For time, t, in hours 0<t<1 a bug is crawling at a velocity.

[Jaskaran]

v, in meters/hour given by v=[(9)/(3+t)].Use deltaT=0.2 to estimate the distance that the bug crawls during this hour. Find an overestimate and an underestimate. Then average the two to get a new estimate...

I already know what the answers, but I don't understand how they got them.

For example, for the underestimate, I would say that because delta is .2, I would evaluate 9/(3+.8) and get 2.368.... but the answer accordingly is 2.515. Am I missing something?

 

[galactus]

Re: For time, t, in hours 0<t<1 a bug is crawling at a veloc

distance is velocity times the change in time. In this case, change in time is .2

The lower estimate can be found by starting at t=.2 and incrementing up by .2 until you get to 1.

\(\displaystyle .2\left(\frac{9}{3+.2}+\frac{9}{3+.4}+......+\frac{9}{3+1}\right)\)

For the upper, start at 0 and go to .8

For the average, add the lower and upper and divide by 2.

Compare to the actual of \(\displaystyle \int_{0}^{1}\frac{9}{3+t}dt\)

 

[Jaskaran]

Re: For time, t, in hours 0<t<1 a bug is crawling at a veloc

Oh!!! Doh! I forgot to account for the width of the intervals :|