The emergent velocity, v, of a liquid flowing from a hole in the bottom of a tank is given by v = Square root(2gh) where g is the acceleration due to gravity and h is the depth of the liquid in the tank. Find the rate of change of v with respect to h when (a) h = 9 , (b) h = 4 .
\( v=\sqrt{2gh} \) can you find \( \large \frac{dv}{dh} \) ?
Would it be \[\frac{ 1 }{ 2 }(2gh)^{-\frac{ 1 }{ 2 }}\] ?
You can take the constant \( \sqrt{2g} \) out before you integrate.
*differentiate.
And have it as \[\sqrt{2g} + \sqrt{h} \] ?
"Take out" is meant to factorize and put it on the left of the differentiation: \( \frac{d\sqrt{2gh}}{dh} \\ =\sqrt{2g}\frac{d\sqrt{h}}{dh} \\ =\sqrt{2g} {\frac{1}{2}}\frac{1}{\sqrt{h}} \\ =\frac{\sqrt{g}}{2h} \)
Ah, that is far more elegant. lol
Do I just input the values now?
No, there is a typo. it should read \( \frac{\sqrt{2g}}{2\sqrt{h}} \) Typo is all on the last line.
That makes much more sense.
Do check answers and be convinced that they are correct before you put it into your answer, i.e. before you "own" it.
I usually do. I was sure I was missing something and have been thinking it though the entire time. I thought you had pulled out the square root of two and somehow turned it into two on the bottom. It made my brain fart. ><;
That's good. Getting the right answer is a temporary pleasure. Knowing how to solve the problem is the real success! Keep up the good work!
So, since that is the first derivative, that represent rate of change, so plug in the force of gravity (32 ft/sec) and the two h's given, right? (Yeah! Or in Isaac Asimov's words: "The true pleasure is in the finding out, rather than the knowing." :D
Exactly! Note: are you working in imperial system or metric? 32.2 ft/s^2 or 9.81 m/s^2 ?
I agree with that too!
Imperial. Mayhaps someday the United States will convert to metric. >.>
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