Differentiation problem

[the flea]

I'm having serious problems solving this. Any help would be greatly appreciated.

http://i44.tinypic.com/n85ug.jpg

[okokokok]

a:
What u will have to do is find the length of the ladders in the small triangles. So try to get the length of the ladder on the left from ladder b and the length of the ladder on the right of ladder a. If u have these lengths u can "flip" them over and have a bigger triangle to solve the length of the ladder AB.

[the flea]

a:
What u will have to do is find the length of the ladders in the small triangles. So try to get the length of the ladder on the left from ladder b and the length of the ladder on the right of ladder a. If u have these lengths u can "flip" them over and have a bigger triangle to solve the length of the ladder AB.

how would this be done whilst only having one value of length and not knowing the angle x?

[masterBB]

AB = b / cos(alpha) + a / sin(alpha)

[the flea]

AB = b / cos(alpha) + a / sin(alpha)

I got that far earlier but hit a brick wall when I couldnt work out what the numerical value for alpha is

E: the differential of that equation is:

(67*sin(alpha)/cos(alpha)^2)-(13*cos(alpha)/sin(alpha)^2)

thats just even more confusing

[okokokok]

I got that far earlier but hit a brick wall when I couldnt work out what the numerical value for alpha is

Let me take a look at it, it's not so easy for me since english is not my first language :) I'll get back to it in a few min

[the flea]

Let me take a look at it, it's not so easy for me since english is not my first language :) I'll get back to it in a few min

thanks for your help.

This is what I have so far which is clearly wrong....

http://i44.tinypic.com/fubazl.jpg

[okokokok]

I'll have to refresh my math a bit :P It's been a while since i have had this kind of math :P

[okokokok]

I didn't really find a solution, but maybe i can help you get some ideas.

http://i39.tinypic.com/23to0ms.jpg

So what i was thinking is if:


ladder DL = tan (z) * 13 ( tan (z) = DL/13)
ladder CL = tan (x) * 67 ( tan (x) = CL/67)
ladder AK = 13 + (tan (x) * 67)
ladder BK = 67 + (tan (z) * 13)

tan (x) = (tan (x) * 67)/(tan (z) * 13)

tan (z) = (tan (x) * (tan (x) *67))/13


And then differentiate the last formula?

It's been too long for me since i had this kind of math :(
Sorry that i couldn't help you more

[the flea]

I didn't really find a solution, but maybe i can help you get some ideas.

http://i39.tinypic.com/23to0ms.jpg

So what i was thinking is if:


ladder DL = tan (z) * 13 ( tan (z) = DL/13)
ladder CL = tan (x) * 67 ( tan (x) = CL/67)
ladder AK = 13 + (tan (x) * 67)
ladder BK = 67 + (tan (z) * 13)

tan (x) = (tan (x) * 67)/(tan (z) * 13)

tan (z) = (tan (x) * (tan (x) *67))/13


And then differentiate the last formula?

It's been too long for me since i had this kind of math :(
Sorry that i couldn't help you more

thanks for the help however I still dont understand what to do :duh:

if it helps anyone trying to solve this the value of x is 30.06765413 and the length AB is 103.3647197 I got the answers because I ran out of time however I can re attempt the question it will just give me different numbers at the start so I am still looking for the solution to this.

[okokokok]

thanks for the help however I still dont understand what to do :duh:

if it helps anyone trying to solve this the value of x is 30.06765413 and the length AB is 103.3647197 I got the answers because I ran out of time however I can re attempt the question it will just give me different numbers at the start so I am still looking for the solution to this.

Well what i tried to do is if u could get Angle z then you would be able to get DL. Since AB splits a 90 degree corner into z and x' (z-angle) you could also get CL. Then if u have those 2 u know what AK and BK is. Then simply use tan(x) = AK/BK and u would have had AB. But maybe i'm completely wrong...

[the flea]

just got home from class now. Still not got any further, I hate this type of question pretty much all other types of math im ok with

[okokokok]

True this is some though math.. :(

[Bam Bam]

AB = 13/sin(x) + 67/sin(90 - x)

This is based on the assumption that the ladder must contact the corner and that neither 0 or 90 degrees is important.

Differentiate and find x where length is minimum. (Slope = 0)

On my calculator it would seem L_max is 103.36ft and x is 30.07

I can post more when Im not doing so from a phone.

-Bam Bam

[the flea]

AB = 13/sin(x) + 67/sin(90 - x)

This is based on the assumption that the ladder must contact the corner and that neither 0 or 90 degrees is important.

Differentiate and find x where length is minimum. (Slope = 0)

On my calculator it would seem L_max is 103.36ft and x is 30.07

I can post more when Im not doing so from a phone.

-Bam Bam

yes that's the correct way to do it. My lecturer went through it in our lesson today since the deadline was last night. Well done on solving it I found it really hard

[Bam Bam]

Sorry I didn't get it done earlier :(

One of my professors today was boring me so I did your problem from memory.
I looked at it last night though and had no idea how to do it then lol :P

-Bam Bam

[tehq]

On my calculator it would seem L_max is 103.36ft and x is 30.07


Got those answers as well using Lmax = (a^(2/3) + b^(2/3) )^(3/2) while phi(Ao) = 0.

[zmon]

I didn't really find a solution, but maybe i can help you get some ideas.

http://i39.tinypic.com/23to0ms.jpg

So what i was thinking is if:


ladder DL = tan (z) * 13 ( tan (z) = DL/13)
ladder CL = tan (x) * 67 ( tan (x) = CL/67)
ladder AK = 13 + (tan (x) * 67)
ladder BK = 67 + (tan (z) * 13)

tan (x) = (tan (x) * 67)/(tan (z) * 13)

tan (z) = (tan (x) * (tan (x) *67))/13


And then differentiate the last formula?

It's been too long for me since i had this kind of math :(
Sorry that i couldn't help you more

tan (y) = sin(y)/cos(y)

and x = 90-z, so tan(z) = sin(90-x)/cos(90-x) = cos(x)/sin(x) = 1/tan(x)

the angle z depends on angle x, you wouldn't get that far with it. you need to express length of ladder as function of x then you can derivate

not to mention that the last two lines don't come out of first four lines.

here is how you do that:

n * sin(x) = a
m * cos(x) = b

n = a/sin(x)
m = b/cos(x)

so length as f(x) = n + m = a/sin(x) + b/cos(x)

derivate of 1/t(x) = -t'(x) / t^2(x)

so a/sin(x) derivate is -acos(x)/(sin(x)^2) and b/cos(x) derivate is bsin(x)/(cos(x)^2)

so you are looking for 0= a cos(x)^3 - b sin(x)^3, or after few adjustments:

tan(x)^3 = a/b
tan(x) = (a/b)^(1/3) (cubic root)

therefore x = approx 30.06

Edit: oops forgot the little drawing with m and n on it
http://oi46.tinypic.com/dp77rq.jpg

[doorsftw]

i cant see the problem...

[Pascal]

Try the cos-rule.

[putonajonny]

Angle = 30.06765414 Degrees
AB = 103.3647197 units

Will take a photo of my working in a sec and upload it, but it seems to be consistent with other people and your answer
http://i.imgur.com/p0y55.jpg

Also post more things like this, maths problems are really fun imo