# Rate of change

1. May 12, 2013

### hbuk

I am trying to complete the following question:

By pumping the air pressure in a tank is reduced by 15%/second. The percentage of air pressure remaining is fiven by the formula.

p=100(0.85)^t

Plot p against t for 0 < t < 30s.

Deternine the rate of change at 5s & 10s. I have plotted the graph.

Verify you anser via calculus.

2. May 12, 2013

### zeralda21

The rate of change is given by the derivative p'(t)=dp/dt where t is the argument(given in seconds). Your are interested in the rate of change at after 5s and 10s, hence you want to calculate p'(5) and p'(10).

You have plotted the graph so you could also find the slope of the tangent(2 lines) at t=5 and at t=10. But the first method is much easier if you know how to differentiate that function.

3. May 12, 2013

### hbuk

thank you, that makes sense.