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# Tutor profile: Juliano R.

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Juliano R.
Math and Physics tutor for 5 years
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## Questions

### Subject:Calculus

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Question:

A cylinder-shaped package with height h and radius r is made using a fixed surface area of $$\sigma=54\pi.$$ At what radius do we maximize the volume?

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Juliano R.

A cylinder has a total surface area composed by the base, the top and the lateral surface. This gives us the following formula: $$\sigma=A_B+A_T+A_L=2\times(\pi r^2)+2\pi r h.$$ Since the surface area is fixed and equal to $$\sigma=54\pi$$, then we have h as a function of r: $$2\pi r^2+2\pi r h=\sigma\implies h(r)=\frac{\sigma}{2\pi r}-\frac{2\pi r^2}{2\pi r}=\frac{\sigma}{2\pi r}-r.$$ Now, the volume of the cylinder is given by: $$V(r)=\pi r^2 h=\pi r^2 \left(\frac{\sigma}{2\pi r}-r\right)=\frac{\sigma r}{2}-\pi r^3.$$ To find the local maxima and/or minima of a function, we find its critical points. In order to do that, we calculate the derivative of V(r) with respect to r and set it to zero.: $$\frac{dV(r)}{dr}=\frac{\sigma}{2}-3\pi r^2=0\implies r=\sqrt{\frac{\sigma}{6\pi}}=\sqrt{\frac{54\pi}{6\pi}}=\sqrt{9}=3.$$ Now that we have the critical point $$r=3$$, we use the 2nd derivative test in order to check if the critical point is a local maxima or minima: $$\left.\frac{d^2V(r)}{dr^2}\right\rvert_{r = 3}=-\left.6\pi r\right\rvert_{r = 3}=-18\pi.$$ Since the 2nd derivative is negative, the critical point $$r=3$$ is a local maximum.

### Subject:Physics

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Question:

Black holes are wondrous and terrifying astrophysical objects that have extreme gravitational pull in their surroundings. A common misconception about them is that they are like vacuum cleaners in space, arbitrarily pulling every object inward, while they are simply very massive objects with very large gravitational pull. In order to demonstrate this concept: a)Knowing that even light can't escape the inside of the black hole, Use the escape velocity formula and assume it to be the speed of light, c, to derive the formula for the radius of the black hole, using Newtonian Physics. This is called the Schwarzschild radius. b)Knowing that, in SI, $$G\approx6.67\times 10^{-11}\,m^3\,kg^{-1}\,s^{-2}$$, $$c\approx3\times 10^8 m\,s^{-1}$$ and that the Sun's mass is $$M\approx 2\times10^{30} kg$$, calculate the Schwarzschild radius of a black hole with the Sun's mass. c)Suppose we instantly replaced the Sun with the black hole calculated in (b). How long would it take for us to notice at Earth? Would we be able to notice any difference in gravitation? Assume an Earth-sun distance of $$d\approx 1.5\times10^{11}m.$$

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Juliano R.

a)We start by using the formula of the escape velocity: $$v_e=\sqrt{\frac{2 G M}{r}},$$ which we can rearrange by taking the square on both sides of the equation: $$v_e^2=\frac{2 G M}{r}\implies r=\frac{2 G M}{v_e^2}.$$ Now, assuming that the escape velocity is the speed of light, c, we have: $$r_s=\frac{2 G M}{c^2},$$ which is the Schwarzschild radius. b)Applying the numbers to formula obtained in (a), we get: $$r_s=\frac{\left(2\times6.67\times10^{-11}\times2\times10^{30} \right)}{(3\times10^8)^2}=\left(\frac{2\times2\times6.67}{9}\right)\times 10^{-11+30-16}\approx 2.96\times10^3m=2.96km$$ Just for comparison, the Sun has a radius of approximately $$6.95 \times 10^8m$$. c)Since the black hole does not emit light like our Sun does, we would see the last light emitted from the Sun when it reached us, and the time it would take for the light to travel from the Sun to Earth would be approximately: $$t\approx\frac{d}{c}=\frac{1.5\times10^{11}}{3\times10^8}=500s=8min20s$$. We would not be able to notice the difference in gravitation, since the black hole would have the same mass as the Sun, so the Earth would continue to orbit this new black hole, instead of the Sun.

### Subject:Differential Equations

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Question:

One of the simplest and most seen ordinary differential equation(ODE) in Physics is the simple harmonic oscillator. A mass $$m$$ is subject to a force done by a spring, $$F(t)= -kx(t)$$, where $$k$$ is the spring's constant and $$x$$ is the displacement from equilibrium. From Newton's Second Law, we have that $$F=m\frac{d^2x(t)}{dt^2}=m\ddot{x}$$, therefore we have an ordinary second-order differential equation: $(m\frac{d^2x(t)}{dt^2}=-kx(t).$)a)Find the general solution of this ODE. b)Solve the following initial value problem(IVP): $(x(0)=x_0,\;\;\;\dot{x}(0)=v_0$)

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Juliano R.

a)In order to solve this 2nd-order ODE, we start by assuming that the solution takes the form of $$x(t)=e^{rt}$$, and therefore we have $$\dot{x}(t)=re^{rt}, \ddot{x}=r^2e^{rt}.$$ Then we obtain the characteristic equation: $$mr^2e^{rt}+ke^{rt}=0\implies e^{rt}(mr^2+k)=0$$ $$\implies mr^2+k=0 \implies r=\pm \sqrt{-\frac{k}{m}}.$$ From this second degree equation, we obtain two complex solutions: $$r_1=+ i\sqrt{\frac{k}{m}}, \;r_2=-i\sqrt{\frac{k}{m}}.$$ For simplicity, let's call $$\omega=\sqrt{\frac{k}{m}}.$$ So the general solution is $$x(t)=c_1e^{r_1t}+c_2e^{r_2t}=c_1e^{i\omega t}+c_2e^{-i\omega t}.$$ Now, in order to put solution in another from, we make use of Euler's Formula: $(e^{\pm i\theta}=\cos{\theta}\pm i\sin{\theta},$)which we can invert in order to obtain sin and cos in terms of complex exponentials: $$\cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2},\,\,\,\sin{\theta}=\frac{e^{i\theta}-e^{-i\theta}}{2i}.$$ so we can rewrite the general solution in the following form: $(x(t)=k_1\cos{\omega t}+k_2\sin{\omega t}.$) b)Using the general solution we have: $$x(0)=x_0\implies k_1\cos{0}+k_2\sin{0}=x_0\implies k_1=x_0.$$ Now the second part of the IVP requires us to calculate $$\dot{x}(t)$$: $$\dot{x}(t)=-k_1\omega \sin{\omega t} + k_2\omega\cos{\omega t},$$ so $$\dot{x}(0)=v_0\implies -k_1\omega \sin{0} + k_2\omega\cos{0}=v_0\implies k_2=\frac{v_0}{\omega}.$$ And the solution is:$(x(t)=x_0\cos{\omega t}+\frac{v_0}{\omega}\sin{\omega t}.$)

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