- #1
yoyo
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Recall that every point (x, y) in the plane is described by its radius-
vector r = xi + yj. A planar curve [gamma] has the following geometric property: at every point on the curve the radius vector and the tangent intersect at a fixed angle [alpha].
1. Derive a first order differential equation dy/dx = f(x, y, [alpha]) describing [gamma]
.
Note that the right-hand side should depend on the parameter [alpha].
For part 1, I drew a circle(not sure if it's suppose to be a circle) on the (x, y) plane. Then i drew a radial vector from the origin to any point. i let the radial vector make an angle [phi] with x-axis. extended the tangent all the way to the x-axis and let the tangent make an angle theta with the x axis.
then i got
tan [phi]= y/x...(1)
tan [theta]= dy/dx...(2)
so [alpha]= [theta] - [phi]
tan [alpha]= tan ([theta] - [phi]) = (tan [theta] - tan [phi])/(1+tan[theta]*tan[phi])...(3)
dont know how to derive the differential from here...and not sure about the rest of the problem...please help...!
2. Solve the equation from Part 1 for the case when [alpha] = pi/2 assuming that [gamma] passes through (1, 0). For which curve is the radius vector always perpendicular to the tangent?
I think the answer is a circle but not sure how to solve it?/
3. Assume that [alpha] = 3pi/4 and the curve passes through (1, 0). Apply Euler's method with step h = 0.5 to trace out the curve for this case.
is this step function? if so how do i go about solving this?
4. Solve the equation from Part 1 for [alpha]= 3pi/4 with (1, 0) as the initial condition. Superimpose the exact solution from this part with the numerical
solution from Part 3. Do you know the name of this curve?
I think in order to plot the curve, i have to switch to polar coordinates
x = r cos [theta]; y = r sin [theta]:
help?
vector r = xi + yj. A planar curve [gamma] has the following geometric property: at every point on the curve the radius vector and the tangent intersect at a fixed angle [alpha].
1. Derive a first order differential equation dy/dx = f(x, y, [alpha]) describing [gamma]
.
Note that the right-hand side should depend on the parameter [alpha].
For part 1, I drew a circle(not sure if it's suppose to be a circle) on the (x, y) plane. Then i drew a radial vector from the origin to any point. i let the radial vector make an angle [phi] with x-axis. extended the tangent all the way to the x-axis and let the tangent make an angle theta with the x axis.
then i got
tan [phi]= y/x...(1)
tan [theta]= dy/dx...(2)
so [alpha]= [theta] - [phi]
tan [alpha]= tan ([theta] - [phi]) = (tan [theta] - tan [phi])/(1+tan[theta]*tan[phi])...(3)
dont know how to derive the differential from here...and not sure about the rest of the problem...please help...!
2. Solve the equation from Part 1 for the case when [alpha] = pi/2 assuming that [gamma] passes through (1, 0). For which curve is the radius vector always perpendicular to the tangent?
I think the answer is a circle but not sure how to solve it?/
3. Assume that [alpha] = 3pi/4 and the curve passes through (1, 0). Apply Euler's method with step h = 0.5 to trace out the curve for this case.
is this step function? if so how do i go about solving this?
4. Solve the equation from Part 1 for [alpha]= 3pi/4 with (1, 0) as the initial condition. Superimpose the exact solution from this part with the numerical
solution from Part 3. Do you know the name of this curve?
I think in order to plot the curve, i have to switch to polar coordinates
x = r cos [theta]; y = r sin [theta]:
help?