Ladder

4 m long ladder touches the cube 1mx1m at the wall. How high reach on the wall?

Correct result:

q1 =  1.362 m
q2 =  3.761 m

Solution:

l=4 f(x)=kx+q f(1)=1 f(x0)=0 k=1/(1x0) q=1k x01=3.7609 x02=1.3622 k1=1/(1x01)=1/(13.7609)0.3622 q1=1k1=1(0.3622)1.36221.362 m l1=q12+x012=1.36222+3.760924l=4 \ \\ f(x)=kx+q \ \\ f(1)=1 \ \\ f(x_{0})=0 \ \\ k=1/(1-x_{0}) \ \\ q=1-k \ \\ x_{01}=3.7609 \ \\ x_{02}=1.3622 \ \\ k_{1}=1/(1-x_{01})=1/(1-3.7609) \doteq -0.3622 \ \\ q_{1}=1-k_{1}=1-(-0.3622) \doteq 1.3622 \doteq 1.362 \ \text{m} \ \\ l_{1}=\sqrt{ q_{1}^2+x_{01}^2 }=\sqrt{ 1.3622^2+3.7609^2 } \doteq 4
k2=1/(1x02)=1/(11.3622)2.7609 q2=1k2=1(2.7609)3.76093.761 m l2=q22+x022=3.76092+1.36222=4k_{2}=1/(1-x_{02})=1/(1-1.3622) \doteq -2.7609 \ \\ q_{2}=1-k_{2}=1-(-2.7609) \doteq 3.7609 \doteq 3.761 \ \text{m} \ \\ l_{2}=\sqrt{ q_{2}^2+x_{02}^2 }=\sqrt{ 3.7609^2+1.3622^2 }=4



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