$r=e^θ $ $($ Assume $0 ≤ θ ≤ 2π.)$

Apparently I keep getting this answer wrong. I dont know if i need to use $n $ in the answer or not...


As $r=\sqrt{x^2+y^2}$ and $\theta=\arctan \frac yx$

Differentiating $$\sqrt{x^2+y^2}=e^{\arctan \frac yx}$$ wrt $x,$

$$\frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}}$$ $$=e^{\arctan \frac yx}\cdot\frac1{1+\left(\frac yx\right)^2}\cdot\frac{\left(x\frac{dy}{dx}-y\right)}{x^2}$$ $$=\sqrt{x^2+y^2}\frac{\left(x\frac{dy}{dx}-y\right)}{x^2+y^2} \text { as } e^{\arctan \frac yx}=\sqrt{x^2+y^2}$$

$$\implies \frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}\frac{\left(x\frac{dy}{dx}-y\right)}{x^2+y^2}$$

$$\implies \frac{dy}{dx}=\frac{x+y}{x-y}$$

For horizontal tangent, $\frac{dy}{dx}=0\implies x+y=0\implies \theta=\arctan (-1)=n\pi-\frac{\pi}4$ where $n$ is any integer.

As $0\le \theta\le 2\pi,n=1,2$

Similarly for vertical tangent, $\frac{dx}{dy}=0\implies x-y=0\implies \theta=\arctan (1)=m\pi+\frac{\pi}4$ where $m$ is any integer.

As $0\le \theta\le 2\pi,m=0,1$


In Cartesian form $y=e^\theta\sin\theta$ and $x=e^\theta\cos\theta$

Now, $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\sin\theta +\cos\theta}{\cos\theta - \sin\theta }$$

Tangent is horizontal when numerator is $0$ which gives $\theta=7\pi/4,3\pi/4$

and Tangent is vertical when denominator is $0$ which gives $\theta=\pi/4, 5\pi/4$


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