Professor Dave again, let’s talk about related rates.
There are plenty of real-world scenarios that calculus can apply to, and now that we’ve
learned about implicit differentiation, we are ready to look at one such application,
the concept of related rates.
Let’s say we are inflating a balloon, and let’s pretend that the balloon is a perfect
sphere, so we essentially have a sphere that is expanding at a constant rate.
Now say we want to describe the rate at which the volume of the sphere is increasing, as
well as the rate at which the radius of the sphere is increasing.
Both of these values are increasing due to the expansion of the balloon, and their rates
of increase are related, hence related rates.
But one of these is probably much easier to physically measure than the other, so rather
than trying to measure both, let’s measure the one that we can, and compute the rate
of change in the other using calculus.
It’s probably easier to measure the change in volume, so let’s say that we do measure
the volume over some period of time, and we find that it is increasing by one hundred
cubic centimeters per second.
How can we use this to find out how fast the radius is increasing at a particular instant,
like the moment when the diameter of the balloon is fifty centimeters?
Well let’s say that volume is V and radius is r.
Then dV over dt, or the change in volume over time, equals one hundred cubic centimeters
per second, as we measured.
But what we want to know is dr over dt, or the change in radius over time, at the moment
that the radius is twenty five centimeters, since the radius is half the diameter.
To see how these values will relate, we need an equation that relates V and r, and that
will be the equation for the volume of the sphere.
When we learned geometry, we saw that this formula is V equals four thirds pi r cubed.
Now as we said, what we are looking for is dr over dt, so let’s differentiate this
with respect to t.
For V, that simply becomes dV over dt, but for the right side we will need to use the chain rule.
This is because t is not present in this equation, the radius is the variable, but the radius
is itself a function of t, which is why we use the chain rule.
So we will have to differentiate this expression with respect to r, and then multiply that
by the derivative of r with respect to t, or dr over dt.
The derivative with respect to r is easy, we just bring the three down here, and change
the three to a two, giving us four pi r squared, and then that must be followed by dr over dt.
Well dr over dt is what we are trying to solve for, because we are interested in how the
radius changes over time, so let’s solve for that.
This means simply dividing both sides by four pi r squared.
That means that dr over dt equals dV over dt, times one over four pi r squared.
We plug in our value for dV over dt, which is one hundred, and the value for r that we
are interested in, which is twenty five.
We evaluate, and we get one over twenty five pi centimeters per second.
Let’s try another example.
Say we have a ladder resting against a wall.
The ladder is ten feet long, and it is sliding away from the wall at a rate of one foot per second.
How fast is the top of the ladder sliding down the wall when the bottom of the ladder
is six feet away from the wall?
Just as with any other example we could look at, we will quickly see that we need to be
able to construct equations that represent the scenario.
Looking at the diagram, we have a ladder with a constant length of ten, and then these values
are the variables x and y, as they both change over time.
So that makes a right triangle with a known hypotenuse and the two legs unknown.
We can relate the variables by the Pythagorean theorem.
That would be x squared plus y squared equals ten squared, or a hundred.
Now as we said, we want to know how these parameters change over time, so let’s differentiate
both sides of this equation with respect to time.
The derivative of x squared with respect to time will be two x times dx over dt.
For y, we get the same thing, two y times dy over dt.
On the right, the derivative of any constant is zero.
Now we want to know about how fast the ladder is sliding down the wall, which is described
by dy over dt, so let’s solve for dy over dt.
We take the term with all the x’s to the other side, and then we divide by two y, and
we get negative x over y times dx over dt.
So all we have to do is plug in what we know, and solve.
We are asking about dy over dt when the ladder is six feet away from the wall.
That means x must be six.
Then, by the pythagorean theorem, we can solve for y at that instant, and that must be eight.
We can also plug in dx over dt, because we said that the ladder was sliding away from
the wall at one foot per second.
That means x is increasing by one foot every second, so dx over dt is positive one.
We simplify, and we get negative three fourths feet per second.
This is a negative value, because y is decreasing over time, as the ladder slides down the wall.
So as we can see, related rates problems can be rather straightforward.
The real challenge lies in understanding what the problem is asking, and being able to first
draw a diagram.
Then we must write equations from that diagram that relate the two variables in question.
Because we are describing rates, these will always be functions of time, and we will be
taking derivatives of functions with respect to time.
This may require the chain rule, as we may be taking the derivative of a function first
with respect to some variable, so we must then multiply by the derivative of that variable
with respect to t, since we are ultimately concerned with terms involving dt.
Once we have done that, we just plug in the information we know, and solve for what we
are looking for, which will always be d something over dt.
Let’s check comprehension.