+0  
 
+1
109
24
avatar+790 

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

 Dec 17, 2018
 #1
avatar+698 
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

 Dec 17, 2018
 #4
avatar+790 
0

unfortunately the reason why i typed it in to desmos is i thought there would be roots to help me solve it but the range is a bit far for me haha 

YEEEEEET  Dec 17, 2018
edited by YEEEEEET  Dec 17, 2018
 #5
avatar+698 
0

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

 

Hope this helps,

 - PM

PartialMathematician  Dec 17, 2018
edited by PartialMathematician  Dec 17, 2018
 #3
avatar+98172 
+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....!!!!

 

 

 

 

cool cool cool

 Dec 17, 2018
 #6
avatar+790 
+1

ahhh ok thank you very much

YEEEEEET  Dec 17, 2018
 #7
avatar+698 
-1

You are welcome! (I didn't exactly do much)

PartialMathematician  Dec 17, 2018
edited by PartialMathematician  Dec 17, 2018

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