find the slope of of the tangent line polar to the curve r=8-3sin(theta) at the point (8-3/sqrt*2, pi/4)

To find the slope of the tangent line to the polar curve \( r = 8 - 3\sin(\theta) \) at the point \( \left(8 - \frac{3}{\sqrt{2}}, \frac{\pi}{4}\right) \), we need to use the formula for the slope of a tangent line to a polar curve.

The formula for the slope of the tangent line to a polar curve \( r = f(\theta) \) is given by:

\[
\text{slope} = \frac{dr/d\theta \cdot \sin(\theta) + r \cdot \cos(\theta)}{dr/d\theta \cdot \cos(\theta) - r \cdot \sin(\theta)}
\]

First, we need to find \( dr/d\theta \).

Given \( r = 8 - 3\sin(\theta) \):

\[
\frac{dr}{d\theta} = -3\cos(\theta)
\]

Next, we evaluate \( r \) and \( dr/d\theta \) at \(\theta = \frac{\pi}{4}\).

\[
r \left( \frac{\pi}{4} \right) = 8 - 3\sin \left( \frac{\pi}{4} \right) = 8 - 3 \cdot \frac{1}{\sqrt{2}} = 8 - \frac{3\sqrt{2}}{2}
\]

\[
\frac{dr}{d\theta} \bigg|_{\theta = \frac{\pi}{4}} = -3\cos \left( \frac{\pi}{4} \right) = -3 \cdot \frac{1}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}
\]

Now, substitute these values into the slope formula:

\[
\text{slope} = \frac{-\frac{3\sqrt{2}}{2} \cdot \sin \left( \frac{\pi}{4} \right) + \left( 8 - \frac{3\sqrt{2}}{2} \right) \cdot \cos \left( \frac{\pi}{4} \right)}{-\frac{3\sqrt{2}}{2} \cdot \cos \left( \frac{\pi}{4} \right) - \left( 8 - \frac{3\sqrt{2}}{2} \right) \cdot \sin \left( \frac{\pi}{4} \right)}
\]

Simplifying:

\[
\sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}
\]

So:

\[
\text{slope} = \frac{-\frac{3\sqrt{2}}{2} \cdot \frac{1}{\sqrt{2}} + \left( 8 - \frac{3\sqrt{2}}{2} \right) \cdot \frac{1}{\sqrt{2}}}{-\frac{3\sqrt{2}}{2} \cdot \frac{1}{\sqrt{2}} - \left( 8 - \frac{3\sqrt{2}}{2} \right) \cdot \frac{1}{\sqrt{2}}}
\]

\[
\text{slope} = \frac{-\frac{3}{2} + \frac{8\sqrt{2} - 3}{2\sqrt{2}}}{-\frac{3}{2} - \frac{8\sqrt{2} - 3}{2\sqrt{2}}}
\]

\[
= \frac{-\frac{3}{2} + \frac{8 - \frac{3}{\sqrt{2}}}{\sqrt{2}}}{-\frac{3}{2} - \frac{8 - \frac{3}{\sqrt{2}}}{\sqrt{2}}}
\]

Simplify further:

\[
= \frac{-\frac{3}{2} + 4\sqrt{2} - \frac{3}{2}}{-\frac{3}{2} - 4\sqrt{2} + \frac{3}{2}}
\]

\[
= \frac{4\sqrt{2} - 3}{-4\sqrt{2} - 3}
\]

Finally, we have:

\[
\text{slope} = \frac{4\sqrt{2} - 3}{-4\sqrt{2} - 3} = -1
\]

Thus, the slope of the tangent line to the polar curve at the given point is \(-1\).

Would you like more details on any step? Here are some other questions you might have:

1. How do you convert polar coordinates to Cartesian coordinates?
2. Can you explain how to find the derivative of a polar function?
3. What is the geometric interpretation of the slope of a tangent line?
4. How do you find the equation of the tangent line at a specific point?
5. What are the common applications of polar coordinates in real-world problems?

**Tip:** Remember, in polar coordinates, the angle \(\theta\) is crucial in determining the position on the curve, so always be careful when evaluating trigonometric functions at specific angles.

Learn how to find the slope of the tangent line to the polar curve r=8-3sin(theta) at a specific point using trigonometric identities and calculus concepts. Detailed step-by-step solution provided.

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