Answer
There are horizontal tangents of that curve at $(1,e)$ and at $(1, \frac 1 e)$, and vertical tangents at $(e, 1)$ and at $(\frac 1 e, 1)$
Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{d(e^{cos(\theta)})}{dt} = -sin(\theta)e^{cos(\theta)}$
$\frac{d(e^{sin(\theta)})}{dt} = cos(\theta)e^{sin(\theta)}$
2. There will be a horizontal tangent when $-sin(\theta)e^{cos(\theta)}= 0$ and $ cos(\theta)e^{sin(\theta)} \neq 0$:
$-sin(\theta)e^{cos(\theta)} = 0$
$(-sin(\theta))(e^{cos(\theta)}) = 0$
$e^{cos(\theta)}$ will never equal $0$.
$-sin(\theta) = 0$
$sin(\theta) = 0$
So, $\theta$ must be equal to $0, \pi$or $2\pi$. (Values between $0$ and $2\pi$)
Check if $ cos(\theta)e^{sin(\theta)} \neq 0$
- The exponential part can never equal $0$, so we can just test $cos(\theta)$.
$cos(0) = 1$
$cos(\pi) = -1$
$cos(2\pi) = 1$
- Therefore, all these values for $\theta$ are valid.
3. There will be a vertical tangent when $cos(\theta)e^{sin(\theta)}= 0$ and $-sin(\theta)e^{cos(\theta)} \neq 0$:
$cos(\theta)e^{sin(\theta)}= 0$
$e^{sin(\theta)}$ will never equal $0$.
$cos(\theta) = 0$
So, $\theta$ must be equal to $\frac {\pi} 2$ or $\frac{3\pi}{2}$
- Check if $-sin(\theta)e^{cos(\theta)} \neq 0$:
The exponential part will can never equal 0, so we can just test $-sin(\theta)$:
$-sin(\frac{\pi} 2) = -1$
$-sin(\frac{3\pi}{2}) = 1$
- Therefore, all these values for $\theta$ are valids.
4. Find the points coordinates:
Horizontal tangents:
$\theta = 0$
$x = e^{sin(0)} = e^0 = 1$
$y = e^{cos(0)} =e^1 =e$
$\theta = \pi$
$x = e^{sin(\pi)} = e^0 = 1$
$y = e^{cos(\pi)} =e^{-1} = \frac{1}{e}$
$\theta = 2\pi$
$x = e^{sin(2\pi)} = e^0 = 1$
$y = e^{cos(2\pi)} =e^1 =e$
Vertical tangents:
$\theta = \frac{\pi} 2$
$x = e^{sin(\frac{\pi} 2)} = e^1 = e$
$y = e^{cos(\frac{\pi} 2)} =e^0 =1$
$\theta = \frac{3\pi} 2$
$x = e^{sin(\frac{3\pi} 2)} = e^{-1} = \frac 1 e$
$y = e^{cos(\frac{3\pi} 2)} =e^0 =1$
