Question

water leaks out of a barrel at a rate proportional to the square root of the depth, D, of the water at that time. Depth is measure in inches and time i measured in hours. (don't worry about units my professor doesn't care) There are three parts I'll post below.

Answer 1

(a) Write a differential equation for the depth of water at time t. Assume the proportionality constant is k>0.
\[\frac{dD}{dt}=-k \sqrt{D} \]

Answer 2

(b) If the depth of the water is initially 36 inches, find D(t). Your answer will have k in it.
\[\int D^{-\frac{1}{2}}dD=\int-kdt\]
\[2D^{\frac{1}{2}}=-kt+C\]
\[D^{\frac{1}{2}}=C-\frac{kt}{2}\]
\[D=C-(\frac{kt}{2})^{2}\]
\[36=C \rightarrow D=36-(\frac{kt}{2})^{2}\]
got it wrong =(

Answer 3

I'm struggling to see my error

Answer 4

Oh, you silly billy.
\[\large\rm \left[C-\frac{kt}{2}\right]^2\quad\ne\quad C^2-\left(\frac{kt}{2}\right)^2\]

Answer 5

Remember that old FOIL business? ;)

Answer 6

darn I forget to write the power of two

Answer 7

I hope you understand what I'm saying.
The fact that you absorbed the square into the C is not the problem.

Answer 8

The problem is:
You should end up with `three terms` when you expand out a 2nd degree binomial.
Not just two terms.

Answer 9

no I get it i separated them when I shouldn't have

Answer 10

oh ok c:

Answer 11

Yah I would input the initial data before squaring.
The algebra works out a little easier that way.
Either is fine though.

You're inputting this to a website or something?

Answer 12

yup

Answer 13

so this way you get \[36-6kt+(\frac{kt}{2})^{2}\]

Answer 14

If you like to expand out the square :) yes!
I think it looks a tad nicer in this form though,\[\large\rm D(t)=\frac{1}{4}\left(12-kt\right)^2\]But yay I think you got it \c:/

Answer 15

Yes! thanks. I'm gonna continue to work out the rest of the problem on here just in case I goof again =)

Answer 16

(c) If the depth of the water drops to 35 inches in 1 hour, how long will it take for the water to leak out of the bucket.
\[35=\frac{1}{4}(12-k(1))^{2}\]
\[\sqrt{140}=12-k\]
\[12-\sqrt{140}=k =0.16784043\]
\[D=\frac{1}{4}(12-t(12-\sqrt{140}))^{2}\]
\[0=\frac{1}{4}(12-12t+t\sqrt{140}))^{2}\]
\[0=(12-12t+t\sqrt{140})^{2}\]
\[0=(284-24\sqrt{140})t^{2}+(24\sqrt{140}-288)t+144\]

Answer 17

\[t=6(6+\sqrt{35})hours\]

Answer 18

71.5 hours

Answer 19

Hmm I would probably square root each side from this step:\[0=(12-12t+t\sqrt{140})^{2}\]Instead of squaring it out >.<

Answer 20

oh I know but i had to show set by step with squaring as I entered it... what a drain

Answer 21

oh haha