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swin2013
Why don't I have to factor out the (1+h)^2? on problem 2, page 2? http://www.math.wisc.edu/~hernande/Teaching/UM/Math115Sec011Winter2011/Quizzes/Quiz3/Solution/Quiz3Sol.pdf
yea but i don't need to square 1+h because why? lol
usually i have to square it... that's why. especially for a slope problem.
I don't understand the question. you are asked to find the average velocity = distance/time the distance traveled in the time between 1 and 1+h (with h=0.1) is 4(1.1)^2 +3 - ( 4*1^2 +3) 4*1.21 + 3 -4 -3 4.84 -4 0.84 the time interval is 0.1 sec, so velocity is 0.84/0.1 = 8.4 sec
in other words, you do square (1+h)
but when i squared (1+h), the 4's and 3's cancel. i get 8.04 when the answer should be 8.4 m/s
in each case, you find how far the particle got by time step 1. Then find where it is at time step 1+h. The difference between those two distances is how far it moved in h secs.
but when i squared (1+h), the 4's and 3's cancel. i get 8.04 when the answer should be 8.4 m/s for case (i), with h = 0.1 1+h is 1.1 1.1*1.1= 1.21 4*1.21= 4.84 subtract 4 0.84 divide by 0.1 to get 8.4 you must have made a mistake doing 1.1 * 1.1 ??
I am not sure how you did your calculation, but you have a "bug" in it, so it would be good to figure out where you are going wrong...
\[4(h ^{2}+2h+1) +3-4(1)^{2}-3\] since t =1?
yes. that looks ok. then 4h^2 +8h +4 +3 -4 -3 = 4h^2+8h if we divide by h (to get velocity) 4h+8 with h=0.1 this is 4*0.1+8 = 0.4+8= 8.4 perfect!
ok that's where i went wrong. i factored out 4h so it would be 4h ( h^2 +2) and canceled out the h's so it would be left with 4(h^2+2)
if you factor out 4h, it should be 4h(h+2) and 4(h+2) after dividing by h
yea that's my mistake... i left the squared. thanks :)
i can't believe it was that miniscule
i know! such a tiny mistake could cost me an answer by one decimal place -_-
sometimes it just doesn't work out...good luck!