand gallons of water, which drains from the bottom of the tank and have an hour, and the valleys in the table show the volume be ofwater remaining in the tank in gallons after team. It's s so that's the problem. We're out of being with part ay if P is the point. Fifteen twenty two fifteen on the graph of V find slopes of the Secret Minds P Q. When Q is the point on the graph with T equals five tickles tentacles twenty equals twenty five anti equals thirty. So there are five different Q points. There s so let's start with the party in order to find a slope, we used the stop formula, which is a ratio of the rise on a graph divided by the run and for our data set from the rise is corresponds to the volume beam. So the change and being ah, over the change in time in this case, that's the run. And if we refer to the problem here, this table shows Dean specific values that we're working with, and we can see here this is Point Pete. AT T equals fifteen minutes. There are two hundred fifty gallons. That is point people. So in order to use this formula for the slope, we're going to be using the data off the remaining parts now as the cue and finding the slope of those seeking lines. For example, let's begin with AH, point peeing, which is at two fifteen. That's the vey Valu minus dovey Valley of the Q Point, which for the first part for first cue is six hundred ninety four at five minutes. So six hundred ninety four p is at fifteen minutes. First cue is at five minutes and again to show exactly what we're doing here. This is P, the point peen, and this is the first point. Q. And doing the math, we get negative four hundred forty for, and that other one we get ten in our scope is negative. Forty for four came, and that's just the first part of a part ay done by hand, going to go to our crunchy over here and then show that now if you have the table values here, this is directly from the table given in the problem. You can calculate the change in V with respect to this point p, and then you can see what that value is for each point. Q. And then you also calculate the change in time. And then again, slope, we use the formula. Rise, overrun, change of your change of tea and these are the corresponding answers for part A. The slope is negative. Forty poor, four point four at five minutes I got a thirty point eight at ten minutes. Native twenty seven point eight twenty minutes, twenty two point two at twenty five minutes and negative sixteen point six, repeating at thirty minutes. Oh, and that is a parting for part being were as to estimate the slope of the tangent line at P, averaging the slope of to seek in lines. And since point peas at fifteen minutes, the two slopes that we want to average are going to be the ones that ten minutes and a twenty minutes, So we're going to be looking at those two slopes and ah, taking that average thirty eight point AIDS added to native twenty seven point eight all the right it by two. That's the average we get an estimate of negative thirty three point three and that is our estimate, right? Averaging slopes of two secret mines around point P and part C. We're going to use a graph to estimate to stop in a different way by finding this look of the tension line at P again using a graph specifically to find this. So here we see this's thie data given to us, the table of values plotted out on a billion versus ten graph, and we're going to use a line here that is tangent at that time equals fifteen minutes. So we see this line perfectly attention. At that point, we make an estimate about where the height of this point is and where, at this point is and we do. The same thing with the slope is we take the rise, which, in this case, it's seven hundred fifty about divided by minus zero zero cerise. And then we do are riding by the run. Curious. Approximately twenty two and a half, right, Tiv minus zero. They should be able to give us the estimate. It's negative. Yeah. Now we're not. We see that this is consistent with our estimate and mark being come again parte si. We found that estimate by using the graft directly and estimating attention line at that point peeing.

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