course MTH271
09/25/09 7:35 pm
Determine the average rate of change of the function y(t) = .5 t 2 + 56 t + -55 between x and x + Dx.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?
[.5 t 2 + 56 t + -55 +.5 (t+Dx) 2 + 56 (t+Dt) + -55]/2
[.5 t 2 + 56 t + -55 +.5 (t+Dt)(t+Dt) +56t + 56 Dt + -55 ]/2
[.5 t 2 + 56 t + -55 +.5(t2 + 2tDT + Dt 2 + 56t + 56 Dt + -55 ]/2
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The average rate of change is (change in y) / (change in t).
Change in y is
[.5t^2 + 2tDT + Dt^2 + 56t + 56 Dt + -55 - (.5 t^2 + 56 t + -55)]
This simplifies quite a bit.
Change in t is (t + `dt) - t = `dt.
So the rate is
[.5t^2 + 2tDT + Dt^2 + 56t + 56 Dt + -55 - (.5 t^2 + 56 t + -55)] / `dt, which then need to be simplified.
[.5 t 2 + 56 t + -55 + .5t2 + tDt + .5Dt2 + 56t + 56Dt + -55] /2
(t2 + 112t + -110 + tDt +5Dt2 +5Dt)/2
(t2 /2) + 56t + - 55 +(tDt/2) + (5Dt2) /2 + (5Dt)/2
P(t) = 760(1+.1)^t
P(t) = 760(1.1)
P(t) = 836
Gotta identify growth rate, which is .1.
Growth factor is 1.1.
Important to know these terms and identify them--among other things it's usually on the test.
&#Your work looks good. See my notes. Let me know if you have any questions. &#