how fast does a tank drain? applied calculus project

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matt757

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dh/dt=kh^1/2 and k is the constant


Suppose that a hole is drilled in the side of a cylindrical bottle and the height of thewater (above the hole) decreases from 10 cm to 3 cm in 68 seconds. Use Equation 2 to find an expression for h(t). Evaluate h(t) for "t"=10,20,30,40,50,60

please help because i am having the hardest time trying to figure out what to do i have the separable equation right, but im am so confused by the question.
 
Last edited:
matt757 said:
dh/dt=kh^1/2 and k is the constant


Suppose that a hole is drilled in the side of a cylindrical bottle and the height of thewater (above the hole) decreases from 10 cm to 3 cm in 68 seconds. Use Equation 2 to find an expression for h(t). Evaluate h(t) for "t"=10,20,30,40,50,60

please help because i am having the hardest time trying to figure out what to do i have the separable equation right, but im am so confused by the question.
Click to expand...
Where is equation 2?

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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im in college and taking calculus this is are project that we are working on and the question that i posted in the first post is question "a" i dont know how to set the problem up after this step
 
Yes, you are good up to that point
h = 14(kt+c)2\displaystyle \frac{1}{4}(k t + c)^2
but lets go back a step first to
2 h1/2 = k t + c.
To get to the final equation in order to evaluate h in terms of numbers we still need two pieces of information. One is the value of c and the other is the value of k. We are given that the height of the water above the hole decreases from 10 cm at a time of t=0 to 3 cm at a time of 68 seconds (plus some start time t0 for both times so we can just start our zero time at t0 and use the equation we have). Thus
210\displaystyle \sqrt{10} = k * 0 + c
and
23\displaystyle \sqrt{3} = k * 68 + c
Can you take it from there?


Notice that had we chosen some other time for t0 (or had it been chosen for us), we have 10 cm at t=t0 and 3 cm at t=t0+68 so we could have rewritten the equation as
2 h1/2 = k (t-t0) + c
to obtain the same results.
 
matt757 said:
what happen to the "H" and would the c=2???
Click to expand...
If h= 10 cm then h12 = 10\displaystyle h^{\frac{1}{2}}\, =\, \sqrt{10} so that at t=0 we have the equation
2 10 = k∗0 + c\displaystyle 2\, \sqrt{10}\, =\, k * 0\, +\, c
Given that equation what is c?
 
matt757 said:
2*101/2=k*0+C
6.32=0+c
6.32=C????
Click to expand...
Right (to two decimal places). Now given that h=3 cm (h1/2=3\displaystyle h^{1/2} = \sqrt{3}) at t=68 we have
23\displaystyle \sqrt{3} = k * 68 + c.
So what is k?
 
2*31/2=68k+6.32
3.46-6.32=68k
-2.86/68=k
-.04=k

so would we do that for 20, 30.. etc intil we got to 60

because the problem say to evaluate for h(t) for t=10,20,30,40,50,60
 
when we find k and c then we can do such things as h=68*-.04+6.32 to find the h for the first evluated of 10 seconds for the first one, then the next one would be 20 and so on..etc
 
matt757 said:
2*31/2=68k+6.32
3.46-6.32=68k
-2.86/68=k
-.04=k

so would we do that for 20, 30.. etc intil we got to 60

because the problem say to evaluate for h(t) for t=10,20,30,40,50,60
Click to expand...
Now you have the equation for h:
h = (1/4) (-0.0421 t + 3.46)2
Now plug in t = 10,20,30,40,50, and 60 to get h at those times
 
h = (1/4) (-0.0421 t + 6.32)2 this would be the final form, and thank you so much i literally spent hours on this question and it ok :)
 
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