A cup of water at an initial temperature of 78^∘C is placed in a room at a constant temperature

step 1 Okay, let's get ready. We have a lot to do in this problem and most of it on a graphing utility.

A cup of water at an initial temperature of 78^∘C is placed...

A cup of water at an initial temperature of $ 78^{\circ}C $ is placed in a room at a constant temperature of $ 21^{\circ}C $. The temperature of the water is measured every 5 minutes during a half-hour period.The results are recorded as ordered pairs of the form $ (t $, $ T) $, where $ t $ is the time (in minutes) and $ T $ is the temperature (in degrees Celsius). $ \left(0, 78.0^{\circ}\right) $, $ \left(5 , 66.0^{\circ}\right) $, $ \left(10, 57.5^{\circ}\right) $, $ \left(15 , 51.2^{\circ}\right) $, $ \left(20 , 46.3^{\circ}\right) $, $ \left(25, 42.4^{\circ}\right) $, $ \left(30 , 39.6^{\circ}\right) $ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points $ \left(t , T\right) $ and $ \left(t, T - 21\right) $. (b) An exponential model for the data $ \left(t, T - 21\right) $ is given by $ T - 21 = 54.4\left(0.964\right)^t $ Solve for $ T $ and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use a graphing utility to plot the points $ \left(t, In\left(T - 21\right)\right) $ and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form $ In\left(T - 21\right) = at + b $. Solve for $ T $, and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the $ y $-coordinates of the revised data points to generate the points $ \dfrac{1}{T - 21} = at + b $. Solve for $ T $, and use a graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

A cup of water at an initial temperature of $ 78^{\circ}C $ is placed in a room at a constant temperature of $ 21^{\circ}C $. The temperature of the water is measured every 5 minutes during a half-hour period.The results are recorded as ordered pairs of the form $ (t $, $ T) $, where $ t $ is the time (in minutes) and $ T $ is the temperature (in degrees Celsius).

$ \left(0, 78.0^{\circ}\right) $, $ \left(5 , 66.0^{\circ}\right) $, $ \left(10, 57.5^{\circ}\right) $, $ \left(15 , 51.2^{\circ}\right) $, $ \left(20 , 46.3^{\circ}\right) $, $ \left(25, 42.4^{\circ}\right) $, $ \left(30 , 39.6^{\circ}\right) $

(a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points $ \left(t , T\right) $ and $ \left(t, T - 21\right) $.

(b) An exponential model for the data $ \left(t, T - 21\right) $ is given by $ T - 21 = 54.4\left(0.964\right)^t $ Solve for $ T $ and graph the model. Compare the result with the plot of the original data.

(c) Take the natural logarithms of the revised temperatures. Use a graphing utility to plot the points $ \left(t, In\left(T - 21\right)\right) $ and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form $ In\left(T - 21\right) = at + b $. Solve for $ T $, and verify that the result is equivalent to the model in part (b).

(d) Fit a rational model to the data. Take the reciprocals of the $ y $-coordinates of the revised data points to generate the points

$ \dfrac{1}{T - 21} = at + b $.

Solve for $ T $, and use a graphing utility to graph the rational function and the original data points.

(e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Transcript

00:01 Okay, let's get ready.

00:01 We have a lot to do in this problem and most of it on a graphing utility.

00:05 So here's my graphing calculator.

00:08 And the first thing i do is go into stat for statistics into edit.

00:12 And then you'll see that i've typed the x coordinates into list one and the y coordinates into list two.

00:18 They're being called lowercase t for time and capital t for temperature.

00:24 And then what we want to do for part a is we also want a list of all the temperatures minus 21.

00:30 And so what i'm going to do is just type in 78 minus 21, 66 minus 21.

00:39 I'm letting the calculator do the subtracting for me.

00:42 I'm just looking at what i have in list two and one by one subtracting 21 from it.

00:50 And there may even be a shortcut method for doing this all at once.

00:55 Luckily, they're not a whole lot of numbers, so it's not going to take too long to do each one, at a time.

01:04 Okay.

01:05 So there we have list three.

01:07 So what we want to do for part a is we want to make a scatter plot using list one and list two, and we also want to make a scatter plot using list one and list three.

01:16 So i'm going to go into my stat plot menu, which is second y equals, going to go into the menu for plot one, i'm going to turn it on.

01:25 By default, it's a scatter plot using list one and list two.

01:28 So i don't need to change anything there.

01:30 Now i'm going to go back into my stat plot menu, go down to choice number two, turn that scatter plot on.

01:37 It's going to be red.

01:39 It's a scatter plot as well, but it's going to use list one and list three.

01:43 So down here i need to change list two to list three, so i can press my second function button and then my three button, and that has l3 as its second function.

01:54 And so now to get a good viewing window, i'm going to go to zoom and choose zoom stat, which is zoom number nine and it should give me a viewing window that works well for this data.

02:05 Okay, we see them both.

02:07 We see in blue the original temperatures and we see in red the temperatures that have had 21 subtracted from them.

02:19 And then for part b, we're given this exponential equation and asked to solve for t.

02:23 So all we need to do is add 21 to both sides.

02:27 And then we're going to graph that using a calculator and we're going to compare it.

02:33 To the original data plot.

02:37 So let's do that.

02:39 Okay, so here we are back in the calculator and we go to y equals and then here we can type in that equation which was 54 .4 times 0 .964 to the power it's t in the book but you have to use an x in your calculator and then go out of the exponent and then have the plus 21 from isolating t in that equation.

03:04 Okay, now now let's just press graph and we should see this graph along with our scatter plots.

03:10 So the question is, how does this compare with the original data scatterplot? it looks like a really good fit.

03:17 It's not perfect, but it looks like a great fit it's passing through almost all of those points.

03:23 Now we're going to move on to part c and we're asked to convert the y coordinates one more time.

03:28 So we go back into stat and into edit.

03:32 And let's go over to list four.

03:34 And this time what we're doing is we're taking the next.

03:36 Natural log of the numbers in list three.

03:39 So we can just do those one at a time.

03:41 Natural log 57, natural log 45, natural log 36 .5, natural log 30 .2, natural log 25 .3, natural log 21 .4, and natural log 18 .6.

04:09 Okay, now we're going to plot this.

04:11 So let's go back into the stat plot window, down to plot three and this time let's change let's turn it on and let's change our y list and use list four so that will be a second function of four and press graph so here we have the things from before and i don't know about you but i don't see another scatter plot so what i'm thinking is that we're going to have to change the window and so what we might want to do is not look at these things three graphs anymore, but just look at the graph from list one and list four.

04:59 So let's go back into y equals.

05:02 Let's turn off the equal sign from that quadratic.

05:05 Let's turn off plot one.

05:07 We can turn the back on later if we need to.

05:09 Turn off plot two.

05:11 And now let's go back into zoom and do another zoom stat and it should use the data from list one and list four.

05:19 So now we can see the scatter plot that we want for part c.

05:23 Notice that it does look linear.

05:26 And so what we want to do is go into stat.

05:30 And once we're in stat, over to calculate.

05:34 And we can choose number four linear regression.

05:36 This is how we're going to find our regression equation.

05:40 Okay, we are going to use list one.

05:42 We are not going to use list two.

05:44 We're going to use list four.

05:45 So change this to l4.

05:48 And then we're probably going to want to graph this.

05:51 So what i'm going to do is store my regression equation in y2.

05:56 And so to get y2, we can press bars over to yvars, choose function, and then choose y2.

06:03 I didn't choose y1 because we already have that other equation in y1 right now, and i don't know if we're going to want to use it again.

06:11 So now let's calculate.

06:18 Okay, what we're going to do is substitute these numbers into the equation of the line, and we end up with something that looks like this.

06:27 I just rounded the numbers to three decimal places.

06:30 So then what we're asked to do is take that equation and solve for t.

06:35 Well, this is interesting because we haven't gotten to solving logarithmic equations in this chapter yet.

06:40 We're going to do that in the next section.

06:43 However, we can work around that if we think about how we can rearrange a logarithmic equation into an exponential equation.

06:50 So remember that natural log is log base e.

06:54 I'm just going to write it that way to make it obvious.

07:02 And then what we can do is remember that if we want to rearrange this into its exponential equation, form, we take our base, e, we raise it to whatever is over here on the right, negative 0 .037t plus 4, or excuse me, 3 .997, and then that should equal t minus 21.

07:22 So we're solving that equation for t, so that means we need to add 21 to both sides, and i'm going to go ahead and rearrange my sides at the same time.

07:30 So i get t equals e to the negative 0 .37t plus 3 .997 plus 21...

Please give Ace some feedback