Daniel K.

asked • 09/09/16

Rates of Change

A container is being filled with water. After t seconds the depth of water, x cm, in the container is increasing at the rate of 1.44tcm/s.

Given that the container is empty when t=0, use calculus to find

a)the value of t when x=18.
 
b)the approx increase in x as t increases from 4 to 4.1.

David S.

tutor
Daniel,
 
The 1.44t would be the first derivative with respect to time of the function x(t).  To do this problem you need to integrate dx/dt = 1.44t and use the initial value of x=0 when t=0 to assign the constant (which end up being zero). Assuming that nothings else is known about the container -- like specific dimensions or a general shape -- and no other information about x(t) is given, that is what I recommend that you do. In my opinion, integration would be a more relevant topic for this question, rather than differentiation.  If you're not sure you got the right equation, you should differentiate it to see if you get 1.44t.
 
Once you have the equation for x(t) you can do part a) by substituting 18 (centimeters, presumably) in place of the x(t) and solving for t.  There're be a square root involved from what I can tell.  Does that help you get started?
 
You can also use the equation you get for x(t) to answer part b) by finding the difference between x(t=4.1) and x(t=4).  I'll check back later to see how you're doing.
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09/09/16

Daniel K.

i can get the answer for a but i dont know how to get the answer for b.
 
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09/09/16

Daniel K.

0.72(4.1)-0.72(4)=0.0072
 
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09/10/16

Daniel K.

0.072*
 
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09/10/16

1 Expert Answer

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David S. answered • 09/10/16

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You look like you started out OK with the 1.44/2 part.  Check what you came up with for x(t) again that you used in part a).  Your answer for b) looks like it would be correct if you squared the values for t, since I had x(t)=0.72t2.  Hopefully you just overlooked this: When you Integrate the variable's exponent Increases; Differentiate, exponent Decreases.
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