# What is the slope of the tangent line of r=2sin(theta/4)*cos(theta/2) at theta=(9pi)/2?

##### 1 Answer
Oct 24, 2016

The slope is $\approx - 0.1$

#### Explanation:

From the reference

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr} \left(\theta\right)}{d \theta} \sin \left(\theta\right) + r \left(\theta\right) \cos \left(\theta\right)}{\frac{\mathrm{dr} \left(\theta\right)}{d \theta} \cos \left(\theta\right) - r \left(\theta\right) \sin \left(\theta\right)}$

I used WolframAlpha to compute the derivative

$\frac{\mathrm{dr} \left(\theta\right)}{d \theta} = \frac{1}{4} \left(3 \cos \left(\frac{3 \theta}{4}\right) - \cos \left(\frac{\theta}{4}\right)\right)$

I used an Excel spreadsheet

I entered the following into cell J1:

=PI()*9/2

I used cell J2 to evaluate $\frac{\mathrm{dr} \left(\frac{9 \pi}{2}\right)}{d \theta}$ by entering the following:

=1/4(3COS(3*J1/4) - COS(J1/4))

I used cell J3 to evaluate $r \left(\frac{9 \pi}{2}\right)$ by entering the following:

=2SIN(J1/4)COS(J1/2)

I used J4 to evaluate the dy/dx and give us the slope, m by entering the following:

=(J2SIN(J1) + J3COS(J1))/(J2COS(J1) - J3SIN(J1))

The value that was returned had 5 decimal places but the second digit was 0 so I rounded to 1 place:

$m \approx - 0.1$