What is the slope of the tangent line of r=-2sin(2theta)+cos(3theta) at theta=(-11pi)/8?

Mar 11, 2018

$\frac{3 \sqrt{2 - \sqrt{2}} - 4 \sqrt{2}}{2}$

Explanation:

To find the slope of a tangent line, take the derivative of the function. Here, $r = f \left(\theta\right) = \cos \left(3 \theta\right) - 2 \sin \left(2 \theta\right)$

We must compute $\frac{\mathrm{dr}}{d \theta}$.

$\frac{d}{d \theta} \left(\cos \left(3 \theta\right) - 2 \sin \left(2 \theta\right)\right)$

Since $\left(f \pm g\right) ' = f ' \pm g '$, we can write:

$\frac{d}{d \theta} \cos \left(3 \theta\right) - 2 \frac{d}{d \theta} \sin \left(2 \theta\right)$

According to the chain rule, (f(g(x))'=f'(g(x))*g'(x)

First, we compute $\frac{d}{d \theta} \cos \left(3 \theta\right)$:

Here, $f \left(\eta\right) = \cos \eta$ where $\eta = 3 \theta$

$\frac{d}{d \eta} \cos \eta \cdot \frac{d}{d \theta} 3 \theta$

$- \sin \eta \cdot 3$

Since $\eta = 3 \theta$:

$- 3 \sin \left(3 \theta\right)$

Next, we compute $\frac{d}{d \theta} \sin \left(2 \theta\right)$:

Here, $f \left(\eta\right) = \sin \eta$ where $\eta = 2 \theta$

$\frac{d}{d \eta} \sin \eta \cdot \frac{d}{d \theta} 2 \theta$

$\cos \eta \cdot 2$

Since $\eta = 2 \theta$, we write:

$2 \cos \left(2 \theta\right)$

So we have:

$\frac{d}{d \theta} \left(\cos \left(3 \theta\right) - 2 \sin \left(2 \theta\right)\right) = - 3 \sin \left(3 \theta\right) + 2 \left(2 \cos \left(3 \theta\right)\right)$

$= 4 \cos \left(2 \theta\right) - 3 \sin \left(3 \theta\right)$

This is the general formula for the slope of the tangent line for $r$. However. we are asked to find the slope at $\theta = - \frac{11 \pi}{8}$. Inputting:

$4 \cos \left(2 \left(- \frac{11 \pi}{8}\right)\right) - 3 \sin \left(3 \left(- \frac{11 \pi}{8}\right)\right)$

$4 \cos \left(- \frac{11 \pi}{4}\right) - 3 \sin \left(- \frac{33 \pi}{8}\right)$

Since $\sin \left(- x\right) = - \sin \left(x\right)$ and $\cos \left(- x\right) = \cos \left(x\right)$, we write:

$4 \cos \left(\frac{11 \pi}{4}\right) + 3 \sin \left(\frac{33 \pi}{8}\right)$

Since $\cos \left(\frac{11 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$ and $\sin \left(\frac{33 \pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$, we write:

$4 \left(- \frac{\sqrt{2}}{2}\right) + 3 \left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right)$

$- 2 \sqrt{2} + \frac{3 \sqrt{2 - \sqrt{2}}}{2}$

$\frac{3 \sqrt{2 - \sqrt{2}} - 4 \sqrt{2}}{2}$

In crude decimals, that is, to four decimal places, $- 1.6804$.

Graphing the function:
graph{cos(3x)-2sin(2x) [-7.017, 0.778, -0.654, 3.244]}

We see that this is true.