The flow of lava from a volcano can be modelled by the function l(t) = 12(2 - 0.8^t) , where *l* is the distance from the crater, in kilometres, and *t* is the time, in hours, since the eruption of the volcano. How fast is the lava travelling down the hillside after 4 h, to the nearest hundredth of a kilometre per hour?

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Given `l(t)=12(2-0.8^t)=24-12*0.8^t`

Differentiating with respect to t,

`(dl)/(dt)=0-12*(0.8)^t*ln(0.8)`

`=12*0.22314*0.8^t`

`=2.67768*0.8^t`

When l is the distance travelled by the lava, `(dl)/(dt)` should be its speed.

So, speed of the lava downhill, after 4 hours

`=2.67768*0.8^4`

=1.096778 kilometres per hour

=1.10 kilometres per hour (nearest to the hundredth).

**Sources:**

The flow of lava from a volcano can be modelled by the function l(t) = 12(2 - 0.8^t), where *l* is the distance from the crater, in kilometres, and *t* is the time, in hours, since the eruption of the volcano. How fast is the lava travelling down the hillside after 4 h, to the nearest hundredth of a kilometre per hour?

Sorry, this is the actual question!!!!!!!!

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