Related Rates: Ladder Problem but with a bureaucratic twist

This problem was inspired by my real life experience:

A 1-meter tall ladder leans against the z-axis. Suppose the base of the ladder follows a path on the xy-plane. The path in terms of x and y is given by the following equation: ${\displaystyle {\frac {\left(x-{\frac {1}{2}}\right)^{2}}{{\frac {3}{2}}-{\sqrt {2}}}}+{\frac {\left(y-{\frac {1}{2}}\right)^{2}}{{\frac {3}{2}}-{\sqrt {2}}}}=1}$. The angle at which the line connecting ${\displaystyle \left({\frac {1}{2}},{\frac {1}{2}}\right)=\left(0.5,0.5\right)}$ with the base of the ladder and to its horizontal radius is changing at 1 radian per second. Suppose it starts moving at ${\displaystyle t=0}$ and the angle is currently 0 radians. By the time ${\displaystyle t={\frac {3\pi }{4}}}$, what is the velocity of the top of the ladder at the point it intersects the z-axis?

Solution:

The equation describing the situation is ${\displaystyle x^{2}+y^{2}+z^{2}=1}$. where base of the ladder is ${\displaystyle \left(x,y\right)}$ and the top of the ladder is located at ${\displaystyle z}$. We then differentiate this equation wrt ${\displaystyle t}$.

This is a rewrite: ${\displaystyle x(t)^{2}+y(t)^{2}+z(t)^{2}=1}$

${\displaystyle 0=2x\left(t\right)x'\left(t\right)+2y\left(t\right)y'\left(t\right)+2z\left(t\right)z'\left(t\right)}$

${\displaystyle z\left(t\right)z'\left(t\right)=-x\left(t\right)x'\left(t\right)-y\left(t\right)y'\left(t\right)}$

Solving for ${\displaystyle z \left( t \right)}$ in the equation describing the situation, we now have:

${\displaystyle z'\left(t\right){\sqrt {1-x\left(t\right)^{2}-y\left(t\right)^{2}}}=-x\left(t\right)x'\left(t\right)-y\left(t\right)y'\left(t\right)}$

${\displaystyle z'\left(t\right)={\frac {\left(-x\left(t\right)x'\left(t\right)-y\left(t\right)y'\left(t\right)\right)}{\sqrt {1-x\left(t\right)^{2}-y\left(t\right)^{2}}}}}$

We now can define everything now:

${\displaystyle x\left(t\right)={\frac {1}{2}}+{\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\cos \left(t\right)}$, ${\displaystyle x\left({\frac {3\pi }{4}}\right)=1-{\frac {1}{\sqrt {2}}}}$

${\displaystyle x'\left(t\right)=-{\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\sin \left(t\right),x\left({\frac {3\pi }{4}}\right)=-{\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\sin \left({\frac {3\pi }{4}}\right)={\frac {1}{2}}-{\frac {1}{\sqrt {2}}}}$,

${\displaystyle y\left(t\right)={\frac {1}{2}}+{\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\sin \left(t\right)}$, ${\displaystyle y\left({\frac {3\pi }{4}}\right)={\frac {1}{2}}+{\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\sin \left({\frac {3\pi }{4}}\right)={\frac {1}{\sqrt {2}}}}$

${\displaystyle y'\left(t\right)={\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\cos \left(t\right),y\left({\frac {3\pi }{4}}\right)={\sqrt {{\frac {3}{2}}-{\sqrt {2}}}}\cos \left({\frac {3\pi }{4}}\right)={\frac {1}{\sqrt {2}}}-{\frac {1}{2}}}$

${\displaystyle z'(t)=-{\frac {\left(1-{\frac {1}{\sqrt {2}}}\right)\left({\frac {1}{2}}-{\frac {1}{\sqrt {2}}}\right)+\left({\frac {1}{\sqrt {2}}}\right)\left({\frac {1}{\sqrt {2}}}-{\frac {1}{2}}\right)}{\sqrt {1-\left(1-{\frac {1}{\sqrt {2}}}\right)^{2}-\left({\frac {1}{\sqrt {2}}}\right)^{2}}}}=-{\frac {\left(1-{\frac {1}{\sqrt {2}}}\right)\left({\frac {1}{2}}-{\frac {1}{\sqrt {2}}}\right)+\left({\frac {1}{\sqrt {2}}}\right)\left({\frac {1}{\sqrt {2}}}-{\frac {1}{2}}\right)}{\sqrt {-1+{\sqrt {2}}}}}=-{\frac {{\frac {3}{2}}-{\sqrt {2}}}{\sqrt {-1+{\sqrt {2}}}}}}$meters per second or ${\displaystyle {\frac {{\frac {3}{2}}-{\sqrt {2}}}{\sqrt {-1+{\sqrt {2}}}}}}$ meters per second in the negative z-axis direction.

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