00:01
In this problem, we're told that the demand for physicians is expected to increase in the future, as shown in this table drawn on the left.
00:09
In part a, we're asked to plot the data, letting t equal zero correspond to 2000.
00:14
So t is going to be the number of years after the year 2000.
00:23
So let's go ahead and plot our data.
00:27
So here is t equals zero, and let's make some marks for...
00:35
So let's redo these marks a little bit.
00:37
So let's say here is 5, 10, 15, 20, 25.
00:42
There's 10, 20, 25, and 15, and 5.
00:48
And on this axis, let's use the scale.
00:52
Let's kind of break.
00:54
And then we'll start at 600, and then 700, 800, 900, 900.
01:03
I'm just choosing to break my scale at 600, because the lowest number that we see is 680.
01:09
So we're going to just have a smaller scale on the y -axis.
01:12
All right, let's plot our point.
01:14
So in 2006, that's t -equals -6.
01:16
We had 680.
01:18
So it's about right there.
01:21
In 2015, that's t -equals 15.
01:24
We're up to 758 .6.
01:27
Oh, maybe right about there.
01:29
T -equals 20 at 2020.
01:31
That's 805.
01:32
So just above the 800 mark.
01:36
Finally, 2025 is 859.
01:39
So it's going to be right up here.
01:45
Now, if we kind of connect the dots, we can see that we do have a nice kind of increasing curve it looks like.
01:56
So it does look like an exponential graph could be a good fit.
02:09
In part b, we're asked to use the data for 2006 and 2015 to actually find an exponential graph that might fit this data.
02:18
So in 2006 and 2015, we want to fit this to the graph of the form c times e to the kt.
02:27
So in 2006, we have f of 6 is equal to 680 .5.
02:34
So we can write that as 680 .5 is equal to c times e to the k times 6.
02:43
Plug in t equals six to here.
02:47
In 2015, we're told that f of 15 is equal to 758 .6.
02:55
This is just coming from the graph.
02:56
So 758 .6 is equal to c times e to the power of k times 15.
03:06
So now what we have here is we have two equations and we have two unknowns.
03:14
We don't know c or k, but we have two equations and we're to try and use these equations to find c and k.
03:22
So the first thing let's do is let's isolate c in these equations.
03:27
So let's start with this one first.
03:31
So 680 .5 divided by e to the power of 6k is equal to c.
03:40
And this one we have 758 .6 divided by e to the k times 15.
03:47
Or let me rewrite that as 15 times k is equal to to c.
03:55
Now we have two c's, we can set them equal to each other, right? so if c is equal to c, then 680 .5 over e to the 6k is equal to 758 .6 over e to the 15k.
04:13
And now we can just rearrange this equation so that we can solve for k.
04:17
So if we multiply both sides by e to the 15k, and if we divide both sides by 680 .5, so maybe i'll write this in blue...