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Please help me to solve these 2 D.E

by yukcream
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yukcream
#1
Aug20-05, 06:33 AM
P: 59
Q1. Let u(x) be a function with the property that the area under the curve between any two points, a, b with a<b, is directly proportional to the different of the functional values at a and b. Obtain a differential equation for u(x).

Q2. A parabolic reflector has the property that a light source placed at its focus produces a parallel beam, or, conversely, parallel rays converge at the focus. Assuming that reflection of light from a curve is determined by the usual laws of reflection for the tangent to the curve at the point of incidence (angle of incidence equals angle of reflection), use the above property to determine a differential equation for the parabola.
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HallsofIvy
#2
Aug20-05, 10:40 AM
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Quote Quote by yukcream
Q1. Let u(x) be a function with the property that the area under the curve between any two points, a, b with a<b, is directly proportional to the different of the functional values at a and b. Obtain a differential equation for u(x).
I assume that the "curve between any two points, a, b with a< b" is the graph of y= u(x). It would have been a good idea to say that.
In order to get a differential equation, fix a and take b= x as variable.
The "area under the curve" is now [itex]\int_a^x u(t)dt[/itex] and that must be equal to u(x)- u(a): [itex]\int_a^x u(t)dt= u(x)- u(a)[/itex]. What do you get if you differentiate both sides of that equation?

Q2. A parabolic reflector has the property that a light source placed at its focus produces a parallel beam, or, conversely, parallel rays converge at the focus. Assuming that reflection of light from a curve is determined by the usual laws of reflection for the tangent to the curve at the point of incidence (angle of incidence equals angle of reflection), use the above property to determine a differential equation for the parabola.
Draw a graph, taking the axis of the parabola to be the x-axis. That way all the "parallel rays" are "y= constant". Let the focus be (a, 0). The slope of the ray from any point (x,y) on the parabola to (a, 0) is, of course, y/(x-a). Use a little geometry to show that the angle that ray makes with the x-axis is exactly twice the angle the incoming ray makes with the tangent line to the parabola. With a simple trig formula (for tan(2&theta;)), show that, with m= slope of tangent line,[itex]\frac{y}{x-a}= \frac{2m}{1+m^2}[/itex].
Of course, m is [itex]\frac{dy}{dx}[/itex].
yukcream
#3
Aug21-05, 01:52 AM
P: 59
thx for ur help!


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