# help with presentation

• Oct 4th 2009, 02:32 PM
eduk8tedidiot
help with presentation
Hello I am a thirty something college student who must present a solution to a word problem in class monday. Please help with each step of the solution. Find the length of the shortest ladder that will reach from the ground, over a wall 8ft high, to the side of a building 1 ft behind the wall. That is minimize the legnth L=L1 plus L2
• Oct 4th 2009, 03:08 PM
galactus
There are various ways to tackle this. I will present a method using trig.

$L=L_{1}+L_{2}$

Then, $L=8csc(t)+sec(t)$ (look at the two triangles made by the ladder, wall, and ground).

$\frac{dL}{dt}=\frac{-8cos^{3}(t)+sin^{3}(t)}{sin^{2}(t)cos^{2}(t)}$

$\frac{dL}{dt}=0$ if $sin^{3}(t)=8cos^{3}(t), \;\ tan^{3}(t)=8$

$tan(t)=2$ which gives the absolute minimum for L because

$\lim_{t\to\ 0^{+}}L=\lim_{t\to \frac{{\pi}^{-}}{2}}L={\infty}$

If $tan(t)=2\Rightarrow csc(t)=\frac{\sqrt{5}}{2} \;\ and \;\ sec(t)=\sqrt{5}$

Hence, $L=8(\frac{\sqrt{5}}{2})+\sqrt{5}=5\sqrt{5} \;\ feet$
• Oct 4th 2009, 09:21 PM
eduk8tedidiot
thanks
is there a way to solve it using derivatives
• Oct 5th 2009, 12:02 AM
mr fantastic
Quote:

Originally Posted by galactus
There are various ways to tackle this. I will present a method using trig.

$L=L_{1}+L_{2}$

Then, $L=8csc(t)+sec(t)$ (look at the two triangles made by the ladder, wall, and ground).

$\frac{dL}{dt}=\frac{-8cos^{3}(t)+sin^{3}(t)}{sin^{2}(t)cos^{2}(t)}$ *

$\frac{dL}{dt}=0$ if $sin^{3}(t)=8cos^{3}(t), \;\ tan^{3}(t)=8$ *

$tan(t)=2$ which gives the absolute minimum for L because

$\lim_{t\to\ 0^{+}}L=\lim_{t\to \frac{{\pi}^{-}}{2}}L={\infty}$

If $tan(t)=2\Rightarrow csc(t)=\frac{\sqrt{5}}{2} \;\ and \;\ sec(t)=\sqrt{5}$

Hence, $L=8(\frac{\sqrt{5}}{2})+\sqrt{5}=5\sqrt{5} \;\ feet$

Quote:

Originally Posted by eduk8tedidiot
is there a way to solve it using derivatives

What do you think the lines marked with * mean ....!?