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Henri F.

Tutor of 8 years & PhD Candidate in Spacecraft Control Engineering

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Calculus

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Question:

Find the area of the region $$R$$, bounded by the curve $$y=x^{-\dfrac{2}{3}}$$, the line $$x=1$$ and the x axis. In addition, find the volume of revolution of this region when rotated $$2\pi$$ radians about the x axis.

Henri F.

Answer:

$(R = \int_1^\infty x^{-\frac{2}{3}}dx=\left[3x^{\frac{1}{3}}\right]_1^\infty=\infty$) The volume of revolution can be computed via: $(V = \pi\int_a^b y^2dx$) Therefore: $(V = \pi\int_1^\infty x^{-\frac{4}{3}}dx=\pi\left[-3x^{-\frac{1}{3}}\right]_1^\infty=3$) This interesting problem shows that a region can have infinite area but its solid revolution can have finite volume.

Aerospace Engineering

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Question:

Find the poles and zeros of the transfer function: $(\frac{s+2}{(s+1)(s^2-0.01s+0.01)}$) and state whether or not the system is stable.

Henri F.

Answer:

Zeros are the roots of the numerator polynomial and poles are the roots of the denominator polynomial. Zero at $$s = -2$$ Pole at $$s = -1$$ Other poles at roots of $$s^2-0.01s+0.01=0$$ $(s = \frac{0.01\pm \sqrt{0.0399}i}{2}$) As at least one of the poles has positive real part, the system will grow over time and hence is unstable.

Differential Equations

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Question:

Find a function, $$\theta(t)$$, that describes the time evolution of a simple, undamped, smoothly hinged, pendulum of arm length $$l$$ and bob-mass $$m$$ in a uniform gravitational field of strength $$g$$ given the pendulum starts from rest at an angle $$\theta_0=\dfrac{\pi}{12}$$. $$\theta(t)$$ represents the angular displacement (in radians) of the pendulum from the downward vertical.

Henri F.

Answer:

Considering the physical system above is fairly simple a forced-based "Newton-style" method could be used to describe the system's time-evolution. Taking moments about the hinge: $( \vec{L} = I\ddot{\theta}$) $(-mgl\sin(\theta)=ml^2\ddot{\theta}$) $(\ddot{\theta}+\dfrac{g}{l}\sin(\theta)=0$) As the system starts at a small angle around the stable equilibrium (pendulum vertically down) a small angle approximation can be used to simplify the equation, namey that for small angles $$\sin(x)\approx x$$. $(\ddot{\theta}+\dfrac{g}{l}\theta=0$) Which yields an Auxiliary Equation of: $(m^2+\dfrac{g}{l}=0\implies m = \pm\sqrt{\dfrac{g}{l}}i$) Thus the general solution is: $(\theta(t)=e^{0t}\left(A\cos\left(\sqrt{\dfrac{g}{l}}t\right)+B\sin\left(\sqrt{\dfrac{g}{l}}t\right)\right)$) $(\theta(t)=A\cos\left(\sqrt{\dfrac{g}{l}}t\right)+B\sin\left(\sqrt{\dfrac{g}{l}}t\right)$) Substituting our initial conditions of $$\theta(0)=\dfrac{\pi}{12}$$ and $$\dot{\theta}(0)=0$$ requires knowledge of $$\dot{\theta}(t)$$, therefore we need to differentiate: $(\dot{\theta}(t)=-A\dfrac{g}{l}\sin\left(\sqrt{\dfrac{g}{l}}t\right)+B\dfrac{g}{l}\cos\left(\sqrt{\dfrac{g}{l}}t\right)$) Subsituting the first condition: $(\theta(0)=A\cos\left(\sqrt{\dfrac{g}{l}}\times0\right)+B\sin\left(\sqrt{\dfrac{g}{l}}\times0\right) = \dfrac{\pi}{12}$) So $$A = \dfrac{\pi}{12}$$ Substituting the second condition: $(\dot{\theta}(0)=-A\dfrac{g}{l}\sin\left(\sqrt{\dfrac{g}{l}}\times0\right)+B\dfrac{g}{l}\cos\left(\sqrt{\dfrac{g}{l}}\times0\right)=0$) So $$B=0$$. Therefore the final time evolution is: $(\theta(t)=\dfrac{\pi}{12}\cos\left(\sqrt{\dfrac{g}{l}}t\right)$) This can be corroborated with simple standard results from Simple Harmonic Theory but the process described here can be used for any linear second order ODE system, even with far more complexity.

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