Topic 9 7B Differentiating Polars

Añadido: 2026-05-12

[00:00:09]all right welcome back everybody we're gonna have some fun in this um lecture we're going to review a lot of things uh that we learned back in 10th grade um so today we're going to be differentiating um the polar form of equations um so it's very similar to the parametric um d y d x is going to be equivalent to d y d theta divided by d x d theta okay and this is um the formula if you really don't have to memorize this i'm just going to take the given function and find the derivative with respect to theta um i'm sorry the derivative of y with respect to theta put it on the top and the derivative of x with respect to theta and put it on the bottom okay so that's what this formula comes out to be so um let's take a look at um these right here very similar to parametrics um you have horizontal tangents when your numerator d d y d theta is equal to zero provided your denominator is not equal to zero at the same time so we've talked about that you have vertical tangents when your denominator d x d theta just like with parametrics is equal to zero um if they're both equal to zero you can make no conclusion about the tangent lines at that theta so let's go ahead and take a look at this particular problem here it says find the slope of the tangent line to the cardioid when theta is pi over three given that r equals one plus sine of theta okay so let's uh go back to um our what we remember about our polars x is equal to r times the cosine of theta so in this particular case the r that's given to us is one plus the sine of theta that's given and we're going to multiply that by the cosine of theta now we're going to have to take the derivative of this we're going to have to do dx d theta the derivative of this with respect to theta and we have a product here if i distribute i end up having a product again so i'm not going to bother distributing that's not going to help okay so let's set up the y as we uh have learned and the videos i had you watch as a review y is equal to the radius times the sine of theta so on the unit circle the radius was one so the y was the sign the x was the cosine so in this particular problem the r that's given to us is one plus the sine of theta and we're going to multiply that by sine theta r times sine theta now on this one if i distribute to get rid of this product that's actually going to help me get rid of the product rule so i'm going to go ahead and distribute and i get sine of theta plus sine squared theta so i won't have to worry about doing the product rule when i find the derivative okay all right so that's the start of it so to find d y d x just like with parametrics it's d y d theta divided by d x d theta so i'm just going to go ahead and find the derivative of this now be careful it's the derivative of this with respect to theta okay so let's go ahead and do that the derivative of sine is cosine and then the derivative of sine squared that's chain rule 2 sine theta to the first times the derivative of sine theta so that's the derivative d y d theta with respect to theta now i'm going to take the derivative of dx with respect to theta or i'm sorry the derivative of x with respect to theta so i'm going to take the derivative of all of this and as i mentioned earlier we have product rule unfortunately so i'm probably going to need some more space so here we go the first times the derivative of the second that's negative sine theta plus the second times the derivative of the first the derivative of one is zero the derivative of sine is cosine so that is the derivative d y d x now what they want us to do is they want us to evaluate that at theta equals pi over 3. so we can go ahead and plug in our pi over 3 right now or we can simplify it a little bit it's completely up to you i'm going to go ahead and simplify it just a little bit i notice on the top they both have a cosine theta in common so i'm going to factor that out okay on the bottom i'm going to go ahead and distribute my negative sine theta negative sine theta minus sine squared theta plus cosine squared theta so i just cleaned it up a little bit before i substitute now some of you are going to say oh sine squared plus cosine squared is equal to 1 well wouldn't that be nice but this is negative sine squared plus cosine squared so that's not going to work so i am now going to evaluate the first derivative when theta is equal to pi over 3. so i cleaned it up a little bit now i can substitute again you can substitute right into here if you would like um but i'm going to do it into this one so we have the cosine of pi over three times one plus two times the sine of pi over three all over 1 i'm sorry negative sine of pi over 3 minus the sine of pi over 3 squared plus the cosine of pi over 3 squared now if this was a free response question you're done because i can type all of this into the calculator into a scientific calculator right but let's assume it's not and let's practice simplifying in case it's a multiple choice so quickly i'm going to go plug in the values from the unit circle if you don't know these values you might want to review your unit circle so the cosine of pi over 3 is not 60 degrees that's what a lot of people like to tell me it's the cosine of 60 degrees that's what they're asking you for okay the cosine of 60 degrees well that's the same as the sine of 30 degrees so again from your unit circle the answer is one half one plus two the sine of this of pi over three the sine of pi over three is the square root of three over two and again that comes through from your unit circle all over the opposite of the sine of pi over three which we already said is the square root of three over two minus again the sine of pi over 3 square root of 3 over 2 squared plus the cosine of pi over 3 which is one half and we're gonna square that as well okay now i'm gonna leave this to you we've got more important things to do i'm gonna leave this to you and see if you can simplify that common denominator squaring all your arithmetic you can come back to that and see if you can simplify that to a negative one okay so least of my concerns right now is that you can simplify that all right so that is the derivative um when it's in polar form okay so let's go to the next page and let's find the points of the cardioid where the tangent line is horizontal or vertical okay well again if the tangent line is horizontal it's just like from the parametrics um that is the numerator of your derivative so if you go back to example number one i'm going to copy down our numerator so horizontal tangent lines are when your numerator your cosine of theta i'm going to write down the factored form so if you write down this one or you write down this one they're one in the same and i know i'm going to have to factor anyway so i'm going to write down the factor form because i have to solve this trig equation and wouldn't you know we're doing that in 10th grade right now okay now i'm going to write down the vertical as well because anytime they're both equal to zero at the same time i'm gonna eliminate that solution because they can't be both zero at the same time so again i'm gonna go back to my denominator i can copy all of this down or i can copy this down right here so i'm going to copy this denominator down from example 1a so that is and you know what i'm gonna reorder it because i don't like negatives first i'm gonna write it as cosine squared theta minus the sine of theta minus sine squared theta equals zero i'm just going to put it in a different order okay so now we have to go back to our tenth grade math and solve these this is already factored so cosine of theta is equal to zero and one plus two sine theta is equal to zero are going to give me my solutions okay so for this one where is the cosine of theta equal to zero well again from your unit circle that's at pi over two and three pi over two this one the sine of theta equals negative one half and if you recall from your unit circle sine is negative in the third and the fourth quadrant with a reference angle of pi over six so that's going to be 7 pi over 6 which is in the third quadrant and 11 pi over 6 which is in the fourth quadrant so those are our four potential horizontal tangent lines occur at those four thetas now let's hope we can eliminate some of them let's see if we can solve this one and anytime they're both equal to zero we can cross them off all right so we have to solve this well this is a quadratic but this doesn't really factor so again going back to 10th grade when you have different trig functions use your pythagorean identities to make them all the same so we have one minus sine squared theta minus the sine of theta so i'm making a substitution for my pythagorean identity cosine squared theta okay so that's going to give me 1 minus sine theta minus 2 sine squared theta equals 0.

[00:14:13]and that will factor but i like to have it in descending order and i like my a value to be positive so i am going to just move all of these to the other side so that's going to give me a positive 2 or multiply both sides by a negative 1.

[00:14:30]however you want to look at it that's going to give me a positive 2 sine squared plus sine minus 1 equals 0. and i like that form better personal preference now i can factor that into two sine theta minus one times sine theta plus one equals 0. when i set this factor equal to 0 i get the sine of theta equals a positive one-half and this one i get the sine of theta equals negative 1.

[00:15:13]i'm going too fast you can always pause it going back to the unit circle sine is positive in the first at pi over six and in the second quadrant at five pi over six and then again unit circle the sine of theta is equal to negative one at three pi over two so i have three possible vertical tangent lines to the curve here and i have four over here but is there ever a time when the numerator is equal to zero and the denominator is equal to zero at the same time and the answer to that is yes here and here so nothing uh you cannot make any conclusions about horizontal and vertical tangent lines when they are both equal to zero at the same time all right so wouldn't it be nice if it just said give me the three angles or what is theta what is the theta when the horizontal tangent line occurs but it doesn't say that it says find the points so we have to go and find the point so let's review a polar point is r comma theta that is a polar point and again in part a they gave you that your r was equal to one plus the sine of theta so we're going to have to go and evaluate that for every single one of these isn't that fun okay so let's start with the pi over two if theta equals pi over two your radius is one plus the sine of pi over two and the sine of pi over two is one so your radius is 2.

[00:17:35]so the first point that has a horizontal tangent line is 2 comma pi over two we have to keep going we have to do the next one when the theta is seven pi over six the r is one plus the sine of seven pi over six the sine of seven pi over six that is in the third quadrant from your unit circle the sign is negative one half which gives you an r value of one half so the second coordinate on your polar graph is one half seven pi over six and again if you need to review plotting points of polars there were some videos um that i posted on my website that you can watch to refresh your memory all right we've got another one 11 pi over six the r value is one plus the sine of eleven pi over six and again sine is this is fourth quadrant so sine is also negative one-half there as well so the third point on the polar graph that has a horizontal tangent line is one-half comma 11 pi over six so those are the three points on your cardioid that has a horizontal tangent line all right now we need to find the two points for the vertical tangent line so we need to find r of pi over six and that's one plus the sine of pi over six which is one plus a half from the unit circle sine of pi over 6 same thing as sine of 30 degrees hopefully you still have that memorized so the radius three halves with a theta of pi over six that point on the polar graph will have a vertical tangent line and we have one more left finding the radius when theta is pi over six so that's one plus the sine i'm sorry five pi over six and again that is positive one half so the other coordinates that will have a horizontal tangent is three halves five pi over six so wasn't that fun now if you want to go ahead and graph this cardioid uh change your graphing calculator to uh polar and type in the one plus sine of theta and graph it and you can see where you would get three horizontal tangents and two vertical tangents on that curve okay all right well you have your calculator out if you wanted to check that you can um we're going to finish example number two which is a calculator problem okay this is another fun one okay it says we have a polar curve here's the equation of our polar curve theta plus cosine of theta okay from zero to two pi if the particles traveling traveling along the polar curve they give us d theta dt and they want us to find dx dt at the instant theta is five pi over six and interpret the meaning okay well if they want us to find dx dt we need to know what x is so what do we know about x again from the review of polars x is equivalent to the radius times the cosine of theta so in this particular problem our radius is theta plus the cosine of theta that's our radius and we're going to multiply that by oops cosine of theta so that is our x so again i'm going to be taking the derivative of x with respect to theta i have a product if i distribute it i will still have a product and then i will have a chain rule so i would probably not distribute it it's going to make it harder if you do all right so oh i apologize we're not going to take the derivative of x with respect to theta we're going to take the derivative of x with respect to time okay so now we're going to find dx dt now these are not t's so we have to be very careful okay this is this is just like um related rates remember your favorite related rates so we're going to go ahead and we're going to find the derivative of this product with respect to time so we have to use product rule so it's going to be the first times the derivative of cosine theta the derivative of cosine is negative sine theta times d theta dt because they want the derivative of x with respect to time so that's first times the derivative of the second plus the second times the derivative of the first so the derivative of theta with respect to time d theta dt plus the derivative of cosine is negative sine so i'm going to actually change that to minus sine theta times d theta dt so there is our first derivative and they want us to evaluate the first derivative i don't see any way to really simplify this so i'm just going to go ahead and evaluate it now it wants us to evaluate the first derivative when when when theta is five pi over six so we have to plug in five pi over six for all theta all right here we go five pi over six plus the cosine of five pi over six times negative sine of five pi over six times d theta dt but don't forget they gave us d theta dt plus cosine of 5 pi over 6 times the given d theta dt minus the sine of five pi over six times the given d theta dt so you're taking the sine of this angle you're not multiplying the angle by a half you're multiplying the sign of the angle by a half so be very careful with that all right well if this was a free response we're done all of that can be typed into the scientific calculator but of course we're going to practice simplifying this because most likely it would not be a free response question it'd be a multiple choice question all right so here we go five pi over six is just five pi over six cosine of five pi over six second quadrant negative square root of 3 over 2. again that comes from the unit circle times the opposite of the sine of five pi over six second quadrant sine is positive so the opposite it would be a negative and in this case five pi over six from the unit circle is one half times the one half plus cosine of five pi over six cosine of five pi over six negative cosine in the second quadrant and that's square root of three over two times the one-half minus sine of five pi over six sine of five pi over six is a positive one half times the one-half that's already there okay so let's simplify this um i'm going to take this and put it in the front that's negative 1 4 times 5 pi over 6 minus the square root of 3 over 2.

[00:29:52]you know simplify that just a little bit okay plus my negative square root of three over two and i'm going to simplify this a little bit okay this is uh one-half or two-fourths minus one-fourth one-half times one-half is one fourth and i know i'm going to subtract them so i'm just going to go ahead and make this a two fourths while i'm at it i'm getting closer so this is a negative 1 4 times 5 pi over 6 minus 3 halves this over here is a negative three halves times one fourth now i'm just going to keep simplifying um [Music] let's go ahead and distribute that's a negative 5 pi over 24 plus the square root of 3 over 8 minus the square root of 3 over 8. oh thank god there's my answer all right you have just evaluated dx dt when theta was five pi over six what did you just calculate dx dt change in x with respect to time what does that mean well dx dt evaluated at theta equivalent to five pi over six what does that mean in the context of this problem i just kind of gave you a hint that represents the change in x so the horizontal rate of change it's a first derivative the horizontal rate of change of the particle as it travels along the polar curve at the instance theta is equal to five pi over six so it's the instantaneous rate of change the instantaneous horizontal rate of change of your particle when theta equals 5 pi over 6.

[00:33:48]that's what you've just calculated okay isn't that wonderful aren't you proud all right so that is the um differentiating uh polar form very similar to differentiating parametrics um takes a lot of time take your time be careful show your work simplify completely a lot of good stuff the end