At what rate is the circumference changing when radius=6cm? The radius is increasing at a rate of 1.5cm/sec.

Question

darkigloo · Answer

So i know :
dr/dt=1.5
r=6
C=2 pi r 
i guess i have to find dC/dt

nincompoop · Answer

differentiation

darkigloo · Answer

can you explain how

nincompoop · Answer

dr/dt

nincompoop · Answer

circumference depends on the radius since pi is constant

nincompoop · Answer

so how would you set up your equation?

darkigloo · Answer

im not sure...would it be
$$\frac{ d }{ dt } [ C = 2 \pi  r]$$

nincompoop · Answer

go on

darkigloo · Answer

$$\frac{ dC }{ dt } = 2 \pi \frac{ dr }{ dt }$$
i'm pretty sure that's incorrect...

nincompoop · Answer

since r is a function of time 
dC/dt = 2pir dr/dt would this be okay?

darkigloo · Answer

why does the r stay if it is being differentiated?

nincompoop · Answer

implicit differentiation

nincompoop · Answer

y'y <--

nincompoop · Answer

r'(t) = ? when r(t) = 6cm

campbell_st · Answer

I would have thought the rate of change in circumference with respect to radius is a constant

and you can use related rates to solve the problem 

$$\frac{dC}{dt} = \frac{dC}{dr} \times \frac{dr}{dt}$$

nincompoop · Answer

C' = 2pi r'

campbell_st · Answer

so 

$$\frac{dC}{dt} = 2\pi \times 1.5 $$

campbell_st · Answer

since if $$C = 2\pi r$$

$$\frac{dC}{dr} = 2\pi$$

darkigloo · Answer

so what happened to r=6cm?

campbell_st · Answer

well the rate of change in circumference is always 3pi  no matter what the radius is... 

but then I'm in Year 11...

campbell_st · Answer

oops should be 

$$\frac{dC}{dt} = 3 \pi   ..cm/ second$$

darkigloo · Answer

hmm...alright. thank you.

campbell_st · Answer

@nincompoop may have something I haven't studied yet... 

whatever implicit differentiation is

darkigloo · Answer

ok. i do believe the answer is 3pi.
2pi(1.5)=3pi

nincompoop · Answer

laughing out loud campbell, i was gonna show that rates can be formulated even without any given data by obtaining the general formula

nincompoop · Answer

I was looking for something that might help explain a cumbersome lesson into simple parts http://www.mathscoop.com/calculus/derivatives/applications/related-rates.php

campbell_st · Answer

ok... sorry to interrupt....

I just see these sorts of questions as related rates... 

sorry again..

nincompoop · Answer

you've been a help in fact

darkigloo · Answer

Thank you both.

nincompoop · Answer

thanks @campbell_st 
I hope the link I posted will help also, @darkigloo 
I had a feeling you're still at a loss with the steps

campbell_st · Answer

Hope I didn't confuse you @darkigloo

darkigloo · Answer

@nincompoop yeah i'm going to take a look at the link. thanks.
@campbell_st don't worry, you didn't confuse me.