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

Answer 1

The slope is #~~ -0.1#

Tangents to a polar curve

From the reference

#dy/dx = ((dr(theta))/(d theta)sin(theta) + r(theta)cos(theta))/((dr(theta))/(d theta)cos(theta) - r(theta)sin(theta))#

I used WolframAlpha to compute the derivative

#(dr(theta))/(d theta) = 1/4(3cos((3theta)/4) - cos(theta/4))#

I used an Excel spreadsheet

I entered the following into cell J1:

=PI()*9/2

I used cell J2 to evaluate #(dr((9pi)/2))/(d theta)# by entering the following:

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

I used cell J3 to evaluate #r((9pi)/2)# 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 ~~ -0.1#
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Answer 2

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 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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