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course Mth173
5/12 3pm
Determine the average rate of change of the function y(t) = .5 t 2 + 56 t + -55 between x and x + `dx.‘dy = .5 (t + ‘dt)^2 + 56(t + ’dt) + -55 - (.5 t 2 + 56 t + -55)
= .5 (t^2 + 2t’dt + ‘dt^2) + 56t + 56’dt + -55 - (.5 t 2 + 56 t + -55)
= .5t^2 + t’dt + .5‘dt^2 + 56t + 56’dt + -55 - .5 t 2 - 56 t + 55
= t’dt + .5‘dt^2 + 56’dt
= ‘dt(t + .5‘dt + 56)
so
the average rate of change of the function y(t) = .5 t 2 + 56 t + -55 between x and x + `dx is
‘dy/’dt = ‘dt(t + .5‘dt + 56) / ((t + ’dt) - t)
= ‘dt(t + .5‘dt + 56) / ‘dt
= t + 56 + .5‘dt
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Good.
It is clear that as `dt approaches zero, this expression approaches t + 56, which as you know is the expression for the derivative of the original function.
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What are the growth rate, growth factor and principal function P(t) for an initial investment of $ 760 which is compounded annually at 10% interest?
growth rate = 10% = .10
growth factor = 1 + .10 = 1.10
p(t) = 760(1.10)^t
&#Good responses. See my notes and let me know if you have questions. &#