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# Use of maths: differentiation - need help finding rate of change watch

1. This is really simple, and I think I've got it, but I just need confirmation as to whether I'm doing this right

If, on a graph starting on the x axis at 8am, I am asked to find the rate of change at 10am do I simply enter t=2 into my dy/dx formula (dy/dx = 3.2 - 0.5t, from f = 42 + 3.2t - 0.25t^2) to make it 3.2 - 0.5(2) = 2.2?
2. Does it say what f is at a particular value of t? You would need that to work out whether to use t=10 or t=2.
3. What's 1 unit in your x axis? If 8am-10am is one unit, then t=1, if 8am-8.30am is one unit, then t=4.

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Updated: April 14, 2011
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