What is the equation of the tangent line of #r=cos(theta+pi/2)*cos(theta-pi) # at #theta=(-pi)/8#?

Answer 1

Simplify before you differentiate.

By the Cosine Sum Formula #cos(theta + pi/2) = cos(theta)cos(pi/2)-sin(theta)sin(pi/2) = 0 -sin(theta) = -sin(theta)# By the Cosine Difference Formula #cos(theta - pi) = cos(theta)cos(pi)+sin(theta)sin(pi) = -cos(theta)+0 = -cos(theta)# Their product is #r = (-sin(theta))(-cos(theta))=sin(theta)cos(theta)# By the Sine Double angle formula, we have #r = sin(theta)cos(theta) = (1/2)sin(2theta)#

To find the slope of the tangent line, use the fact that

#dy/dx=((dr)/(d(theta))sin(theta)+rcos(theta))/((dr)/(d(theta))cos(theta)-rsin(theta)#
Finally, determine the point (x, y) corresponding to the point that you are interested in, where #theta = -pi/8#. Hint: #x = rcos(theta) and y = rsin(theta)#. Once you have a point and the slope, use the Point-Slope Form of a line to obtain its equation.
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Answer 2

To find the equation of the tangent line to the polar curve ( r = \cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) ) at ( \theta = -\frac{\pi}{8} ), we first find the derivative of ( r ) with respect to ( \theta ), denoted as ( \frac{dr}{d\theta} ). Then we evaluate ( r ) and ( \frac{dr}{d\theta} ) at ( \theta = -\frac{\pi}{8} ) to get the slope of the tangent line.

[ r = \cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = \frac{d}{d\theta}\left[\cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right)\right] ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) - \cos\left(\theta + \frac{\pi}{2}\right) \cdot (-\sin\left(\theta - \pi\right)) ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) + \cos\left(\theta + \frac{\pi}{2}\right) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) + \cos\left(\theta + \frac{\pi}{2}\right) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\sin\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) + \cos\left(\theta + \frac{\pi}{2}\right) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\cos\theta \cdot \cos\left(\theta - \pi\right) + (-\sin\theta) \cdot \sin\left(\theta - \pi\right) ]

[ \frac{dr}{d\theta} = -\cos\theta \cdot \cos\theta + (-\sin\theta) \cdot (-\sin\theta) ]

[ \frac{dr}{d\theta} = -\cos^2\theta - \sin^2\theta ]

[ \frac{dr}{d\theta} = -1 ]

The slope of the tangent line at ( \theta = -\frac{\pi}{8} ) is ( -1 ). Using the point-slope form of the equation of a line, we have:

[ y - y_1 = m(x - x_1) ]

[ y - r(-\frac{\pi}{8}) = -1(\theta - (-\frac{\pi}{8})) ]

[ y + \frac{\sqrt{2}}{2} = -(\theta + \frac{\pi}{8}) ]

[ y = -\theta - \frac{\pi}{8} - \frac{\sqrt{2}}{2} ]

So, the equation of the tangent line to the polar curve ( r = \cos\left(\theta + \frac{\pi}{2}\right) \cdot \cos\left(\theta - \pi\right) ) at ( \theta = -\frac{\pi}{8} ) is ( y = -\theta - \frac{\pi}{8} - \frac{\sqrt{2}}{2} ).

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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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