Question #dbcb0

1 Answer
May 7, 2015

The answer is f(theta)=e^(theta) (and f(theta)=e^(-theta), depending on how you decide to measure the angle, see below).

You can think about this in terms of dot products of appropriate vectors in rectangular coordinates. Let r=f(theta) be the unknown polar function you want to find. Then x=f(theta)cos(theta) and y=f(theta)sin(theta). As theta varies, this corresponds to a varying position vector for the point P that is vec(r)=f(theta)cos(theta)hat(i)+f(theta)sin(theta)hat(j).

The velocity vector vec(v)=vec(r)' is tangent to the resulting parametric curve (where the angular coordinate theta is the parameter). Therefore, the dot product vec(r)\cdot vec(v)=|| vec(r) || || vec(v) || cos(alpha), where you want alpha to be the constant angle between vec(r) and vec(v): alpha=pi/4=45^{\circ} so that cos(alpha)=1/sqrt(2) and vec(r)\cdot vec(v)=1/sqrt(2)|| vec(r) || || vec(v) ||.

Now, by the Product Rule,

vec(r)\cdot vec(v)=(f(theta)cos(theta)hat(i)+f(theta)sin(theta)hat(j))\cdot ((f'(theta)cos(theta)-f(theta)sin(theta))hat(i)+(f'(theta)sin(theta)+f(theta)cos(theta))hat(j)).

Through expansion of this dot product and the use of the trigonometric identity cos^{2}(theta)+sin^{2}(theta)=1, this reduces to vec(r)\cdot vec(v)=f(theta)f'(theta). Moreover, the same standard trigonometric identity can be used to derive the fact that, assuming f(theta)\geq 0, we have || vec(r) || || vec(v) ||=f(theta)\sqrt((f(theta))^2+(f'(theta))^2). (Check these claims!)

The condition vec(r)\cdot vec(v)=1/sqrt(2)|| vec(r) || || vec(v) || thus reduces to, after cancelation of f(theta) (assuming f(theta)>0), to f'(theta)=1/sqrt(2)\sqrt((f(theta))^2+(f'(theta))^2). Squaring both sides, rearranging, and multiplying both sides by 2 ultimately gives (f'(theta))^2=(f(theta))^2. This is really a differential equation for f(theta), which in Leibniz notation would be written as (\frac{dy}{d\theta})^2=y^2. Certainly y=f(theta)=e^{theta) is one solution of this equation that satisfies y(0)=1. The function y=f(theta)=e^{-theta} also satisfies the given initial value problem.

Is it possible one of these solutions is extraneous (not a solution of the original problem)? It's certainly possible since we've squared both sides of an equation in our work.

The answer, in a sense, depends on how you measure the angle alpha. If it is measured as the angle between the vectors vec(r) and vec(v) in the standard way (put their base points at the same location), then the second answer turns out to be extraneous.

I'll assume the desired angle is the angle between the two vectors. Based on this assumption, the unique answer is f(theta)=e^{theta}.