Calculus

Applications

Now that you've learned how to find the derivatives of basic (and not-so-basic) functions, let's try putting this knowledge to use in the real world.

In the problems below, you'll be asked to compare the rate of change in two related values. For example, if your mom buys three candies everyday and gives two to you and one to your sister, what would the change in your candies be relative to the change in your sister's candies?

We can set up a derivative for that!

Let y=your candies\text{Let }y=\text{your candies}
Let x=your sister’s candies\text{Let}\ x=\text{your sister's candies}

And now, we can set up an equation to represent yy in terms of x:x:

y=2x.y=2x.

Since everytime your sister gets a candy, you get two, you get twice the candies. Great! Let's find the derivative.

d 2xdx=21x0=2.\frac{d\ 2x}{dx}=2\cdot1x^0=2.

That was so simple- it's just two!Let's try another example.

Suppose we have a bubble into which air constantly flows in at a rate of 12π μL12\pi\ \mu L per second (s).\left(s\right). Let's try to find the rate of change in the radius of the bubble (which we know will always be spherical.) A sphere has a few properties, including volume, surface area, and radius. Which of them is being directly manipulated when air flows into the bubble?

That would be volume, because microliters (μL)\left(\mu L\right) is a unit of volume. Now that we know what properties we are watching, let's write some equations.

v=43πr3v=\frac{4}{3}\pi r^3
dvds=12π\frac{dv}{ds}=12\pi

Notice that we went straight to the derivative for the second equation. This is because we haven't been given enough information to write an equation - we don't know how many microliters the bubble started with before air began flowing in. However, we do know the slope of the equation, and derivative is just another word for slope.

Now, let's find the derivative of the other function:

d43πr3dr=43π  dr3dr=43π3r2=4πr2.\frac{d\frac{4}{3}\pi r^3}{dr}=\frac{4}{3}\pi\ \cdot\ \frac{dr^3}{dr}=\frac{4}{3}\pi\cdot3r^2=4\pi r^2.

Food for thought: isn't that the surface area formula? Make a theory. Maybe prove it.

Okay, so now we know the following values:

dvdr,dvds\frac{dv}{dr},\frac{dv}{ds}

And we want to know what is

drds.\frac{dr}{ds}.

Hmm... I'm getting the chain rule vibe. Are you?

We can setup an equation with all three of these values, thanks to the chain rule!

dvds=dvdrdrds\frac{dv}{ds}=\frac{dv}{dr}\cdot\frac{dr}{ds}

Substitute what we already know:

12π=4πr2drds12\pi=4\pi r^2\cdot\frac{dr}{ds}
3r2=drds.\frac{3}{r^2}=\frac{dr}{ds}.

And we are done! Now, try the following problem on your own.

The width of a rectangle increases at 4cms4 \frac{cm}{s} and the height at 3 cms.3\ \frac{cm}{s}. What is the rate of change on the area with respect to ss when h=5h=5 and w=6w=6?

Hint 1

The information we have is:

dhds=3\frac{dh}{ds} = 3
dwds=4\frac{dw}{ds} = 4

We must find

dwhds=?\frac{dwh}{ds} = ?
Hint 2

What are some of the methods of solving a derivative?

Solution 1

We can use the product rule to evaluate this derivative:

dwhds=h(dwds)+w(dhds).\frac{d wh}{ds} = h(\frac{dw}{ds})+ w(\frac{dh}{ds}).

Substitute:

dwhds=h(4)+w(3).\frac{d wh}{ds} = h(4)+ w(3).

And, when we plug in the given values:

54+63=38.5\cdot 4 + 6\cdot 3 = \boxed{38}.

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