# Another related rates problem

• Dec 7th 2008, 12:24 PM
fattydq
Another related rates problem
Since I was still struggling with the last related rates problem I posted, I thought I'd post a very simple one and I think that will help make my understanding of the topic better.

Each side of a square is increasing at a rate of 2 cm/s. At what rate is the area of the square increasing when the area of the square is 36 cm^2?

Now I know ds/dt is 2 and that I'm trying to find dA/dt. I also already know the solution, and how it came about, as it's in my notes, but I don't understand HOW the solution was derived. It basically ends up being dA/dt=2s*ds/dt but what I don't understand is WHY and HOW to get from knowing your given info, to knowing the equation you should be using to solve this problem is dA/dt=2s*ds/dt.
• Dec 7th 2008, 12:36 PM
galactus
Area of square is $A=x^{2}$

Differentiate:

$\frac{dA}{dt}=2x\cdot\frac{dx}{dt}$

Plug in your knowns to find dA/dt.

We are given dx/dt. What is x when the area is 36?.

• Dec 7th 2008, 12:47 PM
fattydq
Can you do me a favor and treat me as if I were someone who had never done a related rates problem? I'm saying I have no idea why you have to go through the process you did to solve the problem. I know the area of a square is x squared obviously, but I don't know why you would differentiate it and why you'd multiply the derivative (2x) by dx/dt

Isn't the derivative of x squared just 2x? Why is the dx/dt tagged on with it?
• Dec 7th 2008, 01:06 PM
galactus
Yes, the derivative of x^2 is 2x. The dx/dt is there because of the change in x with respect to time. That is the change in x. Afterall, that is what is changing.

Differentials are normally covered right before related rates. That is why.

Think of it as differentiating x implicitly with respect to time.

You will probably run into the famous 'ladder sliding down the wall' problem.

Then, we have Pythagoras. Since the ladder is sliding, the top is going down and the bottom is going out. They are both changing.

So, we get $D^{2}=x^{2}+y^{2}$

If we differentiate: $2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$

See?. But in this case, dD/dt = 0 because the length of the ladder is constant and the derivative of a constant is 0. The ladder does not change length as it slides down the wall.
• Dec 7th 2008, 02:06 PM
fattydq
So when doing related rates problem, you always have to use differentials rather than just finding the derivative? In other words the A=x^2 would always become dA/dt=2x*dx/dt in a related rates problem rather than just dA/dt=2x? Sorry, maybe I'm just not thinking along the right lines here.
• Dec 7th 2008, 02:16 PM
galactus
Yes, because we're dealing with rates of change. That is why it's called 'related rates'.

We can use the Chain Rule to find dy/dx implicitly. So, we use it to find the rates of change of two or more related variables that are changing with respect to time.

For instance, when water is draining from, say, a conical tank. the volume, radius and height of the water are all functions of time.

Here are some guidelines:

1. Identify all the given quantities and ones to be found.

2. Write an equation involving the variables whose rates of change either are given or have to be found.

3. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time.

4. Next, sub into the equation all known values for the variables and their rates of change. Then, solve for the rate of change you need.

The dx/dt, dy/dt, d something/dt are the rates of change with respect to time. Thus the dt in the bottom.
• Dec 7th 2008, 02:29 PM
fattydq
Thanks for the replies, I feel like this has really helped, I'm doing some other problems right now and I seem to get the idea.
• Dec 7th 2008, 02:33 PM
galactus
Post one if you like and show your progress with it. I will try to guide you along. I like related rates problems. They were always one of my favorite part of calc for some reason(Nerd).
• Dec 7th 2008, 04:18 PM
fattydq
haha alright, one I'm working on now is At noon, ship A is 70 km west of ship B. Ship A is sailing south at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 6:00 PM?

I know that C (C being the unknown leg of the triangle, ie the distance between the ships at 6 PM) is 250 by the pythagorean theorem, and I thought I would use

C*dC/dt=A*dA/dt+B*dB/dt to solve the problem, but apparently my answer is incorrect. I have a feeling it has something to do with the fact that the ships are moving the opposite direction, and thus one of the legs is changing by dA/dt+dB/dt, but not sure how I would incorporate that into my work?
• Dec 7th 2008, 04:55 PM
galactus
Here's diagram. It's not that bad.

The combined rates of the boats can be represented as da/dt.

$D^{2}=a^{2}+b^{2}$

$2D\frac{dD}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$

$\frac{da}{dt}=\frac{dx}{dt}+\frac{dy}{dt}$

$\frac{db}{dt}=0$ because b is a constant 70.

Use the respective rates of the boats to find how far they have traveled in 6 hours. That is D=250, which you have.

You can also use it to find a. a=15(6)+25(6)=240

Solve for dD/dt