c3- differentiation

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?

$\frac{dA}{dt}=\frac{dr}{dt} \times \frac{dA}{dr}$. We want to find $\frac{dA}{dt}$ and you're given $\frac{dr}{dt}$. Does that help?

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.

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 ...
then radius =

[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.

sorry, yes it is per second ...

LINK

http://madasmaths.com/archive_maths_booklets_standard_topics_various.html

FILE NAME

related_rates_of_change

I just checked no error in mine

Got it this time haha Cheers