Question

Can someone help me with this question??

https://skydrive.live.com/redir?resid=5F146673C0A3A322!737

Answer 1

Oh it also asks to sketch??

Answer 2

To find the minimum of the function\[E(v) = av^3 \frac{ L }{ v - u }\]we have to differentiate it with v as variable and assuming a, u and L are constants.
Using the product rule, we get:\[\frac{ dE }{ dv } = 3av^2\frac{ L }{ v - u } + av^3*-\frac{ L }{ (v - u)^2 } \]Now solve \[\frac{ dE }{ dv } = 0\]Factoring out gives:\[\frac{ av^2L }{ v - u }\left( 3 - \frac{ v }{ v - u } \right) = 0\]
The left factor gives v = 0 as solution, but this is not what we need.
Setting the right factor zero gives:\[\frac{ v }{ v - u } = 3\]so\[v = 3v - 3u\]Then\[2v = 3u \]gives \[v = \frac{ 3 }{ 2 } u\]
If you would draw a "sign scheme" of dE/dv (don't know the right term for this in English), you'd see the sign going from - to + at v = 3/2 u. This means that E has a minimum there!

Hope this helps!

ZeHanz

Answer 3

Here is a drawing in Geogebra. In it, a = 0.1, L = 100 and u = 5.
Point A has x=coordinate 7.5, which is 1.5 times u.

ZeHanz