course Mth 173
Quiz 1course Mth 173
Justin Funk Mth 173Problem #9
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9) y=.028(t)^2-1.1(t)+68
(t)9.2= 6.5489 (t)9.2+.1 = 6.64258
avg. rt change = 6.5489/6.64258= .98589/sec
It looks like you are dividing one value of y by the other.
You should be dividing change in y by change in t.
The change in t is .1. What is the change in y? What therefore is the requested average rate of change?
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y=.028(t)^2-1.1(t)+68
(t)9.2= 6.5489 (t)9.2+.01=6.55827
avg. change = .99857/sec
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y=.028(t)^2-1.1(t)+68
(t)9.2= 6.5489 (t)9.2+.001=6.54984
avg. change= .999856/sec
@ t=9.2 sec its rate will be 6.55
There is a single function y ' (t) which is evaluated to find the instanteous rate. What is that function and what is its value at the midpoint of the time interval?
Please resubmit this and include a statement of the problem.
A couple of suggestions:
It looks like you are dividing the value of y by the value of t. In general this is not an important result, though in certain applications it can be.
However this does not give an average rate of change. The average rate of change of y with respect to t is change in y / change in t. A single value of y, or a single value of t, does not constitute a change in that quantity.
Also be sure to include enough significant figures in your y values to guarantee that the difference, and hence the rates you obtain, have a sufficient number of significant figures to allow comparisons with other rates.
Please see my notes as submit another revision. Be sure to include a copy of the original problem.