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

The slope is

Tangents to a polar curve

From the reference

I used WolframAlpha to compute the derivative

I used an Excel spreadsheet

I entered the following into cell J1:

=PI()*9/2

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

=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:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the slope of the tangent line of the polar curve ( r = 2 \sin(\frac{\theta}{4}) \cos(\frac{\theta}{2}) ) at the given point ( \theta = \frac{9\pi}{2} ), we can differentiate the equation of the polar curve with respect to ( \theta ) and then evaluate it at ( \theta = \frac{9\pi}{2} ).

Differentiating ( r ) with respect to ( \theta ) using the product rule:

[ \frac{dr}{d\theta} = \frac{d}{d\theta} \left(2 \sin\left(\frac{\theta}{4}\right) \cos\left(\frac{\theta}{2}\right)\right) ]

[ = 2 \cos\left(\frac{\theta}{4}\right) \cos\left(\frac{\theta}{2}\right) \cdot \frac{1}{4} \cdot \sin\left(\frac{\theta}{4}\right) - 2 \sin\left(\frac{\theta}{4}\right) \sin\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} \cdot \cos\left(\frac{\theta}{2}\right) ]

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

Now, evaluate ( \frac{dr}{d\theta} ) at ( \theta = \frac{9\pi}{2} ):

[ \frac{dr}{d\theta} \bigg|_{\theta = \frac{9\pi}{2}} = \frac{1}{2} \cos\left(\frac{\frac{9\pi}{2}}{4}\right) \cos\left(\frac{\frac{9\pi}{2}}{2}\right) \sin\left(\frac{\frac{9\pi}{2}}{4}\right) - \sin\left(\frac{\frac{9\pi}{2}}{4}\right) \sin\left(\frac{\frac{9\pi}{2}}{2}\right) \cos\left(\frac{\frac{9\pi}{2}}{2}\right) ]

[ = \frac{1}{2} \cos\left(\frac{9\pi}{8}\right) \cos\left(\frac{9\pi}{4}\right) \sin\left(\frac{9\pi}{8}\right) - \sin\left(\frac{9\pi}{8}\right) \sin\left(\frac{9\pi}{4}\right) \cos\left(\frac{9\pi}{4}\right) ]

[ = \frac{1}{2} \cos\left(\frac{9\pi}{8}\right) \cdot 0 \cdot \sin\left(\frac{9\pi}{8}\right) - \sin\left(\frac{9\pi}{8}\right) \cdot (-1) \cdot \cos\left(\frac{9\pi}{4}\right) ]

[ = 0 - \sin\left(\frac{9\pi}{8}\right) \cdot (-1) \cdot 0 ]

[ = 0 ]

Therefore, the slope of the tangent line at ( \theta = \frac{9\pi}{2} ) is ( \frac{dr}{d\theta} \bigg|_{\theta = \frac{9\pi}{2}} = 0 ).

By signing up, you agree to our Terms of Service and Privacy Policy

- What is #(-2,9)# in polar coordinates?
- What is the slope of the tangent line of #r=theta/3+sin((3theta)/8+(5pi)/3)# at #theta=(11pi)/8#?
- What is the distance between the following polar coordinates?: # (2,(5pi)/2), (4,(3pi)/2) #
- What is the arc length of the polar curve #f(theta) = cos(3theta-pi/2) +thetacsc(-theta) # over #theta in [pi/12, pi/8] #?
- What is the polar form of #(-2,3)#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7