Rate of change
Example
Satellite burning up on re-entry into the Earth's atmosphere
A satellite is dropping back to Earth. The distance it has fallen, \(D\) metres, since first spotted is given by \(D(t) = 100t + 5{t^2}\), where \(t\) is the length of time in seconds since first spotted.
How fast was the satellite travelling after 10 seconds?
Solution
The speed is the rate of change between the distance and the time.
Remember to calculate a rate of change, we differentiate.
\(D(t) = 100t + 5{t^2}\)
\(D\textquotesingle(t) = 100 + 10t\)
When \(t = 10\),
\(D\textquotesingle (t) = 100 + 10(10)\)
\(D\textquotesingle (t) = 200m/s\)
The satellite was travelling at 200 m/s after 10 seconds.
