DEpartment of physics

In the last lecture, we discussed the history of Calculus of variation and proved the Euler's equations. We also use the Euler's equation to prove that the shortest distance between two points on a plane is a straight line. in this lecture, we will do the Brachistochrone Problem, (Shortest Time).  Also we will use the Euler's Equations to find the shape of a wire, by revolving which give the surface area. We do this problem in two cooordinate systems. We will discuss how crucial it is to choose the coordinate system. We also discuss the importance of putting the Euler's Equation in the second form.

In the last lecture, we discussed how to find the conditions for extremizing an integral quantity, such as the distance between two points, time interval etc. The solutions for extremizing the quantity $$J=\int_{x_1}^{x_2} f\left{y(x),y'(x);x \right} dx $$ as $$ \frac{d}{dx} \frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y} =0 $$ This is called the Euler equation. We applied the Euler's equation to show that the shortest path between two points is a straight line. EXAMPLE: BRACHISTOCHRONE PROBLEM: Bachistro mean short and chrone means time in Greek. That means, here we are trying to find the shortest time between two incidents. We may be looking for the quickest path a car will be moving from point 1 to point 2, so that we can build a smooth roler coaster along that path. Let;s try to get an equation for time, which needs to be minimized. $$ t= \int_{x_1}^{x_2} \frac{ds}{v}$$ where ds is the elemental length and v is the instantaneous velocity within the length segment ds. Now we can write an equation for v considering the conservation of energy. We are basically designing a surface, such that the time of travel is the shortest. (OK, in the free fall, some other quantity will be extremized, not the time). But no matter what, as long as the surface is smooth, we can use the conservation of energy. $$\frac{1}{2} mv^2 = mgy $$ $$ v= \sqrt{2 g y } $$ We know from the previous problem that ds can be written as $$ds=\sqrt{dx^2 + dy^2}$$ Let's write time: $$t= \int_{x_1}^{x_2} \frac{ds}{v}$$ $$ t= \int_{x_1}^{x_2} \frac{\sqrt{dx^2 + dy^2}}{\sqrt{2g y}}$$ OK, here I am going to use a trick, which is different from what we have done earlier. Earlier, I have used x as the independent variable. Now I am going to use y as my independent variable. Just because y appears in my integral quantity. Don't be panic of doing random things in different calculations. We will be familliar with the systems. Also for the purpose of exams, I will guide you if such decision is needed. Anyway, that means we take x is a function of y, Not that y is a function of x. Now I need to write the time as as integral quantity. $$ t= \int_{x_1}^{x_2} \frac{\sqrt{\frac{dx}{dy}^2 + y}}{\sqrt{2g y}} dy $$ That is : $$t=\int_{x_1}^{x_2} \sqrt{\frac{1+x'(y)^2}{2gy}} $$ Now, we are ready to apply Euler Equation. But remember that, we made y as our independent variable. So the Euler's equation read as: $$\frac{\partial f}{\partial x}-\frac{d}{dx}\frac{\partial f}{\partial x'} =0 $$ Our function f is: $$ \sqrt{\frac{1+x'(y)^2}{2gy}} $$ We immediately notice: $$ \frac{\partial f}{\partial x}=0$$ (Well... That is the reason we used y as the independent variable. Now we get: $$\frac{d}{dx}\frac{\partial f}{\partial x'} =0 $$ which means, $$\frac{\partial f}{\partial x'}=\text{constant}=b$$ $$\frac{\partial f}{\partial x'}=\text{constant}=\frac{1}{\sqrt{y}}\frac{1}{\sqrt{1+x'^2}}2x'=b$$ $$\frac{x'^2}{y(1+x'^2)}=\text{Constant=\frac{1}{2a} $$ which gives, $$x'=\sqrt{\frac{y}{2a-y}}$$ Now in order to find x(y), we need to integrate this equation: $$\frac{dx}{dy}=\sqrt{\frac{y}{2a-y}}$$ $$ x=\int \sqrt{\frac{y}{2a-y}}dy $$ Let's substitute: $$y=a(1-\text{Cos} \theta )$$ which can also be written as: $$y= a 2 Sin^2 \frac{\theta}{2} $$ $$dy = 2 a Sin \frac{\theta}{2} Cos \frac{\theta}{2} $$ $$dy = 2 a Sin \frac{\theta}{2} Cos \frac{\theta}{2} $$ Let's substitute those in the above equation: $$x=\int \sqrt{\frac{2a Sin^2 \frac{\theta}{2} }{2a - 2a Sin^2 \frac{\theta}{2}} }2 a Sin \frac{\theta}{2} Cos \frac{\theta}{2} d\theta $$ which gives $$x=\int 2a Sin^2 \frac{\theta}{2} d\theta $$ $$x=\int a(1-Cos\theta) d\theta $$ which finally gives: $$x=a (\theta - Sin theta) + Constant $$ EXAMPLE: SURFACE GENERATED BY REVOLVING A LINE CONNECTING TWO POINTS: Consider a surface generated by revolving a line connecting two fixed points (x1,y1) about an axis coplanar with the two points. Find the equation of the line connecting the points such that the surface area generated by the revolution (i.e. the area of surface evolution) us a minimum. Now let's assume that the line connecting (x1,y1) and (x2,y2) passing around the y-axis, and coplanar with the two points. Let's consider an elemental area, which is formed by rotating a ds area around the y axis. $$dA=2 \pi x ds = 2 \pi x \sqrt{dx^2 + dy^2} $$ $$dA=2 \pi x \sqrt{1+y'^2} dx $$ Now by integrating this quantity, we can achieve the total area. $$A=\int_{x_1}^{x_2} 2 \pi x \sqrt{1+ y'^2} dx $$ In order to find the extremum area, we use the Euler Equation: $$f=2 \pi x \sqrt{1+y'^2} $$ Using the Euler Equation: $$\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}=0 $$ Here : $$ \frac{\partial f}{\partial y}=0 $$ $$ \frac{\partial f}{\partial y'}=2 \pi x \frac{1}{2}\frac{2y'}{\sqrt{1+y'^2}} $$ Then we get: $$\frac{xy'}{\sqrt{1+y'^2}}=\text{Constant}=c $$ $$y'=\frac{c}{\sqrt{x^2=c^2}} $$ $$dy=\frac{c}{\sqrt{x^2=c^2}} dx $$ By integrating, we can find the y(x), basically the shape of a wire that revolves to give the minimum surface area of revolution. HOMEWORK PROBLEM: Fermat's principle states that light travels in the quickest path. That means, if light travels in a given medium, it will travel with a constant velocity, thus takes a stright line. Because shortest distance means the quickest path. Now, consider a light beam passing from one medium to another. Refractive index of the first medium is n1 and the second is n2. Use the Fermat principle to derive the law if refraction: $$n_1 \text{Sin} \theta _1 = n_2 \text{Sin} \theta _2 $$

Notes can be found here

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