related rates help!

[sammccord12]

1. A pebble is dropped into a calm pool of water, causing ripples in the form of concentric circles. If each ripple moves out from the center at 2 ft/sec, at what rate is the total area of disturbed increasing after 6 seconds?

So, for this one I have no idea how to begin. I feel like I would need the radius to do this problem, but it isn't given so I'm confused.


2. A sperical balloon is being inflated so that its volume is increasing at the rate of 2pi ft^3 / sec. At what rate does the surface area change when the radius is 6 inches?

I did this one but I think it's wrong so I need some help.
My work:
dv/dt = 3656p in^3 /sec
r = 6 in
dr/dt =?
dv/dt = 4pi*r^2 * dr/dt
dr/dt = (dv/dt)/4pi*r^2
dr/dt = 24

Thanks in advance for the help!

 

[Steven G]

1. A pebble is dropped into a calm pool of water, causing ripples in the form of concentric circles. If each ripple moves out from the center at 2 ft/sec, at what rate is the total area of disturbed increasing after 6 seconds?

So, for this one I have no idea how to begin. I feel like I would need the radius to do this problem, but it isn't given so I'm confused.

A pebble is dropped into a calm pool of water, causing ripples in the form of concentric circles--Think circles
If each ripple moves out from the center at 2 ft/sec---dr/dt = 2ft/sec
at what rate is the total area of disturbed increasing after 6 seconds--find dA/dt when t=6seconds

Ok, I agree with you that you need the radius. You really need to see a ripple with your own eyes at least once. Since the water is calm there is no circle and the radius of this circle is 0. Then the pebble hits the water and then there is a very very small circle, then a bigger circle and then an even bigger circle.... all with the same center. AND you are told that dr/dt is 2ft/sec. So what will the radius be after 6 seconds, ie what is the radius when t=6???!!

 

[Steven G]

2. A sperical balloon is being inflated so that its volume is increasing at the rate of 2pi ft^3 / sec. At what rate does the surface area change when the radius is 6 inches?

I did this one but I think it's wrong so I need some help.
My work:
dv/dt = 3656p in^3 /sec
r = 6 in
dr/dt =?
dv/dt = 4pi*r^2 * dr/dt
dr/dt = (dv/dt)/4pi*r^2
dr/dt = 24

Thanks in advance for the help!

You say you think it is wrong yet I do not see an answer. An answer would be in the form dA/dt=...

dv/dt =3456p in³ /sec, not 3656p in³ /sec
Let A = surface area. So A = 4pir² and therefore dA/dt =8pirdr/dt. Now you want dA/dt when r=6in. So you need to find dr/dt

So you start to think what information did you not use (this is textbook math after all!). You did not use dv/dt.

V = (4/3)pir³ so dv/dt = 4pir²*dr/dt=3456p in³ /sec. Can you continue from here??

 

[MarkFL]

1. A pebble is dropped into a calm pool of water, causing ripples in the form of concentric circles. If each ripple moves out from the center at 2 ft/sec, at what rate is the total area of disturbed increasing after 6 seconds?

So, for this one I have no idea how to begin. I feel like I would need the radius to do this problem, but it isn't given so I'm confused.

Another way to approach this (so you don't need to compute \(r\)):

The area of a circle is:

[MATH]A=\pi r^2[/MATH]
We are told the radius \(r\) is increasing at time \(t\) at a given constant rate, where \(r(0)=0\), hence:

[MATH]r(t)=2t[/MATH]
Thus:

[MATH]A(t)=\pi(2t)^2=4\pi t^2[/MATH]
So, to answer the question:

[MATH]\left.\d{A}{t}\right|_{t=6}=\,?[/MATH]

 

[Steven G]

Another way to approach this (so you don't need to compute \(r\)):

The area of a circle is:

[MATH]A=\pi r^2[/MATH]
We are told the radius \(r\) is increasing at time \(t\) at a given constant rate, where \(r(0)=0\), hence:

[MATH]r(t)=2t[/MATH]
Thus:

[MATH]A(t)=\pi(2t)^2=4\pi t^2[/MATH]
So, to answer the question:

[MATH]\left.\d{A}{t}\right|_{t=6}=\,?[/MATH]

I never thought of doing it that way before. Thanks!!!!