# Thread: applications of differentiation

1. ## applications of differentiation

A bushfire burns out A hectares of land, t hours after it started, according to the rule:

A = 90t^2 - 3t^3

After how long is the rate of change equal to 756 hectares per hour?
__________________________________________________ _______________

In order to answer this question I found the derivative of A = 90t^2 - 3t^3, which is dV/dt = 180t -9t^2

And then I tried to substitute 756 to equate the derivative of A.
180t - 9t^2 = 756
-9t^2 + 180t -756 = 0

But I got different answers when I tried to factorise the above. Using a calculator, I got t = -0.356466 or t = 23.5647 but...

t = 6 or t = 14 hours...can somebody show where I went wrong? Could somebody show me the complete working out? Any help will be appreciated.

2. Originally Posted by Mr Rayon
A bushfire burns out A hectares of land, t hours after it started, according to the rule:

A = 90t^2 - 3t^3

After how long is the rate of change equal to 756 hectares per hour?
__________________________________________________ _______________

In order to answer this question I found the derivative of A = 90t^2 - 3t^3, which is dV/dt = 180t -9t^2

And then I tried to substitute 756 to equate the derivative of A.
180t - 9t^2 = 756
-9t^2 + 180t -756 = 0

But I got different answers when I tried to factorise the above. Using a calculator, I got t = -0.356466 or t = 23.5647 but...

t = 6 or t = 14 hours...can somebody show where I went wrong? Could somebody show me the complete working out? Any help will be appreciated.
$A=90t^2-3t^3$

$\frac{dA}{dt}=180t-9t^2$

$\frac{dA}{dt}=756$

$180t-9t^2=756$

$9t^2-180t+756=0$

$t^2-20t+84=0$

$(t-6)(t-14)=0$

$t=6$ hours or $t=14$ hours