Acrescimos ditos-Metodos de aproximações

Related rates problems analyze the rate at which functions change for certain instances in time. These problems generally involve two or more functions where you relate the functions themselves and their derivatives, hence the name "related rates."

This is a concept that is best explained by example. Here is a classic.

A circle's radius is initially $8$ meters and grows at a constant rate of $2$ meters per second. At time $t = 4$, what is the rate at which the area of the circle is changing?

We let $r = radius$ and $A = area$. Since both variables are functions of time, we know that

$r(0) = 8$

$\frac{dr}{dt} = 2$

We are asked to find $\frac{dA}{dt}$ at time $t = 4$. This involves differentiating the formula for the area of a circle to find the rate at which said area changes. The desired formula is

$A = \pi \cdot r^2$

For clarity, indicate that $A$ and $r$ are functions of time:

$A(t) = \pi \cdot r(t)^2$

Now differentiate the equation:

$\frac{dA}{dt} = 2\pi \cdot r(t) \cdot \frac{dr}{dt}$

We know that $\frac{dr}{dt} = 2$, so

$\frac{dA}{dt} = 4\pi r(t)$

Our only unknown is $r(t)$ at $t = 4$. However, as the initial radius is $8$ meters and the radius grows at a constant rate of $2$ meters per second, after $4$ seconds the radius will be $16$ meters. Thus $r(4) = 16$:

$\frac{dA}{dt} = 4\pi \cdot 16 = 64\pi$

"Related rates" problems, as the heading suggests, deal with functions that can be expressed in terms of one another. We might know that $y = f(x)$, where both $y$ and $x$ are functions of $t$. The function $f$ is often a well-known law of physics or geometry, and may not have been stated in the problem! Thus, your first step in these problems should always be to try to think of any relevant equations relating the variables, then take their derivatives with respect to the independent variable, which is often time.

Let's try this on an example. A 25m ladder is propped against a wall. Before you see the rest of the problem, try to think of a useful equation.

If we let the base of the ladder's distance to the wall be $x$ and the top's height be $y$, then $x^2 + y^2 = 625$.

In a related rates problem, you are typically given some information about the behavior of the variables and possibly their derivatives and asked to find a missing value. Going back to our example, let's add some more details.

Pretty much all related rates problems can be solved in this way! simply find any relevant equations, then take their derivatives with respect to the independent variable.

Related rates are an application of implicit, or partial, differentiation. These are dependent and related rates because the final answer will rely on two or more variables, so the rates are dependent on or related to each other.

The chain rule states that if $z=f(x,y)$ and $x=a(t)$, $y=b(t)$ then $\dfrac{dz}{dt}=f_x(x_0, y_0)a'(t)+f_y(x_0, y_0)b'(t)$. Another way to communicate this is $\dfrac{dz}{dt}=(\dfrac{∂z}{∂x})(\dfrac{dx}{dt})+(\dfrac{∂z}{∂y})(\dfrac{dy}{dt})$.

Using the chain rule and information given in a problem, we are able to determine missing information.

Example Problem: The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 20 inches and the height is 20 inches?

The item being solved for is the rate of change of the volume of the cone when $h=20 \text{in}$ and $r=20 \text{in}$, so the derivative with respect to the volume. The information given is $\dfrac{dr}{dt}=3 \text{in}^3/\text{sec}$ and $\dfrac{dh}{dt}=-3 \text{in}^3/\text{sec}$. These are our $a'(t)$ and $b'(t)$ values (or our $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ values).

Because the volume of a right circular cone is $V=π{r^2}\dfrac{h}{3}$, the partial or implicit derivatives will be taken with respect to the variables h and r and solved for.

$\dfrac{∂V}{∂r}=2πr\dfrac{h}{3}$

$\dfrac{∂V}{∂h}=π{r^2}\dfrac{1}{3}$

In order to find the values to actually plug into our initial equation $\dfrac{dz}{dt}=(\dfrac{∂z}{∂x})(\dfrac{dx}{dt})+(\dfrac{∂z}{∂y})(\dfrac{dy}{dt})$, we need to plug in the r and h values provided to us by the problem into the partial derivatives. These are $r=20\text{in}$ and $h=20\text{in}$.

$\dfrac{∂V}{∂r}=2π(20\text{in})\dfrac{20 in}{3}=\dfrac{800π}{3}\text{in}$

$\dfrac{∂V}{∂h}=π{(20\text{in})^2}\dfrac{1}{3}=\dfrac{400π}{3}\text{in}$

Now it's time to plug these values into our main equation which will provide us with the rate of change of the volume of the right circular cone when $r=20\text{in}$ and $h=20\text{in}$.

$\dfrac{dz}{dt}=(\dfrac{∂z}{∂x})(\dfrac{dx}{dt})+(\dfrac{∂z}{∂y})(\dfrac{dy}{dt})$

$\dfrac{dV}{dt}=\dfrac{800π}{3}\text{in}^2(3\text{in/sec})+\dfrac{400π}{3}\text{in}^2(-3\text{in/sec})\\ =800π\text{in}^3/\text{sec}-400π\text{in}^3/\text{sec}\\ =400π\text{in}^3/\text{sec}$ so the rate of change of the volume of the right circular cone when $r=20\text{in}$ and $h=20\text{in}$ is $400\pi\text{in}^3/\text{sec}$.

(Video) Related Rates: Ships and Distance

by PatrickJMT

Here's a video by patrickJMT showing you how to solve related rate problems. Despite having some nice alliteration, related rates can be rough. They're word problems, they're puzzles, they're a pain. Hopefully though, Patrick the Math Savior can lead us through.

Before we begin he gives us a nice acronym in order to help us remember the general outline in which we solve these problems: DREDS. Clearly, there's some foreshadowing going on. DREDS stands for Diagram, Rates, Equation, Derivative, and Substitute.

1. Diagram: Come up with a diagram to represent the problem.
2. Rates: Label your rates.
3. Equation: Find an equation.
4. Derivative: Take a derivative.
5. Substitute: Substitute in specific information.

The problem that he looks at in this video is

At noon, ship $A$ is $100\ km$ west of ship $B$. Ship $A$ is sailing south at $35\ km/h$ and ship $B$ is sailing north at $25\ km/h$. How fast is the distance between the ships changing at 4:00 P.M.?

So our first step is to draw a diagram to represent this problem. So we know that at noon there's a ship $A$ and a ship $B$ $100\ km$ west of it.

Next we know that $A$ is sailing south, and $B$ is sailing north.

So here comes an important concept when solving these types of problems. Any quantities that vary with time you want to label them with variables. So certainly this distance $A$ is getting bigger as the ship travels further south, so we can label their distance traveled as $x$. Similarly, we can say ship $B$'s distance is $y$.

They also tell us the rate at which these ships are changing. It says ship $A$ is sailing at $35\ km/h$ and ship $B$ is sailing at $25\ km/h$. You can write this as the change in the distance $x$ with respect to time, $\frac{dx}{dt}$, or similarly with the change in distance $y$ instead.

At some time later we will want to know this distance between these two ships. This distance will be the diagonal line between the two. Let's call it $z$.

So what this problem is asking us to find is the change in the distance between the two ships with respect to time. In other words (or letters I guess), $\frac{dz}{dt}$.

We have our diagram now, and we have our rates defined. The next step is to come up with an equation relating $x$, $y$, and $z$. Okay? How do we do this? Here's the trick; we can redraw our diagram as a right triangle.

We now have a right triangle! And you know what that means!? PYTHAGOREAN THEOREM! $$100^2+(x+y)^2=z^2$$

Our next step is to take the derivative with respect to time. This means we need to use implicit differentiation and solve for $\frac{dz}{dt}$. $$\begin{align} 100^2+(x+y)^2&=z^2\\ 0+2(x+y)\cdot\color{red}{[x+y]'}&=2z\cdot\color{red}{[z]'}\\ 2(x+y)\bigg(\frac{dx}{dt}+\frac{dy}{dt}\bigg)&=2z\frac{dz}{dt}\\ \frac{2(x+y)(\frac{dx}{dt}+\frac{dy}{dt})}{2z}&=\frac{dz}{dt}\\ \frac{(x+y)(\frac{dx}{dt}+\frac{dy}{dt})}{z}&=\frac{dz}{dt} \end{align}$$

We've finally made it to the last step, substitution. So we already had $\frac{dx}{dt}$ and $\frac{dy}{dt}$ given to us in the problem. These were $35\ km/h$ and $25\ km/h$ respectively. $$\frac{(x+y)(35+25)}{z}=\frac{dz}{dt}$$ We don't have $x$, $y$, or $z$ yet but we can figure these out with our given information. These ships started moving at noon and we want to find the distance between them at 4 P.M., which is 4 hours later. If ship $A$ is traveling $35\ km/h$ how many kilometers will it travel in 4 hours? Well, $35*4=140\ km$. The same can be done to find the distance ship $B$ traveled: $25*4=100\ km$. So let's put these into our equation. $$\frac{(140+100)(35+25)}{z}=\frac{dz}{dt}$$ What about $z$? We can actually find $z$ from our Pythagorean Theorem equation. $$\begin{align} z^2&=100^2+(x+y)^2\\ z&=\sqrt{100^2+(x+y)^2} \end{align}$$ So plugging this into our equation we get: $$\frac{(140+100)(35+25)}{\sqrt{100^2+(x+y)^2}}=\frac{dz}{dt}$$ And we already figured out what $x$ and $y$ are so we can just plug those values in again and then solve! $$\begin{align} \frac{(140+100)(35+25)}{\sqrt{100^2+(140+100)^2}}&=\frac{dz}{dt}\\ \frac{(240)(60)}{\sqrt{10000+(240)^2}}&=\frac{dz}{dt}\\ \frac{14400}{\sqrt{10000+57600}}&=\frac{dz}{dt}\\ \frac{14400}{\sqrt{67600}}&=\frac{dz}{dt}\\ \frac{14400}{260}&=\frac{dz}{dt}\\ 55.385&\approx\frac{dz}{dt} \end{align}$$

$$\text{At 4 P.M. the distance between the ships is changing }55.385\ km/h\text{.}$$

(Video) Related Rates: Cones

by PatrickJMT

Here's another video by patrickJMT where he shows you how to solve a related rates problem involving cones.

The problem says:

Gravel is being dumped from a conveyor belt at a rate of $20\ ft^3/min$ and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is $10\ ft$ high?

We begin by drawing our diagram. In this problem we have a cone whose base diameter and height are equal.

The problem tells us that a gravel is being added at a rate of $20\ ft^3/min$. This is the change in volume of the cone with respect to time. If you didn't know this was referring to volume, the indicator is the cubed feet. Volume is measured with cubed units.

Finally, it wants us to find the change in height with respect to time. This means we're looking for $\frac{dh}{dt}$.

The equation we use should be pretty obvious for this problem: the volume for a cone. $$V=\frac13\pi r^2h$$ In this problem the diameter is the same length as height so we can rewrite the radius in terms of height in our equation. $$\boxed{\begin{align} 2r&=d\\ 2r&=h\\ r&=\frac h2 \end{align}}$$ $$\begin{align} V&=\frac13\pi\bigg(\color{red}{\frac{h}2}\bigg)^2h\\ V&=\frac13\pi\frac{h^2}4h\\ V&=\frac{\pi}{12}h^3 \end{align}$$

Now we take the derivative with respect to time and solve for our unknown, $\frac{dh}{dt}$. $$\begin{align} \frac{dV}{dt}&=\frac{\pi}{12}3h^2\cdot\color{red}{[h]'}\\ \frac{dV}{dt}&=\frac{\pi}{12}3h^2\frac{dh}{dt}\\ \frac{dV}{dt}&=\frac{\pi}{4}h^2\frac{dh}{dt}\\ \frac{4}{\pi h^2}\frac{dV}{dt}&=\frac{dh}{dt} \end{align}$$

Finally, we need to substitute in values. We know that $\frac{dV}{dt}=20$. We also know that the height is $10\ ft$ or $h=10$. $$\begin{align} \frac{4}{\pi (10)^2}\cdot20&=\frac{dh}{dt}\\ \frac{80}{100\pi}&=\frac{dh}{dt}\\ \frac{8}{10\pi}&=\frac{dh}{dt}\\ 0.255&\approx\frac{dh}{dt} \end{align}$$

$$\text{When the height is }10\text{ feet, it is increasing by }0.255\ ft/min\text{.}$$

(Video) Related Rates: Point on a Graph

by PatrickJMT

Here's another video by patrickJMT on related rates. This one deals with a point moving on a graph.

The problem says:

A point is moving on the graph of $3x^3+4y^3=xy$. When the point is at $P=(\frac17,\frac17)$, its $y$-coordinate is increasing at a speed of $9$ units per second. What is the speed of the $x$-coordinate at that time and in which direction is the $x$-coordinate moving?

So we have an equation: $$3x^3+4y^3=xy$$ We know that at the point $P=(\frac17,\frac17)$ the $y$-coordinate is increasing at a speed of 9 units per second, or: $$\frac{dy}{dt}=9$$ Finally, what we want to know is the speed of the $x$-coordinate, or: $$\frac{dx}{dt}=?$$

So we can start by taking the derivative of the equation with respect to time and solving for our unknown, $\frac{dx}{dt}$. $$\begin{align} 3x^3+4y^3&=xy\\ 9x^2\cdot\color{red}{[x]'}+12y^2\cdot\color{red}{[y]'}&=\color{red}{[x]'}y+x\color{red}{[y]'}\\ 9x^2\frac{dx}{dt}+12y^2\frac{dy}{dt}&=y\frac{dx}{dt}+x\frac{dy}{dt}\\ 9x^2\frac{dx}{dt}-y\frac{dx}{dt}&=x\frac{dy}{dt}-12y^2\frac{dy}{dt}\\ \frac{dx}{dt}(9x^2-y)&=\frac{dy}{dt}(x-12y^2)\\ \frac{dx}{dt}&=\frac{\frac{dy}{dt}(x-12y^2)}{9x^2-y} \end{align}$$

Now all we need to do is substitute in values. We know that $x=y=\frac17$ and we know that $\frac{dy}{dt}=9$. $$\begin{align} \frac{dx}{dt}&=\frac{9(\frac17-12(\frac17)^2)}{9(\frac17)^2-\frac17}\\ \frac{dx}{dt}&=\frac{9(\frac17-\frac{12}{49})}{\frac{9}{49}-\frac17}\\ \frac{dx}{dt}&=\frac{9(\frac7{49}-\frac{12}{49})}{\frac{9}{49}-\frac7{49}}\\ \frac{dx}{dt}&=\frac{9(-\frac{5}{49})}{\frac{2}{49}}\\ \frac{dx}{dt}&=-\frac{\frac{45}{49}}{\frac{2}{49}}\\ \frac{dx}{dt}&=-\frac{45}{2} \end{align}$$

So the $x$-coordinate is changing at a speed of $\frac{45}{2}$ units per second. Notice that we did not include the negative symbol for the speed. This is because speed is always positive. It is the absolute value of velocity. But because of this negative sign we know that it is decreasing at a speed of $\frac{45}{2}$.

(Video) Related Rates: Trigonometry

by PatrickJMT

Here's another video by patrickJMT where he looks at a related rates problem which involves trig.

The problem states:

In the right angle triangle, the angle $\theta$ is increasing at a constant rate of $6$ radians per hour. At what rate is the side of length $x$ increasing when $x=6$ feet?

So we know that theta is increasing at a constant rate of 6 radians per hour. $$\frac{d\theta}{dt}=6$$ What we want to find is the rate at which the side $x$ is increasing when it's at 6 feet. $$\frac{dx}{dt}=?\ \text{ when }\ x=6$$

So how can we relate $x$ and $\theta$? Well, we can use sine to do that. Remember that sine of an angle is the opposite over the hypotenuse. $$\begin{align} \sin\theta&=\frac{x}{10}\\ 10\sin\theta&=x \end{align}$$ So now all we have to do is take the derivative with respect to time. $$\begin{align} 10\sin\theta&=x\\ 10\cos\theta\cdot\color{red}{[\theta]'}&=\frac{dx}{dt}\\ 10\cos\theta\frac{d\theta}{dt}&=\frac{dx}{dt} \end{align}$$

We want to find $\cos\theta$ now. Remember that cosine is the adjacent over the hypotenuse. We can find the adjacent by using the Pythagorean Theorem, since we know the side $x$ is $6$. $$\begin{align} a^2+6^2&=10^2\\ a^2+36&=100\\ a^2&=64\\ a&=8 \end{align}$$

So the cosine of theta is $$\cos\theta=\frac{8}{10}=\frac45$$

Plugging this all in we get: $$\begin{align} 10\cdot\frac45\cdot6&=\frac{dx}{dt}\\ 48&=\frac{dx}{dt} \end{align}$$

$$\text{When the side }x\text{ is at }6\text{ feet it is increasing at a rate of }48\ ft/h\text{.}$$

(Video) Related Rates: Baseball!

by PatrickJMT

Here's another video by patrickJMT dealing with related rates.

The problem states:

A baseball diamond is a square with side $90$ feet. If a batter hits the ball and runs towards first base with a speed of $20$ feet per second, at what speed is his distance from second base decreasing when he is halfway to first base?

So we have a baseball diamond and a batter who's halfway to first base.

Let's denote his distance to second base as $z$ and his distance from home plate as $x$.

We are given that the batter runs towards first base with a speed of $20$ feet per second. What we want to find is the speed at which his distance from second base is decreasing.

Since we know the length of a side to be $90$ feet, and the runner is halfway to first base, we can conclude that $x=45$ feet.

To come up with an equation to use, we can actually pull out the right triangle that's being made.

Since we denoted the distance between the batter and home to be $x$, the distance between the batter and first is $90-x$.

Now, by using the Pythagorean Theorem, we can get our equation. $$(90-x)^2+90^2=z^2$$ Which we can then take the derivative of with respect to time. $$\begin{align} 2(90-x)\cdot\color{red}{[x]'}+0&=2z\cdot\color{red}{[z]'}\\ 2(90-x)\cdot-\frac{dx}{dt}&=2z\frac{dz}{dt}\\ -\frac{2(90-x)}{2z}\frac{dx}{dt}&=\frac{dz}{dt}\\ -\frac{90-x}{z}\frac{dx}{dt}&=\frac{dz}{dt} \end{align}$$

We can begin substituting in our known values. Remember that $z$ is found with the Pythagorean Theorem. $$\begin{align} -\frac{90-x}{z}\frac{dx}{dt}&=\frac{dz}{dt}\\ -\frac{90-x}{\sqrt{(90-x)^2+90^2}}\frac{dx}{dt}&=\frac{dz}{dt}\\ -\frac{90-45}{\sqrt{(90-45)^2+90^2}}\cdot20&=\frac{dz}{dt}\\ -\frac{45}{\sqrt{2025+8100}}\cdot20&=\frac{dz}{dt}\\ -\frac{900}{\sqrt{10125}}&=\frac{dz}{dt}\\ -4\sqrt{5}&\approx\frac{dz}{dt} \end{align}$$

$$\text{When the batter is halfway to first base,}\\ \text{his speed is decreasing by }4\sqrt{5}\ ft/sec\text{.}$$

(Video) Related Rates: Circumference

by PatrickJMT

Here's another video by patrickJMT going over a related rates word problem. This one deals with the change in a circles circumference.

Th problem states:

A pebble is dropped into a pool of water, generating circular ripples. The radius of the largest ripple is increasing at a constant rate of $6$ inches per second. What is the increase in the circumference of the ripple after $3$ seconds have passed?

Let's start by making a diagram.

We have our pebble dropped in the pond and it creates ripples which increase as the time increases. Let's suppose that, from where the pebble was dropped out to the largest ripple, it's a radius of $r$.

We also know the radius of the largest ripple is increasing at a constant rate of 6 inches per second. Which means the radius is changing with respect to time.

What we want to find is the change in the circumference of the ripple; the change in circumference with respect to time.

Now that we have a diagram and the rates, we can go ahead and create an equation. Since we're looking for the change in circumference, the equation for circumference should be a safe bet. $$C=2\pi r$$

Next, let's take the derivative with respect to time. $$\frac{dC}{dt}=2\pi\frac{dr}{dt}$$ Notice that we have no need for the $t=3$ information described in the problem (we know this because there's no $t$ in our derivative equation). So the change in circumference with respect to time is independent of time. We can now finish this problem by substituting in the specific information. $$\begin{align} \frac{dC}{dt}&=2\pi\color{red}{(6)}\\ \frac{dC}{dt}&=12\pi \end{align}$$

$$\text{The circumference of the circle is increasing at a rate of }12\pi\ in/sec\text{.}$$

(Video) Related Rates: Sliding Ladder

by PatrickJMT

Here's another video by patrickJMT showing you how to solve a related rates problem involving a sliding ladder.

The problem states:

A $20$ foot ladder is being pulled away from a wall at a rate of $5$ feet per second. What is the rate of change in the area beneath the ladder when the ladder is $8$ feet from the wall?

Let's draw our diagram to represent this.

The ladder is being pulled away from the wall as time progresses. That means its height will change with respect to time so we should label it with a variable, $h$. Similarly, the base will be pulling out from the wall, meaning it will change with respect to time. So we need to label the base with a variable, $b$. Finally, the length of the ladder is not changing as time goes on. It is constant. So we can label it with a constant, $20$.

Furthermore, we know the ladder is being pulled away at a rate of $5$ feet per second. This translate to a change in base with respect to time, $\frac{db}{dt}$. We also want to find the rate of change of the area below the ladder, $\frac{dA}{dt}$.

Since we're looking for the change in area, we should probably start with the equation for the are of a triangle. $$A=\frac12bh$$ But before we take the derivative take a look at what would happen with our variables. The $A$ and $b$ would give us rates that we either want to find or have information for. But the $h$ would give is $\frac{dh}{dt}$. We don't have any information about this in the problem. So, we need to come up with a way to re-express $h$. Do we have any other formulas related to triangles that could help us? Hell yeah we do! Pythagorean Theorem. $$\boxed{\begin{align} b^2+h^2&=20^2\\ h^2&=400-b^2\\ h&=\sqrt{400-b^2} \end{align}}$$

Now, let's plug this into our area equation. $$A=\frac12b\sqrt{400-b^2}$$ Since it's all in terms of base now, and we have a rate for base, we can take the derivative. $$\begin{align} \frac{dA}{dt}&=\frac12(\color{red}{[b]'}\sqrt{400-b^2}+b\color{red}{[\sqrt{400-b^2}]'})\\ &=\frac12\bigg(\frac{db}{dt}\sqrt{400-b^2}+b\cdot\frac{1}{2\sqrt{400-b^2}}\cdot\color{red}{[400-b^2]'}\bigg)\\ &=\frac12\bigg(\sqrt{400-b^2}\frac{db}{dt}+\frac{b}{2\sqrt{400-b^2}}\cdot-2b\cdot\color{red}{[b]'}\bigg)\\ &=\frac12\bigg(\sqrt{400-b^2}\frac{db}{dt}+\frac{-2b^2}{2\sqrt{400-b^2}}\frac{db}{dt}\bigg)\\ \frac{dA}{dt}&=\frac12\frac{db}{dt}\bigg(\sqrt{400-b^2}-\frac{b^2}{\sqrt{400-b^2}}\bigg)\\ \end{align}$$

So now we just need to plug in the rate of change of the base, $\frac{db}{dt}=5$. We also know that the rate of change of the area is happening when the ladder is $8$ feet away from the wall, $b=8$. $$\begin{align} \frac{dA}{dt}&=\frac12\frac{db}{dt}\bigg(\sqrt{400-b^2}-\frac{b^2}{\sqrt{400-b^2}}\bigg)\\ \frac{dA}{dt}&=\frac12(5)\bigg(\sqrt{400-(8)^2}-\frac{(8)^2}{\sqrt{400-(8)^2}}\bigg)\\ \frac{dA}{dt}&=\frac52\bigg(\sqrt{400-64}-\frac{64}{\sqrt{400-64}}\bigg)\\ \frac{dA}{dt}&=\frac52\bigg(\sqrt{336}-\frac{64}{\sqrt{336}}\bigg)\\ \frac{dA}{dt}&\approx37.097 \end{align}$$

$$\text{The rate of change of the area beneath the ladder}\\ \text{when the base is }8\ ft\text{ is }37.097\ ft^2/sec\text{.}$$

(Video) Related Rates: Traveling Cars

by PatrickJMT

Here's the final video by patrickJMT on related rates. This one deals with cars traveling from an intersection.

The problem states:

Two cars leave an intersection at the same time, one headed east and the other north. The eastbound car is moving at $30\ mph$ while the northbound car is moving at $60\ mph$. Twenty minutes later, what is the rate of change in the perimeter of the right triangle created using the two cars and the intersection?

As always, we draw our diagram.

Since the distance from each car to the intersection is changing with time, we need to label these distances with variables, say $x$ and $y$.

Since there's a right triangle being made by these two cars, we can make the distance between them be a line. This is also changing with respect to time, so we can label it with a variable $z$.

Now, we can start labeling our rates. It says the eastbound car is traveling at $30\ mph$. This means $\frac{dx}{dt}=30$. Similarly, $\frac{dy}{dt}=60$ because the northbound car is traveling at $60\ mph$. Finally, what we're looking for is the rate of change in the perimeter, or $\frac{dP}{dt}$, after $20\ min$ has passed. Since this time is in minutes and our rates are in hours, we need to rewrite the time in terms of hours, $t=\frac{20}{60}=\frac13$.

So now let's use the equation for the perimeter. $$P=x+y+z$$ Notice, though, that if we took the derivative immediately we would get $\frac{dz}{dt}$ which we don't have any information for. So, like in the previous example, we need to rewrite $z$ in terms of $x$ and $y$. And, like before, we do it using the Pythagorean Theorem. $$\boxed{\begin{align} x^2+y^2&=z^2\\ \sqrt{x^2+y^2}&=z \end{align}}$$

Let's fix our equation now. $$P=x+y+\sqrt{x^2+y^2}$$

Now that everything is in terms of $x$ and $y$, and we have rates for those, we can take the derivative. $$\begin{align} P&=x+y+\sqrt{x^2+y^2}\\ \frac{dP}{dt}&=\frac{dx}{dt}+\frac{dy}{dt}+\frac{1}{2\sqrt{x^2+y^2}}\cdot\color{red}{[x^2+y^2]'}\\ \frac{dP}{dt}&=\frac{dx}{dt}+\frac{dy}{dt}+\frac{1}{2\sqrt{x^2+y^2}}\cdot(2x\cdot\color{red}{[x]'}+2y\cdot\color{red}{[y]'})\\ \frac{dP}{dt}&=\frac{dx}{dt}+\frac{dy}{dt}+\frac{1}{2\sqrt{x^2+y^2}}\cdot\bigg(2x\frac{dx}{dt}+2y\frac{dy}{dt}\bigg)\\ \frac{dP}{dt}&=\frac{dx}{dt}+\frac{dy}{dt}+\frac{1}{\sqrt{x^2+y^2}}\cdot\bigg(x\frac{dx}{dt}+y\frac{dy}{dt}\bigg) \end{align}$$

Finally, we can plug in all our information, but notice that we don't have values for $x$ and $y$. Here is where we use the fact that $t=\frac13\ hr$. The car that was going east was traveling at $30\ mph$. In $\frac13$ of an hour it traveled $x=\frac{30}{3}=10$ miles.

Similarly, if the other car is traveling north at $60\ mph$, after $\frac13$ of an hour it will have traveled $y=\frac{60}{3}=20$ miles.

Let's plug this all in. $$\begin{align} \frac{dP}{dt}&=\frac{dx}{dt}+\frac{dy}{dt}+\frac{1}{\sqrt{x^2+y^2}}\cdot\bigg(x\frac{dx}{dt}+y\frac{dy}{dt}\bigg)\\ \frac{dP}{dt}&=30+60+\frac{1}{\sqrt{10^2+20^2}}\cdot[10(30)+20(60)]\\ \frac{dP}{dt}&=90+\frac{300+1200}{\sqrt{100+400}}\\ \frac{dP}{dt}&=90+\frac{1500}{\sqrt{500}}\\ \frac{dP}{dt}&\approx157.082 \end{align}$$

$$\text{The rate of change of the perimeter of the triangle is }157.08\ mi/hr\text{.}$$