What is the equation of the line that is normal to the polar curve #f(theta)= cos(pi-2theta)+thetasin(5theta-pi/2) # at #theta = pi/2#?

Answer 1

#y = 2/(5pi)x + 1#

We want to figure out two things: the slope of the curve and the position of the point. Therefore, we can find a line that crosses perpendicular to that point.

The point is easy. Plugging in #theta = pi/2#, we get #f(pi/2) = cos(0) + pi/2 sin(2pi) = 1#, i.e. a point at a radius of 1 and an angle of #pi/2#, which is the point (0, 1). Therefore, this is our y-intercept.
Next, we find the slope. We can do this by calculating #(df)/(d theta)# and using it in the following: #y = rsintheta -> dy = sintheta dr + rcostheta d theta = y/r dr + xd theta# #x = rcostheta -> dx = costheta dr - rsintheta d theta = x/r dr - yd theta #
In other words, #(dy)/(dx) = (dy)/(d theta) / (dx)/(d theta) = (y/r cdot (dr)/(d theta) - x)/(x/r cdot (dr)/(d theta) - y)#
Since our point is at (0, 1), this simplifies significantly to #(dy)/(dx) =((dr)/(d theta))/(- 1) = -(dr)/(d theta)#
Now what is #(dr)/(d theta)#? We calculate using the formula given:
#f(theta) = cos(pi - 2 theta) + theta sin(5theta - pi/2)# #f'(theta) = 2 sin(pi - 2 theta) + sin(5theta - pi/2) + 5thetacos(5theta - pi/2) # #f'(pi/2) = 2 sin(0) + sin(2pi) + 5 cdot pi/2 cdot cos(2pi) = (5pi)/2 #
Therefore, the slope of the curve (in Cartesian coordinates) is #-(5pi)/2#.
Orthogonal slopes are determined using the negative reciprocal, i.e. the slope for the line is #2/(5pi)#. From the slope and the intercept, we get the equation
#y = 2/(5pi)x + 1#
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Answer 2

To find the equation of the line normal to the polar curve ( f(\theta) = \cos(\pi - 2\theta) + \theta \sin(5\theta - \frac{\pi}{2}) ) at ( \theta = \frac{\pi}{2} ), follow these steps:

  1. Compute the derivative ( f'(\theta) ).
  2. Evaluate ( f'(\theta) ) at ( \theta = \frac{\pi}{2} ) to find the slope of the tangent line.
  3. The slope of the normal line will be the negative reciprocal of the slope of the tangent line.
  4. Use the point ( (\theta, f(\theta)) ) where ( \theta = \frac{\pi}{2} ) to find the equation of the normal line using the point-slope form.

This process will yield the equation of the line normal to the polar curve at ( \theta = \frac{\pi}{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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