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Thread: Differentiation problem

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    Default Differentiation problem

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


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    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.
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    Quote Originally Posted by okokokok View Post
    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?

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    AB = b / cos(alpha) + a / sin(alpha)
    Working on: Tithe Farmer

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    Quote Originally Posted by masterBB View Post
    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

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    Quote Originally Posted by the flea View Post
    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
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    Quote Originally Posted by okokokok View Post
    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....


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    I'll have to refresh my math a bit :P It's been a while since i have had this kind of math :P
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    I didn't really find a solution, but maybe i can help you get some ideas.



    So what i was thinking is if:

    Simba Code:
    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
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    Quote Originally Posted by okokokok View Post
    I didn't really find a solution, but maybe i can help you get some ideas.



    So what i was thinking is if:

    Simba Code:
    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

    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.

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    Quote Originally Posted by the flea View Post
    thanks for the help however I still dont understand what to do

    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...
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    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

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    True this is some though math..
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    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

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    Quote Originally Posted by Bam Bam View Post
    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

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    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

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    Quote Originally Posted by Bam Bam View Post

    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.

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    Quote Originally Posted by okokokok View Post
    I didn't really find a solution, but maybe i can help you get some ideas.



    So what i was thinking is if:

    Simba Code:
    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
    Last edited by zmon; 05-22-2012 at 12:59 AM.
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    i cant see the problem...

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    Try the cos-rule.

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    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


    Also post more things like this, maths problems are really fun imo
    Last edited by putonajonny; 05-28-2012 at 05:53 PM.

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