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???