Question

@ganeshie8

The number of tons of paper waiting to be recycled at a 24 h recycling
plant can be modelled by the equation p(t)=0.5 sin((pi/6)t)+4
where t is the time, in hours, and is the number of tons waiting to
be recycled.
a) Use the equation to estimate the instantaneous rate of change in tons
of paper waiting to be recycled when this rate is at its maximum. To
make your estimate, use each of the following centred intervals:
i) 1 h before to 1 h after the time when the instantaneous rate of
change is at its maximum

Answer 1

How do I determine when the Max Instantaneous rate of change occurs?

Answer 2

oh i forgot the equation

Answer 3

its p(t)=0.5 sin((pi/6)t)+4

Answer 4

Would the Max instantaneous rate of change occur when the slope of the instantaneous rate of change is the steepest?

Answer 5

I'm thinking since it says to estimate the instantaneous rate of change we are suppose to find the average rate of change from t-1 to t+1
assuming t is where we have our max instantaneous rate of change
then find the max of the resulting function

Answer 6

yeah i think the teacher wants him to figure out where the max occurs visually and estimate the instantaneous rate of change at that point

Answer 7

see if this graph helps
http://gyazo.com/38e581194095afb7de1665561286990c

Answer 8

okay so at x=6 is when the instantaneous would approximately equal the max right?

Answer 9

Yes! thats the first time the instantaneous rate is max after t=0

Answer 10

okay cool, thanks guys :D

Answer 11

hey wait, but the graph is decreasing

Answer 12

okay then i can just choose at x=12 since its increasing?

Answer 13

Very clever ! yes :)

Answer 14

Awsome Thanks again for all the help!

Answer 15

may be just write out that the absolute value of instantaneous rate is maximum at t=6 and t=12 etc.. but the graph is decreasing at t=6, the instantaneous rate of change is negative at t=6 and since the question wants MAXIMUM instantaneous rate of change, it wants positive...

Answer 16

oh and when i'm calculating rate of change, i choose the point +- 1 hour from the max right?

Answer 17

so when t=11 and t=13

Answer 18

yep use t = 12-1 and t = 12+1

Answer 19

okay thanks

Answer 20

I got and 0.25 tonnes/ hour, which is right in the book :)

Answer 21

just for fun
\[\frac{f(t+1)-f(t-1)}{(t+1)-(t-1)} \\ \frac{(.5\sin(\frac{\pi}{6}(t+1))+4)-(.5\sin(\frac{\pi}{6}(t-1))+4)}{2} \\ \frac{.5(\sin(\frac{\pi}{6}t+\frac{\pi}{6})-\sin(\frac{\pi}{6}t-\frac{\pi}{6}))}{2} \\ \frac{1}{4} (\sin(\frac{\pi}{6}t+\frac{\pi}{6})-\sin(\frac{\pi}{6}t-\frac{\pi}{6})) \\ \frac{1}{4}(\sin(\frac{\pi}{6}t)\cos(\frac{\pi}{6})+\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}t)-\sin(\frac{\pi}{6}t)\cos(\frac{\pi}{6})+\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}t)) \\ \frac{1}{4} (\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}t)+\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}t)) \\ \frac{2}{4}\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}t) \\ \frac{1}{2} \sin(\frac{\pi}{6})\cos(\frac{\pi}{6}t) \\ \frac{1}{2} \frac{1}{2} \cos(\frac{\pi}{6}t) \\ \frac{1}{4} \cos(\frac{\pi}{6}t) \]
the max of .25cos(pi/6 *t) is .25
to find what what t is when the instanteous rate of change of .25
we do
\[\frac{1}{4}\cos(\frac{\pi}{6}t)=\frac{1}{4} \\ \cos(\frac{\pi}{6}t)=1 \\ \frac{\pi}{6}t=2 n \pi \\ t=\frac{6}{\pi}2 n \pi=12n \text{ where } n \text{ is an integer }\]
that was just for fun

Answer 22

O.O no wonder you were taking so long to reply lol

Answer 23

Wow! that looks a lot better to justify than the graph as the max value of cosine is pretty straightforward to find out xD

Answer 24

I thought that is what it wanted us to do
when he said some about looking at the interval from an hour before to an hour after

Answer 25

but you guys got the answer in like 2 seconds from observation
:)

Answer 26

yeah i feel they want something analytic like that after seeing above method, but it looks like a new method which you invented on the fly hmm never seen/thought about it before but it looks great !

Answer 27

well yeah it something I mentioned on the fly
and it seemed to work
and i thought it could work since it said to estimate the instantaneous rate of change

Answer 28

that is a secant line going through the points (t+1,f(t+1)) and (t-1,f(t-1))
where t is assumed to give us the max instantaneous rate change

Answer 29

I think it works for any interval :
\[\frac{f(t+\delta)-f(t-\delta)}{2\delta} \\~\\

\frac{1}{2\delta } \sin (\frac{\pi \delta}{6})\cos(\frac{\pi}{6}t)
\]

Answer 30

max value = \(\large \lim\limits_{\delta \to 0 }\frac{1}{2\delta } \sin (\frac{\pi \delta}{6})\)

Answer 31

Oh yeah i see where it comes from now, your method is equivalent to average value limit definition of derivative XD

Answer 32

https://bell232.wikispaces.com/file/view/MHF4U+N2609.pdf
this looks ton easier

Answer 33

they just found the period