Last time we learned how to use a bunch of formulas that make calculating derivatives
easier than they would be if we always had to use the definition of the derivative. Today,
we’re going to be talking about one of the most commonly studied applications of derivatives,
the dreaded related rates problem. You might be wondering why we’re focusing
in on such a specific application of derivatives when there are so many to explore. Well, I
can't cover all of them in this overview series, but Related Rates felt like it deserved a
special place in the series because it’s so common in calculus courses, and because
it gives so many students so much trouble. Today, we’re going to conquer it.
IMPLICIT DIFFERENTIATION Before we can jump fully into Related Rates,
we need to understand the concept of implicit differentiation. We already know from previous
sections that we have no problem calculating a derivative if the left-hand side of our
function is y or f(x), and the right-hand side is something in terms of x. But what
do we do if we don’t have that kind of clean separation, where we can’t get all the y’s
on the left, separate from the x’s on the right? That’s where implicit differentiation
comes in. With implicit differentiation problems where
our variables can’t be separated, we’ll be treating the x variable in the same way
we always have. We’ll take it’s derivative according to the derivative rules we learned
last time. What’s different is how we treat y. Before, our functions have always been
y equals, and then something in terms of x, and when we took the derivative we'd change
that y into y prime. With implicit differentiation, we'll treat y like x, such that the derivative
of y would be 1, in the same way the derivative of x would be 1. In other words, you'll treat
y exactly the same way you treat x. The only difference is that every time you take the
derivative of y, you'll multiply by dy/dx. But that’s it, that’s the only difference.
We’ll see in a second how this becomes super important when we start working with Related
Rates problems, but let's take a look at the application of implicit differentiation on
this equation. As I mentioned, we'll take the derivative of x in the same way we always
have. So, when we run across 2x here, we know by the Power Rule that the derivative is 2.
In the second term, we come to a y. If this were an x, the derivative of 3e^x would be
3e^x. Since it's a y, the derivative is taken the same way, and it's 3e^y, but because it's
a y, we have to multiply by dy/dx. Following the same pattern on the right-hand side, we
come to e^x+y. Remember that the derivative of e^x is e^x, so the derivative of e^x+y
is e^x+y. However, we need to use Chain Rule and multiply by the derivative of x+y. We
know the derivative of x is 1. The derivative of y is also 1, but when we take its derivative,
we have to multiply by dy/dx. Now that the hard part is done, all that's left to do is
simplify and solve for dy/dx. RELATED RATES
Generally, your steps with Related Rates problems will be:
1. Identify the information you’ve been given, and exactly what you need to find.
2. Set up your equation and plug in any variables that you can. Remember that you may need more
than one. 3. Use implicit differentiation to find the
derivative of your equation and plug in any remaining pieces of information, such that
your only remaining variable is the one you needed to find.
4. Solve for that variable. Let’s take a look at a common example to
start getting a grasp on how this works. SPHERICAL BALLOON
Here’s the problem: Air is being pumped into a spherical balloon at 10 cm^3/min. Calculate
the rate at which the radius of the balloon is increasing when the diameter is 15 cm.
The first thing we know is that the balloon is spherical, which means we'll need to work
with the equation for the volume of a sphere. We know how fast air is being added to the
balloon, or put another way, we know how fast the volume of the balloon is increasing. This
is the rate of change of the volume, so which is not V, it's the derivative of V, or dV/dt.
We also know the diameter at the point in time in question. Because our equation is
in terms of radius, not diameter, we'll convert the diameter to a value for radius.
We need to find the rate at which the radius is increasing. Rate of change of a variable
is that variable's derivative, so we're actually looking for dr/dt.
Now that we know what we have and what we need, we'll differentiate both sides of our
volume equation. Remember, treat every variable like you've always treated x when dealing
with regular derivatives. This time though, every time you take the derivative of a term
that involves a variable, you need to multiply by the derivative of that variable. So, when
you take the derivative of V, just like taking the derivative of x, you get 1, but you have
to multiply by dV/dt. Similarly, when you take the derivative of r cubed, you'll get
3 r squared, but you have to multiply by dr/dt. Once we've used implicit differentiation to
calculate the derivative, we can plug in the values we wrote down before, leaving ourselves
only with the value we needed to solve for, dr/dt. Getting dr/dt alone on the left-hand
side gives us our final answer for the rate of change of the radius of the balloon when
the diameter is 15 cm as air is being pumped in at 10 cubic cms per minute.
LEANING LADDER The leaning ladder example is another very
common related rates problem. Despite the fact that we'll be working with the Pythagorean
Theorem as our main equation, instead of the volume of a sphere like in the last example,
the steps are the same. We'll start by writing down what we have and what we need. We'll
set up our equation, find the derivative with implicit differentiation, then plug in and
solve for the missing variable. Here's the problem: A 41 foot ladder is leaning
against a vertical wall. The top of the ladder is sliding down the wall while its bottom
slides along the ground away from the wall at 4 ft/second. Calculate how fast the top
of the ladder is moving when it is 9 ft above the ground.
The first thing we should realize is that we know two sides of the triangle formed by
the ladder, the wall and the ground. The ladder is 41 feet long, and the top of it is 9 feet
off the ground at the point in time we're interested in. We can use the Pythagorean
Theorem to find that the length of the third side along the ground is 40.
Our equation will be the Pythagorean Theorem, and we'll plug in for the length of the ladder
since it never changes. Then we'll differentiate both sides of our equation using implicit
differentiation. Remember that we'll have to multiply by da/dt when we take the derivative
of a, and we'll multiply by db/dt when we take the derivative of b. We know that a,
the distance between the wall and the base of the ladder, is 40. We know that da/dt,
the rate at which a is changing, is 4 ft/sec. Finally, we know that b, the distance between
the ground and the top of the ladder, is 9, so we'll plug all those in, leaving only db/dt.
Then we'll solve for db/dt, which is the rate at which the ladder is sliding down the wall
when the top of it is 9 feet above the ground. Even though we've only covered two examples
here, the method behind every related rates problem is the same. If you follow the steps
we mentioned earlier, these problems won't give you so much trouble.
Next time we’ll dive into another application of derivatives, optimization, before we turn
toward the second half of calculus, integrals. I’ll see you then.