Can someone work this problem ASAP?

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warwick

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A wall 8 m high stands 4 m away from a building. What is the length of the shortest ladder that will lean over the wall and touch the building? Use as independent variable the angle theta that the ladder makes with the ground.
 
start with a sketch.

let x = horizontal distance from the foot of the ladder to the 8' wall

L = length of the ladder

\(\displaystyle \L \theta\) = angle the ladder makes with the ground

using your carefully labeled sketch, you should be able to see the following two trig relationships ...

\(\displaystyle \L \tan{\theta} = \frac{8}{x}\)

\(\displaystyle \L \cos{\theta} = \frac{x+4}{L}\)

using these two equations, you can get L in terms of \(\displaystyle \L \theta\) ...

\(\displaystyle \L L = 4(\frac{2+\tan{\theta}}{\sin{\theta}})\)

now, find \(\displaystyle \L \frac{dL}{d \theta}\) and minimize L.
 
skeeter said:
start with a sketch.

let x = horizontal distance from the foot of the ladder to the 8' wall

L = length of the ladder

\(\displaystyle \L \theta\) = angle the ladder makes with the ground

using your carefully labeled sketch, you should be able to see the following two trig relationships ...

\(\displaystyle \L \tan{\theta} = \frac{8}{x}\)

\(\displaystyle \L \cos{\theta} = \frac{x+4}{L}\)

using these two equations, you can get L in terms of \(\displaystyle \L \theta\) ...

\(\displaystyle \L L = 4(\frac{2+\tan{\theta}}{\sin{\theta}})\)

now, find \(\displaystyle \L \frac{dL}{d \theta}\) and minimize L.
Click to expand...

I got the tan theta = x/8, but I used tan theta = y/4. No wonder I wasn't getting anywhere with another variable. I knew I had to write everything in terms of x. Ugh. :roll: I hate it when I overcomplicate things. You should've seen what I got using x AND y with theta as the independent variable. Haha.
 
\(\displaystyle \L\\L_{1}+L_{2}=L\)

\(\displaystyle \L\\L_{1}=8csc({\theta})\)

\(\displaystyle \L\\L_{2}=4sec({\theta})\)

\(\displaystyle \L\\L=8csc({\theta})+4sec({\theta})\)

\(\displaystyle \L\\\frac{dL}{d{\theta}}=-8csc({\theta})cot({\theta})+4sec({\theta})tan({\theta})=\frac{-8cos({\theta})}{sin^{2}({\theta})}+\frac{4sin({\theta})}{cos^{2}({\theta})}=\frac{-8cos^{3}({\theta})+4sin^{3}({\theta})}{sin^{2}({\theta})cos^{2}({\theta})}\)

\(\displaystyle \L\\\frac{dL}{d{\theta}}=0\) when \(\displaystyle 4sin^{3}({\theta})=8cos^{3}({\theta})\)

If \(\displaystyle \L\\tan^{3}({\theta})=2\) which give sthe absolute minimum.

You have enough to find theta. Use it in the original function to find the length.

Finish up?.
 
galactus said:
\(\displaystyle \L\\L_{1}+L_{2}=L\)

\(\displaystyle \L\\L_{1}=8csc({\theta})\)

\(\displaystyle \L\\L_{2}=4sec({\theta})\)

\(\displaystyle \L\\L=8csc({\theta})+4sec({\theta})\)

\(\displaystyle \L\\\frac{dL}{d{\theta}}=-8csc({\theta})cot({\theta})+4sec({\theta})tan({\theta})=\frac{-8cos({\theta})}{sin^{2}({\theta})}+\frac{4sin({\theta})}{cos^{2}({\theta})}=\frac{-8cos^{3}({\theta})+4sin^{3}({\theta})}{sin^{2}({\theta})cos^{2}({\theta})}\)

\(\displaystyle \L\\\frac{dL}{d{\theta}}=0\) when \(\displaystyle 4sin^{3}({\theta})=8cos^{3}({\theta})\)

If \(\displaystyle \L\\tan^{3}({\theta})=2\) which give sthe absolute minimum.

You have enough to find theta. Use it in the original function to find the length.

Finish up?.
Click to expand...

You'll have to finish it up. I went your route but didn't see the way through.
 
It's right in front of you.

If \(\displaystyle \L\\tan^{3}({\theta})=2\), then solve for theta.

Sub it into the function for L to find the total length.
 
That's what I got. Maybe keep your answer in the form

\(\displaystyle \L\\L=4(2^{\frac{2}{3}}+1)^{\frac{3}{2}}\)

Looks nicer.

Now, try skeeter's method to see if you get the same thing.

A double check can't hurt.
 
galactus said:
That's what I got. Maybe keep your answer in the form

\(\displaystyle \L\\L=4(2^{\frac{2}{3}}+1)^{\frac{3}{2}}\)

Looks nicer.

Now, try skeeter's method to see if you get the same thing.

A double check can't hurt.
Click to expand...

Actually, I did. I figured it would be easier than plugging in csc and sec. Heh.
 
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