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Hi, everyone. welcome back to integralcalc. We’re going to be talking about related
rates today and what kind of problems you can use related rates to figure out. We’re
going to be doing two examples. The first one here is the volume of a sphere. The second
one is going to be a quantity and price function. Related rates basically allow you to determine
how two variables change together. So for example with the volume of a sphere, you can
use related rates to determine how quickly the radius is increasing as the volume of
the sphere increases or you can use related rates to determine how quickly the volume
of the sphere is decreasing at a volume of 200cu.cm when the radius is decreasing at
a rate of 10cm/s or something like that. With quantity and price, you can use related rates
to determine how quickly the quantity is changing over time when the price is increasing at
a rate of 2 dollars/hr; things like that. So let’s go ahead and do a couple of examples.
First of all, before we go forward, there are three steps with related rates. The first
one being, identify which third independent variable both of the variables in your problem
depend upon. Second, differentiate both sides of the equation with respect to that third
independent variable and third, plug in the information that you're given to solve for
the remaining variable. Let’s go ahead and do a couple of examples.
The first example we have is this volume of the sphere here. Like I said, we need to determine
that third independent variable upon which both volume and radius depend. So our problem
here is: Suppose that air is being pumped into a spherical balloon so that it's volume
is increasing at a rate of 400 cu.cm/s. How fast is the radius of the balloon increasing
when it is 100cm? So you can tell by reading that problem, they're asking you about the
rate of change of the volume and the radius of the sphere over time. So time is our third
independent variable upon which both of these depend. With related rates, that third independent
variable is very often time. So look out for that.
Now that we've determined that the third independent variable is time, we need to differentiate
both sides of these equation with respect to time. So we'll call time t and differentiate
both sides. When we have a single variable like this, just v, differentiating with respect
to t comes out just simply dv/dt equals. Since we're differentiating we take the derivative
here, so we multiply the 3, the exponent, out in front and the 3s will end up canceling
so we get 4pi r^2 and again because we're differentiating, we can add this which means
we took the derivative and it's dr/dt. It’s always the variable on that side of the equation
over that third independent variable so we have v/t and r/t.
So we've differentiated both sides of the equation and now what we do is plug in the
information that we were given. You will notice here, we have 3 variables that we can plug
in information for. We have r which is a standard variable. That’s the radius but we also
have dv/dt and dr/dt, which if you remember, the derivative of any function just means
the rate of change or the slope of the function so this dv/dt means the rate at which the
volume is changing over time, whether increasing or decreasing. And this dr/dt mean the rate
at which the radius is changing over time. So the information that we're given in this
problem, we are supposed to suppose that the volume is increasing at a rate of 400 cucm/s.
Because this is the rate at which the volume is changing, we go ahead and plug in 400 because
the volume is increasing at a rate of 400cucm/. And then we have 4pi here which we leave alone.
It asks us to determine how fast the radius is increasing when the radius is 100cm. So
we're given the radius, 100cm. So we go ahead and plug in 100 for r and we're left with
dr/dt which makes sense because we're asked to solve for the rate at which the radius
of the balloon is increasing and this is the rate at which the radius changes over time.
So that's good. We plugged in the other two, we're left with
this third so now we can solve for the rate of change of the radius. So let's go ahead
and simplify. We’ll have 400 equals, 100 squared is 10,000 so 40000 pi times dr/dt.
And now we're going to go ahead and divide both sides by 40000pi and we'll get 400/40000pi
equals dr/dt. So this will be 1/100pi. This is the rate at which the radius is increasing
over time when the volume is increasing at a rate of 400ucm/s and right at that moment
when the radius is 100cm long. That’s how we solve that problem.
Let’s go ahead and try the demand function over here. Again, remember our three steps.
Determine the independent third variable on which quantity and price depend upon. Differentiate
both sides with respect to that variable and then plug in the information we're given in
the problem to solve. So the problem asks us if the item is currently selling at 100
dollars/unit and the quantity supplied is decreasing at a rate of 80 units/week, find
the rate at which the price of the product is changing. So again we have time. They’re
asking how the rates are changing over time so time is that third independent variable.
We'll call it t, again and we go ahead and differentiate both sides with respect to t.
We'll have this single variable over here, the derivative is just dq/dt equals, lets
go ahead and take the derivative of this 4000e to the - 0.01p. We will have 4000 times negative
0.01 e to the negative 0.01p. That’s the derivative of this part here and then of course,
because we took the derivative, we can add to this side over here, dp/dt. So again just
like the last problem, we're left with three variables here. The rate at which the quantity
changes over time, price and the rate at which price changes over time.
So let’s refer back to the information that we were given in the problem. If the item
is currently selling for 100 dollars/unit. So the price is currently 100 dollars/unit
so we can go ahead and plug in 100 for price. And the quantity supplied is decreasing at
a rate of 80 units/week. So again, this is the rate at which quantity changes and the
rate of which price changes. They’ve told us that the quantity is decreasing at a rate
of 80 units/week. so since it's decreasing, it's very important that we put in a negative
sign in front of quantity over here. So we have negative 80 because the quantity is decreasing
at 80 units and then we can go ahead and write in the rest of the information here. Remember
we said price was 100 and then they say find the rate at which the price of the product
is changing. So we know we did it right because they're asking for the rate at which the rate
changes over time so that is the variable that we're solving for. Here’s our function.
All we have to do is simplify. Let’s go ahead and do that now.
We’ll have negative 80 = -40e^-1 dp/dt and to further simplify, we'll divide. We have
negative 80/(-40e^-1). We divided both sides by -40e^-1. And that leaves dp/dt on it's
own so if we simplify both of these, because there's a negative exponent on the e here,
we need to move it on the top to make that exponent positive; so we'll end up with 80e/40
which is the same as 2e = dp/dt and because e is a constant, it's like 2.7 something,
you could calculate the decimal. Looks like it's about 5.44 but that means that the rate
at which the price is changing over time when the price is 100 dollars/unit and the rate
at which the quantity is decreasing per week is by 80 units. At that point, price is increasing
by about 5 dollars and 44 cents per unit over time. So that's how we solve for related rates
problems. I hope that helped you guys. I'll see you next time. Bye!