STEP Project — 2017.3 Question 5

Since we have $x = r cos 𝜃$ and $y = r sin 𝜃$ , and $r = f (𝜃)$ , we have

\begin{array}{l} \frac{d x}{d 𝜃} & = \frac{d r}{d 𝜃} \cdot cos 𝜃 + r \cdot \frac{d cos 𝜃}{d 𝜃} \\ = f^{'} (𝜃) cos 𝜃 - f (𝜃) sin 𝜃, \end{array}

and

\begin{array}{l} \frac{d y}{d 𝜃} & = \frac{d r}{d 𝜃} \cdot sin 𝜃 + r \cdot \frac{d sin 𝜃}{d 𝜃} \\ = f^{'} (𝜃) sin 𝜃 + f (𝜃) cos 𝜃, \end{array}

Therefore,

\begin{array}{l} \frac{d y}{d x} & = \frac{\frac{d y}{d 𝜃}}{\frac{d x}{d 𝜃}} \\ = \frac{f^{'} (𝜃) sin 𝜃 + f (𝜃) cos 𝜃}{f^{'} (𝜃) cos 𝜃 - f (𝜃) sin 𝜃} \\ = \frac{f^{'} (𝜃) tan 𝜃 + f (𝜃)}{f^{'} (𝜃) - f (𝜃) tan 𝜃} . \end{array}

For the two curves, we must have

\frac{d y}{d x} | f \cdot \frac{d y}{d x} | g = - 1

for them to meet at right angles. Therefore,

\begin{array}{l} \frac{f^{'} (𝜃) tan 𝜃 + f (𝜃)}{f^{'} (𝜃) - f (𝜃) tan 𝜃} \cdot \frac{g^{'} (𝜃) tan 𝜃 + g (𝜃)}{g^{'} (𝜃) - g (𝜃) tan 𝜃} & = - 1 \\ (f^{'} (𝜃) tan 𝜃 + f (𝜃)) \cdot (g^{'} (𝜃) tan 𝜃 + g (𝜃)) & = - (f^{'} (𝜃) - f (𝜃) tan 𝜃) \cdot (g^{'} (𝜃) - g (𝜃) tan 𝜃) \\ f^{'} (𝜃) g^{'} (𝜃) (1 + {tan}^{2} 𝜃) + f (𝜃) g (𝜃) (1 + {tan}^{2} 𝜃) & = 0 \\ f^{'} (𝜃) g^{'} (𝜃) + f (𝜃) g (𝜃) & = 0 . \end{array}

We have $f (- \frac{π}{2}) = 4$ . Let

g_{a} (𝜃) = a (1 + sin 𝜃) .

Therefore,

g_{a}^{'} (𝜃) = a cos 𝜃,

and we have

f^{'} (𝜃) (a cos 𝜃) + f (𝜃) a (1 + sin 𝜃) = 0,

and therefore

\frac{d f (𝜃)}{d 𝜃} cos 𝜃 = - f (𝜃) (1 + sin 𝜃) .

By separating variables we have

\frac{d f (𝜃)}{f (𝜃)} = - \frac{d 𝜃 (1 + sin 𝜃)}{cos 𝜃} .

Notice that

- \frac{1 + sin 𝜃}{cos 𝜃} = - \frac{(1 - sin 𝜃) (1 + sin 𝜃)}{(1 - sin 𝜃) cos 𝜃} = - \frac{cos 𝜃}{1 - sin 𝜃} = \frac{cos 𝜃}{sin 𝜃 - 1},

integrating both sides gives us

ln f (𝜃) = ln | sin 𝜃 - 1 | + C = ln (1 - sin 𝜃) + C,

which gives

f (𝜃) = A (1 - sin 𝜃) .

Since $f (- \frac{π}{2}) = 4$ , we have $2 A = 4$ and $A = 2$ , therefore $f (𝜃) = 2 (1 - sin 𝜃)$ .

xyrrr = = = 142+((11s+−inss𝜃inin 𝜃𝜃))