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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: In the twelfth
lecture, we're going to talk about maxima and minima. Let's finish up what
we did last time. We really only just started
with maxima and minima. And then we're going to talk
about related rates. So, right now I want to give you
some examples of max-min problems. And we're going to
start with a fairly basic one. So what's the thing about
max-min problems? The main thing is that we're
asking you to do a little bit more of the interpretation of
word problems. So many of the problems are expressed
in terms of words. And so, in this case, we have
a wire which is length 1. Cut into two pieces. And then each piece
encloses a square. Sorry, encloses a square. And the problem - so
this is the setup. And the problem is to find the
largest area enclosed. So here's the problem. Now, in all of these cases, in
all these cases, there's a bunch of words. And your job is typically
to draw a diagram. So the first thing you want to
do is to draw a diagram. In this case, it can be
fairly schematic. Here's your unit length. And when you draw the diagram,
you're going to have to pick variables. So those are really the
two main tasks. To set up the problem. So you're drawing a diagram. This is like word problems of
old, in grade school through high school. Draw a diagram and name
the variables. So we'll be doing a
lot of that today. So here's my unit length. And I'm going to choose the
variable x to be the length of one of the pieces of wire. And that makes the other
piece 1 - x. And that's pretty much the whole
diagram, except that there's something that we
did with the wire after we cut it in half. Namely, we built two little
boxes out of it. Like this, these are
our squares. And their side lengths are
x / 4 and (1 - x) / 4. So, so far, so good. And now we have to think, well,
we want to find the largest area. So I need a formula for area
in terms of variables that I've described. And so that's the last thing. I'll give the letter a as
the label for the area. And then the area is just
the square of x/4 + the square of 1 - x. Whoops, that strange
2 got in here. Over 4. So far, so good. Now, The instinct that you'll
have, and I'm going to yield to that instinct, is we should
charge ahead and just differentiate. Alright? That's alright. We'll find the critical
points. So we know that those are
important points. So we're going to find
the critical points. In other words, we take the
derivative we set, the derivative of a with
respect to x = 0. So if I do that differentiation,
I get the, well, so the first one,
x^2 / 16, that's 8. Sorry. That's x / 8, right? That's a derivative of this. And if I differentiate this, I
get well, the derivative of 1 - x ^2 is 2 ( 1 - x)( a - 1). So it's - (1 - x) / 8. So there are two minus signs in
there, I'll let you ponder that differentiation, which
I did by the chain rule. Hang on a sec, OK? Just wait until we're done. So here's the derivative. Is there a problem? STUDENT: [INAUDIBLE] PROFESSOR: Right, so there's
a 1/16 here. This is x^2 / 16. And so it's 2x / 8,
over 16, sorry. Which has an 8. That's OK. Alright, so now, This is equal
to 0 if and only if x = 1 - x. That's 2x = = 1, or in
other words x = 1/2. Alright? So there's our critical point. So x = 1/2 is the
critical point. And the critical value, which
is what you get when you evaluate a at 1/2, is (1/2)
/ 4, that's 1/8. So that's (1/8)^2 + (1/8)^2
which = 1/32. So, so far, so good. But we're not done yet. We're not done. So why aren't we done? Because we haven't checked
the end points. So let's check the end points. Now, in this problem,
the end points are really sort of excluded. The ends are between
0 and 1 here. That's the possible lengths
of the cut. And so what we should really be
doing is evaluating in the limit, so that would be the
right-hand limit as x goes to 0 of a. And if you plug in x = 0, what
you get here is 0 + (1/4)^2. Which is 1/16. And similarly, at the other
end, that's 1 - 1 from the left we get (1/4)^2 + 0,
which is also 1/16. So, what you see is that the
schematic picture of this function, and isn't even so far
off from being the right picture here, is that it's level
here as 1/16 and then it dips down and goes up. Right? This is 1/2, this is
1, and this level here is a half that. This is 1/32. So we did not find, when we
found the critical point we did not find the largest
area enclosed. We found the least
area enclosed. So if you don't pay attention
to what the function looks like, not only will you about
half the time get the wrong answer, you'll get the absolute
worst answer. You'll get the one which
is the polar opposite from what you want. So you have to pay a little
bit of attention to the function that you've got. And in this case it's
just very schematic. It dips down and goes up, and
that's true of pretty much most functions. They're fairly simple. They maybe only have
one critical point. They only turn around once. But then, maybe the critical
point is the maximum or maybe it's the minimum. Or maybe it's neither,
in fact. So we'll be discussing that
maybe some other time. So what we find here
is that we have the least area enclosed. Enclosed is 1/32. And this is true when x = 1/2. So these are equal squares. And most when there's
only one square. Which is more or less the
limiting situation. If one of the pieces
disappears. Now, so that's the first
kind of example. And I just want to make one more
comment about terminology before we go on. And I will introduce it with
the following question. What is the minimum? So, what is the minimum? Yeah. STUDENT: [INAUDIBLE] PROFESSOR: Right. The lowest value of
the function. So the answer to that
question is 1/32. Now, the problem with this
question and you will, so that refers to the minimum value. But then there's this other
question which is where is the minimum. And the answer to
that is x = 1/2. So one of them is the minimum
point, and the other one is the minimum value. So they're two separate
things. Now, the problem is that
people are sloppy. And especially since you usually
find the critical point first, and the value
that is plugging in for a second, people will stop short
and they'll give the wrong answer to the question,
for instance. Now, both questions are
important to answer. You just need to have
a word to put there. So this is a little
bit careless. When we say what is
the minimum, some people will say 1/2. And that's literally wrong. They know what they mean. But it's just wrong. And when people ask
this question, they're being sloppy. Anyway. They should maybe be a little
clearer and say what's the minimum value. Or, where is the
value achieved. It's achieved at, or where is
the minimum value achieved. "Where is min achieved?",
would be a better way of phrasing this second question. So that it has an unambiguous
answer. And when people ask you for
the minimum point, they're also - so why is it that we
call it the minimum point? We have this word, critical
point, which is what x = 1/2 is in critical value. And so I'm making those same
distinctions here. But there's another notion of
a minimum point, and this is an alternative if you like. The minimum point is the
point (1/2, 1/32). Right, that's a point
on the graph. It's the point - well, so that
graph is way up there, but I'll just put it on there. That's this point. And you might say min there. And you might point to this
point, and you might say max. And similarly, this one
might be a max. So in other words, what this
means is simply that people are a little sloppy. And sometimes they mean
one thing and sometimes they mean another. And you're just stuck with this,
because there'll be some authors who will say one thing
and some people will mean another and you just have to
live with this little bit of annoying ambiguity. Yeah? STUDENT: [INAUDIBLE] PROFESSOR: OK, so that's
a good - very good. So here we go, find the
largest area enclosed. So that's sort of a trick
question, isn't it? So there are various - that's
a good thing to ask. That's sort of a trick
question, why? Because according to the rules,
we're trapped between the two maxima at something
which is strictly below. So in other words, one answer to
this question would be, and this is the answer that I would
probably give, is 1/16. But that's not really true. Because that's only
in the limit. As x goes to 0, or
as x goes to 1 -. And if you like, the
most is when you've only got one square. Which breaks the rules
of the problem. So, essentially, it's
a trick question. But I would answer
it this way. Because that's the most
interesting part of the answer, which is that it's 1/16
and it occurs really when one of the squares disappears
to nothing. So now, let's do another
example here. And I just want to illustrate
the second style, or the second type of question. Yeah. STUDENT: [INAUDIBLE] PROFESSOR: The question is,
since the question was, what was the largest area, why did
we find the least area. The reason is that when we go
about our procedure of looking for the least, or the most,
we'll automatically find both. Because we don't know
which one is which until we compare values. And actually, it's much more
to your advantage to figure out both the maximum and minimum
whenever you answer such a question. Because otherwise you won't
understand the behavior of the function very well. So, the question. We started out with one
question, we answered both. We answered two questions. We answered the question of
what the largest and the smallest value was. STUDENT: Also, I'm wondering
if you can check both the minimum [INAUDIBLE] approaches
[INAUDIBLE]. PROFESSOR: Yes. One can also use, the question
is, can we use the second derivative test. And the
answer is, yes we can. In fact, you can actually also
stare at this and see that it's a sum of squares. So it's always curving up. It's a parabola with a positive
second coefficient. So you can differentiate
this twice. If you do you'll get
1/8 + another 1/8 and you'll get 1/16. So the second derivative
is 1/16. Is 1/4. And that's an acceptable
way of figuring it out. I'll mention the second
derivative test again, in this second example. So let me talk about
a second example. So again, this is going to
be another question. STUDENT: [INAUDIBLE] PROFESSOR: The question is, when
I say minimum or maximum point which will I mean. STUDENT: [INAUDIBLE] PROFESSOR: So I just repeated
the question. So the question is,
when I say minimum point, what will I mean? OK? And the answer is that for the
purposes of this class I will probably avoid saying that. But I will say, probably, where
is the minimum achieved. Just in order to avoid that. If I actually sat at I often am
referring to the graph, and I mean this. And in fact, when you get your
little review for the second exam, I'll say exactly that
on the review sheet. And I'll make this very clear
when we were doing this. However, I just want to prepare
you for the fact that in real life, and even me when
I'm talking colloquially, when I say what's the minimum point
of something, I might actually be mixing it up with this
other notion here. So let's do another example. So this is an example to get
us used to the notion of constraints. So we have, so consider
a box without a top. Or, if you like, we're going to
find the box without a top. With least surface area
for a fixed volume. Find the box without a top with
least surface area for a fixed volume. The procedure for working this
out is the following. You make this diagram. And you set up the variables. In this case, we're going to
have four names of variables. We have four letters that
we have to choose. And we'll choose them
in a kind of a standard way, alright? So first I have to tell
you one more thing. Which is something that we could
calculate separately but I'm just going to give
it to you in advance. Which is that it turns
out that the best box has a square bottom. And that's going to get rid of
one of our variables for us. So it's got a square bottom,
and so let's draw a picture of it. So here's our box. Well, that goes down like this,
almost. Maybe I should get it a little farther down. So here's our box. Let's correct that just a bit. So now, what about the
dimensions of this box? Well, this is going to be
x, and this is very foreshortened, but
it's also x. The bottom is x by x, it's
the same dimensions. And then the vertical
dimension is y. So far, so good. Now, I promised you two
more letter names. I want to compute the volume. The volume is, the base is x
^2, and the height is y. So there's the volume. And then the area, the area is
the area of the bottom, which is x ^2, that's the bottom. And then there are
the four sides. And the four sides are
rectangles of dimensions xy. So it's 4xy. So these are the sides. And remember, there's no top. So that's our setup. So now, the difference between
this problem and the last problem is that there are two
variables floating around, namely x and y, which
are not determined. But there's what's called
a constraint here. Namely, we've fixed the
relationship between x and y. And so, that means that we can
solve for y in terms of x. So y = v / x ^2. And then, we can plug that
into the formula for a. So here we have a which is
x ^2 + 4x ( v / x^2). Question. STUDENT: [INAUDIBLE] PROFESSOR: The question is,
will you need to know this intuitively? No. That's something that I would
have to give to you. I mean, it's actually true
that a lot of things, the correct answer is something
symmetric. In this last problem, the
minimum turned out to be exactly halfway in between
because there were sort of equal demands from
the two sides. And similarly, here, what
happens is if you elongate one side, you get less - it actually
is involved with a two variable problem. Namely, if you have a rectangle
and you have a certain amount of length
associated with it. What's the optimal thing
you can do with that. But I won't, in other words,
the optimal rectangle, the least perimeter rectangle,
turns out to be a square. That's the little sub-problem
that leads you to this square bottom. But so that would have been a
separate max-min problem. Which I'm skipping, because I
what to do this slightly more interesting one. So now, here's our formula for
a, and now I want to follow the same procedure as before. Namely, we look for the
critical point. Or points. So let's take a look. So again, a is (x
^2 + 4v) / x. And A' = 2x - (4v / x^2). So if we set that equal to
0, we get 2x = 2v / x ^2. So 2x^3. How did that happen
to change into 2? Interesting, guess
that's wrong. OK. So this is x ^ 3 = 2v. And so
x = (2 ^ 1/3)( v ^1/3). So this is the critical point. So we are not done. Right? We're not done, because we don't
even know whether this is going to give us the worst
box or the best box, from this point of view. The one that uses the most
surface area or the least. So let's check the ends,
right away. To see what's happening. So can somebody tell me what
the ends, what the end values of x are? Where does x range from? STUDENT: [INAUDIBLE] PROFESSOR: What's the smallest
x can be, yeah. STUDENT: [INAUDIBLE] PROFESSOR: OK, the claim was
that the largest x could be root a, because somehow there's
this x ^2 here and you can't get any further
past than that. But there's a key feature
here of this problem. Which is that a is variable. The only thing that's fixed in
the problem is v. So if v is fixed, what do you
know about x? STUDENT: [INAUDIBLE] PROFESSOR: x > 0, yeah. The lower end point,
that's safe. Because that has to do
geometrically with the fact that we don't have any boxes
with negative dimensions. That would be refused by the
Post Office, definitely. Over and above the empty
top, which they wouldn't accept either. STUDENT: [INAUDIBLE] PROFESSOR: It's true that x
< square root of v / y. So that's using this
relationship. But notice that y = v / x ^2. So 0 to infinity, I just
got a guess there over here, that's right. Here's the upper limit. So this is really important
to realize. This is most problems. Most
problems, the variable if it doesn't have a limitation,
usually just goes out to infinity. And infinity is a very important
end for the problem. It's usually an easy end
to the problem, too. So there's a possibility that if
we push all the way down to x = 0, we'll get a better box. It would be very strange box. A little bit like our
vanishing enclosure. And maybe an infinitely
long box, also very inconvenient one. Might be the best box. We'll have to see. So let's just take a look
at what happens. So we're looking at a, at 0 +. And that's x^2 + 4v
/ x with x at 0 +. So what happens to that? Notice right here, this
is going to infinity. So this is infinite. So that turns out
to be a bad box. Let's take a look at
the other end. So this is x ^2 + 4v / x,
x going to infinity. And again, this term
here means that this thing is infinite. So the shape of this thing,
I'll draw this tiny little schematic diagram over here. The shape of this thing
is like this, right? And so, when we find that one
turnaround point, which happened to be at this strange
point 2/3 , (2 ^ 1/3)( v ^ 1/3), that is going
to be the minimum. So we've just discovered
that it's the minimum. Which is just what we
were hoping for. This is going to be
the optimal box. Now, since you asked earlier and
since it's worth checking this as well, let's also check
an alternative justification. So an alternative to checking
ends is the second derivative test. I do not recommend the
second derivative test. I try my best, when I give you
problems, to make it really hard to apply the second
derivative test. But in this example, the function is simple
enough so that it's perfectly OK. If you take the derivative
here, remember, this was whatever it was, 2x
- (4v / x ^2). If I take the second
derivative, it's 2 + (8v / x ^3). And that's positive. So this thing is concave up. And that's consistent
with its being, the critical point is a min. Is a minimum point. See how I almost said,
is a min, as opposed to minimum point. So watch out. Yes. STUDENT: [INAUDIBLE] PROFESSOR: You're one
step ahead of me. The question is, is this the
answer to the question or would we have to give y and
a and so on and so forth. So, again, this is something
that I want to emphasize and take my time with right now. Because it depends, what kind
of real life problem you're answering, what kind of
answer is appropriate. So, so far we've found
the critical point. We haven't found the
critical value. We haven't found the dimensions
of the box. So we're going to spend a little
bit more time on this, exactly in order to address
these questions. So, first of all. The value of y. So, so far we have x =
(2 ^ 1/3)( v ^ 1/3). And certainly if you're going
to build the box, you also want to know what
the y value is. The y value is going
to be, let's see. Well, it's v / x ^2, so that's
v / ((2 ^ 1/3)( v ^ 1/3) ^2, which comes out to be (2
^ - 2/3)( v ^ 1/3). So there's the y value. On top of that, we could figure
out the value of a. So that's also a perfectly
reasonable part of the answer. Depending on what one is
interested in, you might care how much money it's going to
cost you to build this box. This optimal box. And so you plug in
the value of a. So a, let's see, is up here. It's x ^2 + 4v / x. So that's going to be ((2 ^
1/3)( v ^ 1/3) ^2 + (4v / (2 ^ 1/3)( v ^ 1/3)). And if you work that all out,
what you get turns out to be 3 ( 2 ^ 1/3)( v ^ 2/3). So if you like, one way of
answering this question is these three things. That would be the minimum point corresponding to the graph. That would be the answer
to this question. But the reason why I'm carrying
it out in such detail is I want to show you that there
are much more meaningful ways of answering
this question. So let me explain that. So let me go through some more
meaningful answers here. The first more meaningful answer
is the following idea simply, what are known as
dimensionless variables. So the first thing that you
notice is the scaling law. That a / v ^ 2/3 is
the thing that's a dimensionless quantity. That happens to be
3 ( 2 ^ 1/3). So that's one thing. If you want to expand the
volume, you'll have to expand the area by the 2/3 power
of the volume. And if you think of the area
as being in, say, square inches, and the volume of the
box as being in cubic inches, then you can see that this is
a dimensionless quantity and you have a dimensionless
number here, which is a characteristic independent
of what a and v were. The other dimensionless
quantity is the y:x. So x : y. So, again, that's inches
divided by inches. And it's (2 ^ 1/3)( v ^ 1/3) /
( 2 ^ - 2/3)( v ^ 1/3), which happens to be 2. So this is actually the best
answer to the question. And it shows you that the
box is a 2:1 box. If this is 2 and this is
1, that's the good box. And that is just the shape,
if you like, and it's the optimal shape. And certainly that,
aesthetically, that's the cleanest answer to
the question. There was a question
right here. Yes. STUDENT: [INAUDIBLE] PROFESSOR: Could you repeat
that, I couldn't hear. STUDENT: I'm wondering if you'd
be able to get that answer if you [INAUDIBLE]
square. PROFESSOR: The question is,
could we have gotten the answer if we weren't told that
the bottom was square. The answer is, yes in 18.02
with multivariable. You would have to have three
letters here, an x, a y, and a z, if you like. And then you'd have to work
with all three of them. So I separated out into one,
there's a separate one variable problem that you
can do for the base. And then this is a second one
variable problem for this other thing. And it's just two consecutive
one variable problems that solve the multivariable
problem. Or, as I say in multivariable
calculus, you can just do it all in one step. Yeah? STUDENT: [INAUDIBLE] PROFESSOR: Why did I divide
x by y, rather than y by x, or in any? So, again, what I was aiming
for was dimensionless quantities. So x and y are measured
in the same units. And also the proportions
of the box. So that's another word for
this is proportions. Are something that's universal,
independent of the volume v. It's something
you can say about any box, at any scale. Whether it be, you know,
something by Cristo in the Common. Maybe we'll get in here
to do some fancy -- STUDENT: [INAUDIBLE] PROFESSOR: The proportions is
with geometric problems typically, when there's a
scaling to the problem. Where the answer is the
same at small scales and at large scales. This is capturing that. So that's why, the ratios are
what's capturing that. And that's why it's
aesthetically the nicest thing to ask. STUDENT: So, what exactly does
the ratio of the area to the volume ratio [INAUDIBLE]
tell us? PROFESSOR: Unfortunately,
this number is a really obscure number. So the question is what
does this tell us. The only thing that I want to
emphasize is what's on the left-hand side here. Which is, it's the area to the
2/3 power of the volume, so it's a dimensionless quantity
that happens to be this. If you do this, for example,
in general with planar diagrams, circumferenced area
is a bad ratio to take. What you want to take
is the square of circumference to area. Because the square of
circumference has the same dimensions; that is, say,
inches squared to area. Which is in square inches. So, again, it's these
dimensionless quantities that you want to cook up. And those are the ones that
will have universal properties. The most famous of these is the
circle that encloses the most area for its
circumference. And, again, that's only true if
you take the square of the circumference. You do the units correctly. Anyway. So we're here, we've
got a shape. We've got an answer
to this question. And I now want to
do this problem. Well, let's put it this way. I wanted to do this problem
by a different method. I think I'll take the
time to do it. So I want to do this problem
by a slightly different method here. So, here's Example 2 by implicit
differentiation. So the same example, but now I'm
going to do it by implicit differentiation. Well, I'll tell you the
advantages and disadvantages to this method here. So the situation is, you have
to start the same way. So here is the starting
place of the problem. And the goal was the minimum
of a with v constant. So this was the situation
that we were in. And now, what I want to do
is just differentiate. The function y is implicitly
a function of x, so I can differentiate the first
expression. And that yields 0 = 2xy
+ (x ^2 )( y'). So this is giving me my implicit
formula for y', So y' = - 2xy / x ^2. Or in other words, - 2y / x. And then I also have
the dA/dx. Now, you may notice I'm not
using primes quite as much. Because all of the variables are
varying, and so here I'm emphasizing that it's a
differentiation with respect to the variable x. And this becomes
2x + 4y + 4xy'. So again, this is using
the product rule. And now I can plug in for
what y' is, which is right above it. So this is 2x + 4y
+ 4x ( - 2y / x). And that's equal to 0. And so let's gather
that together. So what do we have? We have 2x + 4y, and
then, altogether, this is 8 - 8y = 0. So that's the same
thing as 2x = 4y. The - 4y goes to
the other side. And so, x / y = 2. So this, I claim, so you have
to decide for yourself. But I claim that
this is faster. It's faster, and also it gets
to the heart of the matter, which is this scale in
variant proportions. Which is basically also nicer. So it gets to the nicer
answer, also. So those are the advantages
that this has. So it's faster, and it
gets to this, I'm going to call it nicer. And the disadvantage is
it did not check. Whether this critical point
is a max, min, or neither. So we didn't quite finish
the problem. But we got to the answer very
fast. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: How would
you check it? STUDENT: [INAUDIBLE] PROFESSOR: Well, so it gives
you a candidate. The answer is - so the
question is, how would you check it? The answer is that for this
particular problem, the only way to do it is to do
something like this. So in other words, it doesn't
save you that much time. But with many, many, examples,
you actually can tell immediately that if the two
ends, the thing is, say, 0, and inside it's positive. Things like that. So in many, many, cases
this is just as good. So now I'm going to change
subjects here. But the subject that I'm going
to talk about next is almost, is very, very closely linked. Namely, I talked about implicit
differentiation. Now, we're going to just
talk about dealing with lots of variables. And rates of change. So, essentially, we're going
to talk about the same type of thing. So, I'm going to tell you
about a subject which is called related rates. Which is really just another
excuse for getting used to setting up variables
and equations. So, here we go. Related rates. And i'm going to illustrate this
with one example today, one tomorrow. So here's my example
for today. So, again, this is going
to be a police problem. But this is going to be a word
problem and - sorry, I'm don't want to scare you, no police. Well, there are police in the
story but they're not present. So, but I'm going to draw it
immediately with the diagram because I'm going to save
us the trouble. Although, you know, the point
here is to get from the words to the diagram. So you have the police, and
they're 30 feet from the road. And here's the road. And you're coming along, here,
in your, let's see, in your car going in this
direction here. And the police have radar. Which is bouncing
off of your car. And what they read off is that
you're 50 feet away. They also know that you're
approaching along the line of the radar at a rate of
80 feet per second. Now, the question is,
are you speeding. That's the question. So when you're speeding, by the
way, up 95 feet per second is about 65 miles per hour. So that's the threshold here. So what I want to do now is
show you how you set up a problem like this. This distance is 50. This is 30, and because it's
the distance to a straight line you know that this
is a right angle. So we know that this is
a right triangle. And this is set out to be a
right triangle, which is an easy one, a 3, 4, 5 right
triangle just so that we can do the computations easily. So now, the question is, how
do we put the letters in to make his problem work, to
figure out what the rate of change is. So now, let me explain
that right now. And we will actually do the
computation next time. So the first thing is, you
have to understand what's changing and what's not. And we're going to use t
for time, in seconds. And now, an important distance
here is the distance to this foot of this perpendicular. So I'm going to name that x. I'm going to give
that letter x. Now, x is varying. The reason why I need a letter
for it as opposed to this 40 is that it's going to have
a rate of change with respect to t. And, in fact, it's related to,
the question is whether dx / dt is faster or slower
than 95. So that's the thing
that's varying. Now, there's something
else that's varying. This distance here
is also varying. So we need a letter for that. We do not need a letter
for this. Because it's never changing. We're assuming the police
are parked. They're not ready to roar out
and catch you just yet, and they're certainly not in motion
when they've got the radar guns aimed at you. So you need to know something
about the sociology and style of police. So you need to know things
about the real world. Now, the last bit is, what
about this 80 here. So this is how fast you're
approaching. Now, that's measured along
the radar gun. I claim that that's d by dt
of this quantity here. So this is d is also changing. That's why we needed a
letter for it, too. So, next time, we'll just put
that all together and compute dx / dt.
