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:
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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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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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