Welcome to a video on Horizontal and Vertical Tangent Lines to Polar Curves. In the last video we discussed how to find dy/dx,when the equation is in polar form, using this formula here. So to determine where a tangent line is horizontal or vertical, we just need to draw some conclusions, about this formula. What I mean by that is dy/ d theta measures the change in y with respect to theta. So if the change in y with respect to theta was zero, we would have a horizontal tangent line. So if a tangent line is horizontal, then dy/d theta is equals to 0. And dx/d theta measures the change in x with respect to theta. And if the change of x with respect to theta is equal to 0, we would have a vertical tangent line. So if the tangent line is verical, then dx/d theta is equal to zero? Let’s go ahead and see how we’re going to apply this. We’re going to determine where the tangent line would be horizontal and vertical to r equals 1 minus sine theta. Let’s start by finding where we have horizontal tangent lines. This will occur where dy/ d theta is equal to 0. So we are going to determine f prime of theta sine theta plus f of theta cosine theta, set it equal to zero, and solve for theta. So dy/ d theta will equal to f prime of theta times sine theta. F prime of theta will be the derivative of 1 minus sine theta. So that will be negative cosine theta, times sine theta, plus, f of theta, 1 minus sine theta, times cosine theta. And remember, to find horizontal tangent lines, we’re only concerned about d y d theta, or the numerator of dy/dx Let’s go ahead and distribute here, set this equal to zero, and see if we can solve this equation. So we’ll have plus cosine theta, minus cosine theta sine theta, equals 0 We have 2 like terms, so we are going to have cosine theta minus 2 cosine theta sine theta. And now we can factor out the common factor of cosine theta, we’ll be left with 1 minus 2 sine theta. So this is equal to 0 when cosine theta is equal to 0 and 1 minus sine theta is equal to zero. If we solve this equation here for sine theta we would subtract one, and then divide by negative 2. We want to know when sine theta is equal to positive one half. And before we do this we should recognize that we’re going to trace out this entire plane curve, on the interval from 0 to 2 pi. If we take a look at the graph of cosine theta, it would look something like this. We know the period is 2 pi. So cosine theta is going to equal zero at pi over 2 and 3 pi over 2. And to determine where sine theta is equal to 1/2, we first need to recognize that sine is going to be positive in quadrant 1 and quadrant 2 And if we draw a 30 degree reference angle in both quadrants, these 2 angles would have a sine ratio of 1/2. So this will be 30 degrees or pi over 6 and this will be 150 degrees or 5 pi over 6. So what we’ll do now is take these 4 angles, set them back into our original equation, to determine what the radius will be. Lets go ahead and do that on the next screen So when theta is pi over 6 we’ll have r equals 1 minus sine pi over 6. Well sine of pi over 6 is 1/2 so r is equal to 1/2. so we have to point, 1/2 comma pi over 6. When theta is equal to 5 pi over 6, so we have r equals 1 minus sine 5 pi over 6.This is also equal to 1/2 so we have 1 minus 1/2 which is 1/2. so we have to point, 1/2,5 pi over 6. When theta equal pi over 2, we have r equals 1 minus sine pie over 2. Sine pi over 2 is equal to 1 so this will be 0. So we’ll have the point 0, pi over 2. That’s gona be at the pole. And then lastly we have 1 minus sine 3 pi over 2. Well the sine of 3 pi over 2 is negative 1 so we have 1 minus negative 1, that will give us 2, so we have the point 2, 3 pi over 2. So from our work we can expect horizontal tangent lines at these 4 points, but lets go check. Notice upon closer inspection, there’s only have 3 points that actually have horizontal tangent lines. We have one at 1/2 pi over 6, 1/2 5 pi over 6 and another horizontal tangent line at 2, 3 pi over 2. Notice at the point 0, pi over 2, at the pole, we do not have a horizontal tangent line So it is important that we check the points on our graph. It is true the if dy/d theta and dx/d theta equal 0, no conclusion can be made about having a vertical or horizontal tangent line. And we’ll see that next when we go to find the vertical tangent lines. Let’s go ahead and do that now. So now we’ll do the second half of this problem where we want to find where we have vertical tangent lines. And we have vertical tangent lines when dx/d theta is equal to 0. So we’re going to find f prime of theta cosine theta minus f of theta sine theta, set it equal to 0 and solve for theta. Let’s go ahead and try it. So we’ll have f prime of theta times cosine theta. f prime of theta is going to be negative cosine theta times cosine theta minus f of theta which is 1 minus sine theta. Times sine theta. Let’s go ahead and simplify this, we have negative cosine squared theta . Distributing here we’lll have sine theta but it’s going to be minus sine theta. And then we’ll have a negative sine squared theta here but we’re going to subtract that so it’s plus sine squared theta. We want to know when this is equal to zero. And notice in its current form, we’re not going to be able to solve this equation but we can get this all in turns of sine’s, if we replace cosine squared with 1 minus sine squared, using the identity sine square theta plus cosine squared theta equals 1. So this will be negative 1 minus sine squared theta minus sine theta plus sine squared theta. Now When we simplify this, notice that we’re going to have a postie sine squared theta and there is another 1 here so we’ll have 2 sine squared theta. Then we’ll have minus sine theta and we’ll also have a minus 1. Now this equation does factor. We”re gonna have 2 sine theta in this position and sine theta in this position. We want the inner and outer some to be negative sine theta. So if we put a negative 1 here and a positive one here, we’ll have negative 2 sine theta plus 1 sine theta does give us negative sine theta. So now we wana set these factors equal to 0 and determine what theta will be. So here we’ll have sine theta equals negative 1/2 and here we’ll have sine theta equals positive 1. So from this equation we know that theta would have to be pi over 2 And then to determine where sine theta is equal to negative 1/2, well we just determined when sine theta was positive 1/2, that was in the first and second quadrant. So now we’ll have a 30 degree reference angle in the third and fourth quadrant. So we’ll have 210 degrees which is 7 pi over 6 and then we’ll also have 330 degrees which is 11 pi over 6 radians. Now let’s go ahead and determine what r wold be using the equation r equals 1 minus sine theta. And let’s do that on the next screen. So when theta is pi over 2 we’ll have r equals 1 minus, we’ll sine pi over 2 is equal to 1, so r is 0. So we have to point 0, pi over 2. Remember we found the same point when we were looking for horizontal tangent lines. For 7 pi over 6 we’ll have 1 minus sine 7 pi over 6, which is negative 1/2 so r we’ll be equal to 3/2. So we’ll have 3/2, 7 pi over 6. And then lastly we’ll have r equals 1 minus sine 11 pi over 6, which again is negative 1/2, so we have 3/2 again. So these are the points where we expect to see vertical tangent lines. Let’s go ahead and check that. So here we have the point, 3/2, 7 pi over 6 where we would have a vertical tangent line. And here’s 3/2, 11 pi/6, were we also have a vertical tangent line. But notice at 0, pi over 2, were at the pole, we do not have a vertical or horizontal tangent line, because we have a sharp point right there. So once again, it is important that we do verify our points by looking at a graph as well as our calculus techniques. I hope you found this video helpful, thank you for watching

buena explicación saludos desde Colombia

Thank you!

Thank you for a superb explanation. Your videos need more exposure. I always tell other students to look for your videos. I am grateful to individuals like yourself who take the time to help others struggling with the concepts in math.

Your demonstrations are very much appreciated.

@maximus5415 I'm getting an error trying to explain why it is correct. I'll send you a message.

love you, lol.

hi

great videos

thx

thank you

Shouldn't there be a vertical tangent since point (0,pi/2) is undefined?

I missed some notes for some reason in class and I couldn't figure out on my own why the (0, pi/2) points weren't included in the answer. This video made everything clear, and now I know to check the points on my test. It was such an easy solution to my puzzlement! Thank you.

This is a bitchload amount of work for one problem

very helpful. better than my textbook's video explanations

Thank you, it helped a lot 😉

+Mathispower4u !! Why can we write a vertical tangent on the sharp point 10:31?? You just said we couldn't draw the tangent there 5:43! My Webassign agrees with you!

Very helpful video. However my textbook says that when you get the same point for both horizontal and tangent line (like the (0,pi/2) in this problem), you have to take the limit of the derivative using L'Hopital's rule. I also wonder if you can skip that and just see if the sharp point is facing left or right and conclude that the tangent is horizontal or if the sharp point is facing up or down the tangent is vertical.

THANK YOUUUUUU

That's what I'm talking about.

Thank you!

Bless up though.

Im confused I thought sin inverse had a range of (-pi/2,pi/2)

very helpful, I really appreciate.. the only thing I can't understand was the quote at the end of the video 😂