# c3- differentiationWatch

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Thread starter 5 years ago
#1
Need some help with the following question:

The radius of a circular disc is increasing at a constant rate.of 0.003cm s^-1. Find the rate at which the areais increasing when the radius is 20cm.

How do I go about doing this? I think you have to use the chain rule but not sure what goes where?
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5 years ago
#2
(Original post by Super199)
Need some help with the following question:

The radius of a circular disc is increasing at a constant rate.of 0.003cm s^-1. Find the rate at which the areais increasing when the radius is 20cm.

How do I go about doing this? I think you have to use the chain rule but not sure what goes where?
. We want to find and you're given . Does that help?
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Thread starter 5 years ago
#3
(Original post by Chlorophile)
. We want to find and you're given . Does that help?
Is it A= pi*r^2
dA/dr= 2pi*r. Then just sub in 20 for r?
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5 years ago
#4
(Original post by Super199)
Is it A= pi*r^2
dA/dr= 2pi*r. Then just sub in 20 for r?
Yes, once you've multiplied the dr/dt by dA/dr.
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Thread starter 5 years ago
#5
(Original post by Chlorophile)
Yes, once you've multiplied the dr/dt by dA/dr.
Right got it cheers
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Thread starter 5 years ago
#6
I have another question if someone can help me with

A viscous liquid is poured on to a flat surface. It forms a circular patch whose area grows at a steady rate of 5cm^2 s^-1. Find, in terms of pi,

a) The radius of the patch 20 seconds after pouring has commenced

b) The rate of increase of the radius at this instance

I think I have dA/dt= 5
Area = pi*r^2

So dA/dr = 2pi*r

I think it's dA/dt= dA/dr * dr/dt

But then I got stuck and don't know how to do part a? Should be able to do part b with the information from part a.

Thanks
0
5 years ago
#7
(Original post by Super199)
I have another question if someone can help me with

A viscous liquid is poured on to a flat surface. It forms a circular patch whose area grows at a steady rate of 5cm^2 s^-1. Find, in terms of pi,

a) The radius of the patch 20 seconds after pouring has commenced

b) The rate of increase of the radius at this instance

I think I have dA/dt= 5
Area = pi*r^2

So dA/dr = 2pi*r

I think it's dA/dt= dA/dr * dr/dt

But then I got stuck and don't know how to do part a? Should be able to do part b with the information from part a.

Thanks
rate of 5 cm2 every minute means every minute the area gets 5 cm2 larger (steadily)
thus in 20 seconds we know the area (assuming t=0, area =0)
then area of circle ...

[Did you look at the resource I pointed out yesterday?]
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Thread starter 5 years ago
#8
(Original post by TeeEm)
rate of 5 cm2 every minute means every minute the area gets 5 cm2 larger (steadily)
thus in 20 seconds we know the area (assuming t=0, area =0)
then area of circle ...

[Did you look at the resource I pointed out yesterday?]
Isn't it per second?

So in 20 seconds you would get 100cm^2?

Then 100=area
Rearrange for r?

I was on my phone yesterday so couldn't use the link. But it doesn't seem to be working, do you mind sending it again? I seem to be getting a 404 error.
0
5 years ago
#9
(Original post by Super199)
Isn't it per second?

So in 20 seconds you would get 100cm^2?

Then 100=area
Rearrange for r?

I was on my phone yesterday so couldn't use the link. But it doesn't seem to be working, do you mind sending it again? I seem to be getting a 404 error.
sorry, yes it is per second ...

FILE NAME

related_rates_of_change

I just checked no error in mine
1
Thread starter 5 years ago
#10
(Original post by TeeEm)
sorry, yes it is per second ...

FILE NAME

related_rates_of_change

I just checked no error in mine
Got it this time haha Cheers
0
5 years ago
#11
(Original post by Super199)
Got it this time haha Cheers
with your question you are on the right track.

there are several questions like the one you describe.
enlarge the PDF to see answer in the photo-frame
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