Related Rates of change (pebble/ripple problem)

[Femme_physics]

You forgot to take care of the round thingies.

I'm not sure what you mean by "round thingies"!

Your mistake is that you said $(6t)^{2}=36t^{2}$ and you also want:
$$\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}$$

When did I ever say that?

I don't see any inclusion of the rate of change of radius in your work.

I did write

R(t) = 6t

 

[I like Serena]

I'm not sure what you mean by "round thingies"!

There!
Round thingies!

You just removed them as if they weren't there.

 

[Femme_physics]

Then I'll do what you told me:

The chain rule goes something like "the derivative of the surrounding function, multiplied by the derivative of the enclosed function".

So 6t is the enclosed function, and the rest of it is the surrounding one. So should be...http://img703.imageshack.us/img703/6706/latestattempt05.jpg

 

[I like Serena]

I was aiming for $A(R)=\pi (6t)^2=\pi \cdot 6^2 t^2 = \pi \cdot 36 t^2$
This is what I meant when I said you should take care of the round thingies.
And like this you won't need the chain rule.

However, it's good practice to learn and apply the chain rule, so this would make a good example on how to do it.
But before I help you do the chain rule, could you do it without the chain rule (with the formula I've just shown)?
That way you will have the answer, and you will know what to expect when using the chain rule.

 

[Femme_physics]

Ooooooooooh! You first off open parenthesis, THEN solve it. I see how your logic makes sense

However, it's good practice to learn and apply the chain rule, so this would make a good example on how to do it.
But before I help you do the chain rule, could you do it without the chain rule (with the formula I've just shown)?
That way you will have the answer, and you will know what to expect when using the chain rule.

Yea, that's easy,

2pi36t

t -> 2

= 144pi

 

[I like Serena]

Right! So back to the chain rule.Let me rewrite what you wrote.

$$\frac d {dt} A(R) = \frac d {dR} A(R) \cdot \frac d {dt} R$$

$$\frac d {dt} A(R) = 2 \pi (6t) \cdot \frac d {dt} (6t)$$

$$\frac d {dt} A(R) = 2 \pi (6t) \cdot 6$$I'm sorry that I changed a lot of what you wrote.
I hope you can make some sense of it. :shy:
Another way to write it down is with what I wrote before:

(u(v))' = u'(v) . v'​

Now we have u=A and v=R.
So:

(A(R))' = A'(R) . R'​

$$(\pi (6t)^2)' = 2 \pi (6t) \cdot (6)$$

 

[1MileCrash]

I did write

R(t) = 6t

That's no inclusion of rate of change of radius, that's just what you do to find the radius to solve for when t = 2. 6t is just a replacement for r in your work. Radius rate of change has it's own separate role, which is clear when you derive with respect to time.

If I didn't ask for the area rate for a certain time, but instead asked you to find area rate of change when radius = 12 and told you the radius was changing at 2000 lightyears per second, you'd get the same answer you've been getting because you have no inclusion of radius rate of change.

 

[Femme_physics]

I realized throughout the topic that what I'm really missing is more fundamental knowledge with respect to implicit differentiation, so right now am going over it

Thanks a bunch ILS, 1Mile^^