BAM!!!
Mr. Tarrou In this example video, usually I am doing a full lesson but in this one we
are just going to do an example of a related rate problem in Calculus.
It is going to be involving putting some water we will say into a cone.
So a conical tank is 10 feet across, of course it could be any fluid, but it is 10 feet across
and 20 feet deep.
If water is flowing into this tank at the rate of 8 cubic feet per minute, find the
rate of change of the depth of the water when the water has reached a fixed depth of 10
feet.
It is not going to be a fixed depth of course because you are going to be putting water
into this tank and it is going to be constantly changing.
But at that instant where the depth is 10 feet how fast will the water be rising.
We need Calculus to figure this out because, well the cone has basically a diameter of
zero at the bottom and increasing to a diameter of 10.
So as that diameter increase the cone is going to be able to hold more volume at some particular
value of depth that you are at.
So the higher up the cone you are going to get as you are pouring the water in at a fixed
rate and not changing the rate at which the water is going into the cone, the rate at
which it is rising is going to slow down.
It is going to be variable.
Ok, so if we are going to do a related rates problem we need to start figuring out all
of the variables that we are given and what variables do we care about.
What variables are we going to trying to figure out at the end of the problem.
Well it says to find the rate of change of the depth of the water when it is 10 feet.
We are going to be looking for a rate of change, that means derivative.
We are going to be looking for dh/dt.
What is the rate of change for the height at some particular point in time, that is
going to be set by the fact the water is 10 feet deep.
Well that is what we don't know.
What else are we given?
We are saying that we are putting water into the tank at 8 cubic feet per minute.
You know, that is volume so the rate of change for the volume dv/dt is going to be in unit
of time measure is of course per minute, is 8 cubic feet per minute.
So 8 ft^3/min.
Kind of getting a little bit messy there.
Seems like that is what we have.
We have got the fixed overall depth of the cone.
We have...
Now I have worked out the fact that the radius is 5 because our volume formula for a cone
does not use diameter it uses radius.
So if the overall diameter is 10 the radius is 5.
Thus we are also going to mark off this depth of 10 feet for the water and some radius which
we don't know at this point for that depth of 10'.
Ok, now what do we do next.
Well, maybe you know but we can also just let the math tell us what we need to do next.
We are talking about rates of change so we are talking about derivatives.
We have volume or v is equal to 1/3 pi r squared h.
So we are thinking ok, derivative.
So dv/dt because we are going to be finding the derivative with respect to time.
You should not just write v prime, it is kind of confusing that way.
This is equal to...
Now my equation here has two variables in it.
Both of these variables are going to be variable.
The radius of which you have for the water is going to be changing as the water increases
in depth.
Of course the height is going to be changing.
So to find the derivative of this function we need to use the product rule.
It is going to be the first factor times the derivative of the second factor plus the second
factor times the derivative of the first factor.
This means that we are going to need to use the Chain Rule.
We are going to take that 2 and multiply it to 1/3, reduce that power by one which is
going to be just r, and then again because we are finding the derivative with respect
to time it is just that... you have to multiply by the derivative of the base.
I don't know why I put that parenthesis there.
Ok, now look at our derivative that we have for volume.
I can clean it up a little bit, but I don't need to worry about doing that because I see
that hmm... well... in my derivative, it is asking for the rate of change for the radius.
I am not given any information about that and that means that I am just going to really
over complicate the problem if I try and work with this formula initially both in terms
of r and h.
Because I care about is finding out how fast the water is rising.
The change in depth or I have labelled as h for height.
So let's rewrite this formula so that we don't have a variable of r in it so we don't have
to worry about that rate which we are not given.
Excuse the interruption.
My dogs were barking so I wanted to wait for them to stop and calm them down.
So we are looking to get this equation just in terms of h.
Well, the sides of the cone are going to allow us to do that.
The cone with the straight sides is giving us a constant slope showing us a relationship
between the radius and the height.
Let's see here.
What is the ratio of the height of the cone, or then the water, is going to be... the height
is going to be over the radius.
You can do r/h.
It does not make any difference when you are setting up equal ratios as long as you are
consistent with both of those fractions or ratios.
So the overall height of the cone is 20 and the radius of the cone is 5.
This relationship is constant again because it is a cone.
The sides are linear functions or lines so the slope is going to be constant.
That is going to allow us to write equation in terms of only h if I solve this for r.
If we cross-multiply we get 5h is equal to 20r.
Divide both sides by 20, trying to solve for r so that I can take the r out and plug in
some kind of expression that just has an h in it.
So we have r is equal to 5 over 20 is 1/4.
Excellent.
Now I can come over here and say well now my volume is equal to 1/3 pi times r which
is equal to 1/4h squared times h.
So now I have volume in terms of just h.
Let's clean this up a little bit.
We have volume is equal to 1/3 pi. 1/4 squared is 1/16th.
Then we have h squared times h.
Cleaning this up a little bit more we have volume is equal to 3 times 16 is going to
be 48 so 1/48 pi.
Let's put that pi on top, now h cubed.
When you multiply like bases you add those exponents so h^2 times h is going to be h^3.
Now when I find the derivative I am only going to have a value for... or a variable of the
volume... or the rate of change so dv/dt.
The derivative of the volume with respect to time or the change in volume with respect
to time.
3 times pi/48 is going to be 3pi/48 times h squared.
Again we are finding the derivative with respect to time, not just a derivative with respect
to h, so we have to multiply by the derivative of that inside function or that base which
is going to be dh/dt.
That simplifies to be 1/16.
So let's just do that.
Ok.
Now that I have taken my volume, I have found the derivative basically implicitly if you
will, not basically I did.
I have the change in volume is equal to pi/16 times the height squared times the dh/dt...
basically the rate of change for the height.
Now I can finally bring in that fixed depth of the water being 10 feet.
Now that I have found the derivative and that allows us to have some variables in here about
rates of change, I can go ahead and plug all these numbers in.
The volume...the water is flowing in at 8 cubic feet per minute.
That is 8 is equal to pi/16, the fixed height we want to find the rate of change of the
depth when the water is 10 feet deep.
We are looking for dh/dt.
I am starting to run out of room and I am getting to a point that I need to...you know...worry
about making sure that I don't make any small calculation errors in my head.
So of... well I can obviously 100 squared is 100... 10 squared is 100 so we get 8 is
equal to 100pi/16 dh/dt.
If we want to make sure that we keep that in exact format, we would take this and we
are going to...right off the top off of my head multiply both sides by the reciprocal.
So dh/dt is equal to 8/1 times...take this fraction and flip it and multiply it to both
sides of the equation ...so times 16...
I could have this reduced but 16/100pi.
If you are looking for again an answer in exact format this is going to be dh/dt equals
80 and 48 is going to be 128/100pi feet per minute.
Of course that 128/100 is reducible.
I have my answer both in .... haha... using your notes is only good if they don't have
any mistakes in them.
I have 2 decimal approximations for the rate of change.
I realized that one of them has to be wrong.
Apparently I should have erased one of them.
128 over 100pi reduces to 32/25pi feet per minute.
That is again the rate of change for the depth of the water when it is 10 feet deep.
The decimal approximation of the answer to three decimal places is .407 feet per minute.
I am Mr. Tarrou.
BAM!!!
Go do your homework:D