Find the values of t for which the height of the water is decreasing.

I tried typing in the graph on desmos expecting a cubic graph but it just gave me a straight line

YEEEEEET Dec 17, 2018

#1**0 **

That is where the graph caught you off guard! The graph is actually not a straight line. Zoom out, and you will see that the graph curves like an "N" at around y = 58!

- PM

PartialMathematician Dec 17, 2018

#5**0 **

So there you have it... the height of the water decreases around \(t = 2.25\).

Hope this helps,

- PM

PartialMathematician
Dec 17, 2018

#3**+4 **

Take the derivative

h ' = 12t^2 - 59t + 72

Set this to 0 and set to 0

12t^2 - 59t + 72 = 0

(4t -9) (3t - 8) = 0

Setting both factors to 0 and solving for x gives

t = 9/4 t = 8/3

Take the second derivative

24t - 59

Plugging 9/4 into this gives - 4....so....we have a max at t = 9/4 sec = 2,25 sec

Plugging 8/3 into this gives 5....so....we have a min at t = 8/3 sec = 2.66 sec

So...the water height is decreasing from 2.25 sec to 2.66 sec

Heres the graph : https://www.desmos.com/calculator/ne2dnm3tks

YEEEEEET.......This does look like a straight line if you don't "zoom in"....very subtle....!!!!

CPhill Dec 17, 2018