What is the equation of the tangent line of #r=sintheta + costheta# at #theta=pi/2#?

Answer 1

For the inward normal, #theta = pi#. For the outward normal in the opposite direction, it is #theta = 3/2pi#, obtained by adding #pi# to # theta#.

Here,#

#r = sin theta + cos theta#
#=sqrt 2((1/sqrt 2)cos theta+(1/sqrt 2)sin theta)#
#=sqrt 2(cos (pi/4) cos theta+sin(pi/4)sin theta)#
#=sqrt 2 cos(theta - pi/4)#

This represents the circle with center at the pole r = 0 and diameter

sqrt2. The radius is #sqrt 2/2 = 1/sqrt 2#. The initial line is along a
diameter , #theta = pi/4#.

The normal at a point on the circle is along the radius to the point.

Thus, the radial line #theta = pi/2# is outward normal to the point
under reference, #(sqrt 2, pi/2)#.

For the inward normal, the equation is

#theta =3/2pi#, for the opposite direction.
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Answer 2

To find the equation of the tangent line of ( r = \sin(\theta) + \cos(\theta) ) at ( \theta = \frac{\pi}{2} ), we first need to find the derivative of ( r ) with respect to ( \theta ), and then evaluate it at ( \theta = \frac{\pi}{2} ).

The derivative of ( r ) with respect to ( \theta ) is given by:

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

Evaluating this derivative gives:

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

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

[ \frac{dr}{d\theta} \bigg|_{\theta=\frac{\pi}{2}} = \cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1 ]

At ( \theta = \frac{\pi}{2} ), the slope of the tangent line is ( -1 ).

Next, we need to find the value of ( r ) at ( \theta = \frac{\pi}{2} ):

[ r\bigg|_{\theta=\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 ]

Therefore, the point of tangency is ( (\theta, r) = \left(\frac{\pi}{2}, 1\right) ).

Now, we have the slope of the tangent line (( m = -1 )) and a point on the line ( (\theta, r) = \left(\frac{\pi}{2}, 1\right) ). We can use point-slope form to find the equation of the tangent line:

[ r - r_1 = m(\theta - \theta_1) ]

[ r - 1 = -1(\theta - \frac{\pi}{2}) ]

[ r - 1 = -\theta + \frac{\pi}{2} ]

[ r = -\theta + \frac{\pi}{2} + 1 ]

So, the equation of the tangent line of ( r = \sin(\theta) + \cos(\theta) ) at ( \theta = \frac{\pi}{2} ) is ( r = -\theta + \frac{\pi}{2} + 1 ).

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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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