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In December 1998, the Mars climate orbiter
was launched in hopes of providing detailed

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information about the atmosphere of Mars.
The launch went well. However the two teams

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that collaborated on the project worked in
different units--one used the metric system,

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and the other the English system, and the
two systems of units were never properly reconciled.

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So while the orbiter was intended to fall
into orbit around Mars, instead it crash landed

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on the surface. So units are pretty important.

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This video is part of the Problem Solving
video series. Problem-solving skills, in combination

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with an understanding of the natural and human-made
world, are critical to the design and optimization

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of systems and processes.

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Hi my name is Ken Kamrin, and I am a mechanical
engineering professor at MIT. Today I want

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to talk to you about using unit analysis in
problem solving. This might seem simple, but

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it is a critical tool for validating your
calculations. So pay attention.

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To understand this example, you should be
familiar with the definition of work.

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After watching this video, you should be able
to

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utilize and apply the key properties of unit
analysis:

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when two quantities are multiplied, their
units also multiply.

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all terms added, subtracted or equated must
have the same units.

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You should also be able to explain how derivatives
and integrals affect units.

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Before we start the main example, let's discuss
how integration and differentiation affect

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units. The first question is: what are the
units of dx? You should think of dx as a very

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small change in x or a "little bit of x",
Recall that a little bit of ice cream, is

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still ice cream. Even if it is a very very
small amount. So the units of dx are precisely

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the units of x. Let's do an example. Let's
consider the position function x(t) with units

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of meters, where t is time, with units of
seconds. Then velocity is the time derivative

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of position v = dx/dt, and has units of meters
per second. Let's look at how the notation

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is consistent with our physical interpretation:
dx has units of meters, and it is divided

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by dt, which has units of seconds. So dx over
dt should have units of meters over seconds.

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Now let's look at acceleration, which has
units of meters per second squared. We know

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that acceleration is the time derivative

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of velocity, which is the second derivative
of position with respect to time. Let's see

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how this plays out with notation... We can
also determine how integration affects units.

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Let's use the example of the integral of velocity
v with respect to time t. We know physically

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that this is position, which has units of
meters. Unit analysis also makes this clear

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because v has units of m/s, and dt has units
of s, so when we multiply, the seconds cancel,

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and we end up with units of seconds It might
be interesting to point out that the integral

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symbol doesn't affect units at all. Just like
the differential did not have units, it is

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just a symbol that represents the limit of
sums. Since every term in the sum has units

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of v times units of t, the sums, hence the
integral sign doesn't affect units at all.

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What we've seen: given a function f(y), differentiating
with respect to y divides the units of f by

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the units of y. Differentiating twice divides
the units of f by the units of y squared.

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Similarly, integrating with respect to y multiplies
the units off by the units of y

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In this example, we'll see how unit analysis
can help us check our calculation of the work

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done by an applied force. A machine in a factory
is programmed to apply a force to a 3.0kg

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object to move it back and forth in the horizontal
direction. The position of the object as a

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function of time is given by the equation
x equals 3.0t minus 4.0 t squared plus 1.0

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t cubed, where x is measured in units of meters,
and t in seconds. Find the work done on the

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object by the force from t=0 to t=2. Note
that work is done when a force is applied

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over a distance on an object. Let x0 be the
position of the object at time t = 0. Let

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x2 be the position at time t = 2. Then the
amount of work done on the object is computed

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by the equation W = integral x2 F · dx. Let's
start by figuring out the units of work. We

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do this by analyzing our equation, which tell
us that the units of work are the units of

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force times the units of dx. If we forget
what the units of force are, we can use unit

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analysis to figure it out. We know that F
= ma, and acceleration is in units of meters

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per second squared, and mass has units of
kilograms, So the unit of force, also known

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as the Newton, is equivalently written as
kilogram-meters per second squared. So the

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unit of force, also known as the Newton, is
equivalently written as kilogram-meters per

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second squared. Recall that the units of dx
are the units of x, which is meters. Thus

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the unit of work is a Newton-meter. So now
we know what units our answer needs to be

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in. So let's start solving our problem. In
order to solve our problem, we need to know

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the force acting on our object. We are given
the mass of the object, the position function

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for the object, and F=ma. Because we can determine
acceleration as the second derivative of position

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with respect to time, we can use our given
information to determine the force. Let's

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go ahead and solve this problem, perhaps incorrectly.
See if you can catch my mistake. Alright,

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so here we have position as a function of
time. So let's differentiate to get the velocity.

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Where I've left off the decimal points. And
we'll go ahead and differential the velocity

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to get the acceleration. And finally to get
the force, we go ahead with F=ma. So there

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is my m. And according to this, our expression
for a is -8 + 6t. So I look at this and I

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think to myself, uhoh, this can't be right,
because this doesn't have units of force.

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We know that force should have units of Newtons,
but it doesn't look like this has the correct

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units. So this entices us to go back into
our problem to figure out what went wrong.

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Looking back up at the first equation for
position, I see immediately that something

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has gone wrong, because position is in units
of meters, and each of these terms don't have

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units of meters. In fact they don't even agree
with each other.

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This tells me there had to be units for each
of these constants that somehow got swept

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under the rug. So let's be rigorous, and put
those units back in. Here we need units of

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meters per second to cancel with the seconds
from time, and leave us with units of meters.

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Note that this means something physical, it
is the initial velocity. Let's go back into

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this term here, and see that I have a t squared,
and I need something with units of meters

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per seconds squared if I am going to get from
this term something with units of meters.

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And lastly, we have this term over here that
has a t cubed. And likewise the missing unit

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must have units of meters per second cubed,
which physically is the unit of jerk, or the

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rate of change of acceleration.

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Now every term has units of meters, and so
this equation makes sense. You may want to

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pause the video here and find the units for
the constants in the formulas for velocity

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and acceleration. Show that these units agree
with the units you find by differentiating

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x(t) Now that you have checked the units of
acceleration, let's put it back in the force

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formula, and see that everything works out
ok. So we got F=ma. M as before equals 3 kilograms.

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And the acceleration written with correct
units is negative 8 point 0 meters per second

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squared plus 6 point 0 t meters per second
cubed. To see that this has the correct units,

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lets expand this formula. -24 kg m per s squared
yep, that's a Newton. + 18t kg m per s cubed.

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Just to check, note that while this doesn't
have units of N, time has units of seconds,

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so
this term is entirely in units of newtons
as well. Now that we've verified the of Force

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is in units of Newtons. we can compute the
work integral. That means we want to evaluate

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the integral: x0 to x2 of F dx. We have an
expression for F, so let's go ahead and put

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that in. I'm going to go ahead and leave off
the units for now. We leave it as an exercise

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for later that you will do that. What do we
get? Integral x0tox2 minus 24 plus 18t dx.

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So we look at this and we realize that we
actually have a slight problem. Because the

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force is written in terms of time, and we
are integrating with respect to x. Fortunately

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this is easy to resolve. Just remember that
a small change in x, dx is equivalently velocity

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times a small change in time. This is easy
to see if you express velocity as dx dt, because

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you multiply by dt, and it is clear that this
gives you a small change in x. So in your

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calculus class this little maneuver is called
a change of variable. So let's go ahead and

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do that. This means our integral can be written
as an integral over time, from time 0 to time

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t=2 of force times velocity times dt. We write
F as ma, and we recall that acceleration is

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the time derivative of velocity. Now we see
how to integrate this, and we find that the

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integral is m times velocity squared divided
by 2 evaluated at the two time end points.

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Plugging in our formula for the velocity,
and evaluating, we get that the answer is

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negative 12. I intentionally left off the
units while doing this computation. Now I

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leave it as an exercise for you to use unit
analysis to check that this integral does

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in fact give you 12 Newton-meters or 12 Joules,
the units for work. In the example, we saw

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that when you are adding numbers with units,
it is important for those numbers to have

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the same units if you want your quantity to
have physical meaning. We also learned that

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the differential quantity dx has the same
units as x. We also saw that it is helpful

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to use unit analysis to check your work at
points during and after the computation. You

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can see now that unit analysis can be a useful
part of your problem solving strategy. It

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is important to understand the properties
of units, and how they are affected by mathematical

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operations. Checking units at the end of a
computation useful to see if solution is reasonable.

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Sometimes, unit analysis can suggest a formula
for an unknown quantity in terms of given

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information. But physical knowledge is necessary
to validate solution completely.