Find the points on the graph of the polar equation where the tangent line is horizontal or vertical. (Assume $0 \leq \theta \leq 2\pi$.) $r = e^\theta$

horizontal tangent

• smaller $\theta$-value $\quad (r, \theta) = \boxed{?}$
• larger $\theta$-value $\quad (r, \theta) = \boxed{?}$

vertical tangent

• smaller $\theta$-value $\quad (r, \theta) = \boxed{?}$
• larger $\theta$-value $\quad (r, \theta) = \boxed{?}$

Neetesh Kumar | January 3, 2025

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This is the solution to Math 1c
Assignment: 10.3 Question Number 29
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Step-by-step solution:

Step 1: Tangent line slope formula in polar coordinates

The slope of the tangent line is given by:

$\frac{dy}{dx} = \frac{r'\sin(\theta) + r\cos(\theta)}{r'\cos(\theta) - r\sin(\theta)}$

To find the points where the tangent line is horizontal or vertical:

1. Horizontal tangent:

   $\frac{dy}{dx} = 0$ occurs when the numerator equals zero:

   $r'\sin(\theta) + r\cos(\theta) = 0$

2. Vertical tangent:

   $\frac{dy}{dx}$ is undefined when the denominator equals zero:

   $r'\cos(\theta) - r\sin(\theta) = 0$

Step 2: Compute $r$ and $r'$ for $r = e^\theta$

For $r = e^\theta$, differentiate with respect to $\theta$:

$r' = \frac{dr}{d\theta} = e^\theta$

Substitute $r = e^\theta$ and $r' = e^\theta$ into the conditions for horizontal and vertical tangents.

Step 3: Solve for horizontal tangents

The condition for horizontal tangents is:

$r'\sin(\theta) + r\cos(\theta) = 0$

Substitute $r = e^\theta$ and $r' = e^\theta$:

$e^\theta\sin(\theta) + e^\theta\cos(\theta) = 0$

Factor out $e^\theta$:

$e^\theta(\sin(\theta) + \cos(\theta)) = 0$

Since $e^\theta > 0$, this simplifies to:

$\sin(\theta) + \cos(\theta) = 0$

Divide through by $\sqrt{2}$:

$\frac{\sin(\theta)}{\sqrt{2}} + \frac{\cos(\theta)}{\sqrt{2}} = 0$

Recognize this as:

$\cos\left(\theta - \frac{\pi}{4}\right) = 0$

Solve for $\theta$:

$\theta - \frac{\pi}{4} = \frac{\pi}{2}, \frac{3\pi}{2} \implies \theta = \frac{3\pi}{4}, \frac{7\pi}{4}$

For each $\theta$, compute $r = e^\theta$:

• At $\theta = \frac{3\pi}{4}$:

  $r = e^{\frac{3\pi}{4}}$

• At $\theta = \frac{7\pi}{4}$:

  $r = e^{\frac{7\pi}{4}}$

Horizontal tangent points:

$\boxed{\left(e^{\frac{3\pi}{4}}, \frac{3\pi}{4}\right)}, \boxed{\left(e^{\frac{7\pi}{4}}, \frac{7\pi}{4}\right)}$

Step 4: Solve for vertical tangents

The condition for vertical tangents is:

$r'\cos(\theta) - r\sin(\theta) = 0$

Substitute $r = e^\theta$ and $r' = e^\theta$:

$e^\theta\cos(\theta) - e^\theta\sin(\theta) = 0$

Factor out $e^\theta$:

$e^\theta(\cos(\theta) - \sin(\theta)) = 0$

Since $e^\theta > 0$, this simplifies to:

$\cos(\theta) - \sin(\theta) = 0$

Divide through by $\sqrt{2}$:

$\frac{\cos(\theta)}{\sqrt{2}} - \frac{\sin(\theta)}{\sqrt{2}} = 0$

Recognize this as:

$\sin\left(\theta - \frac{\pi}{4}\right) = 0$

Solve for $\theta$:

$\theta - \frac{\pi}{4} = 0, \pi \implies \theta = \frac{\pi}{4}, \frac{5\pi}{4}$

For each $\theta$, compute $r = e^\theta$:

• At $\theta = \frac{\pi}{4}$:

  $r = e^{\frac{\pi}{4}}$

• At $\theta = \frac{5\pi}{4}$:

  $r = e^{\frac{5\pi}{4}}$

Vertical tangent points:

$\boxed{\left(e^{\frac{\pi}{4}}, \frac{\pi}{4}\right)}, \boxed{\left(e^{\frac{5\pi}{4}}, \frac{5\pi}{4}\right)}$

Final Answer:

Horizontal tangents:

Smaller $\theta$-value: $\boxed{\left(e^{\frac{3\pi}{4}}, \frac{3\pi}{4}\right)}$

Larger $\theta$-value: $\boxed{\left(e^{\frac{7\pi}{4}}, \frac{7\pi}{4}\right)}$

Vertical tangents:

Smaller $\theta$-value: $\boxed{\left(e^{\frac{\pi}{4}}, \frac{\pi}{4}\right)}$

Larger $\theta$-value: $\boxed{\left(e^{\frac{5\pi}{4}}, \frac{5\pi}{4}\right)}$

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