The previous section defined polar coordinates, leading to polar functions. We investigated plotting these functions and solving a fundamental question about their graphs, namely, where do two polar graphs intersect?
We now turn our attention to answering other questions, whose solutions require the use of calculus. A basis for much of what is done in this section is the ability to turn a polar function (r=f( heta)) into a set of parametric equations. Using the identities (x=rcos heta) and (y=rsin heta), we can create the parametric equations (x=f( heta)cos heta), (y=f( heta)sin heta) and apply the concepts of Section 9.3.
Polar Functions and ( frac{dy}{dx})
We are interested in the lines tangent a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equations. In each of these contexts, the slope of the tangent line is (frac{dy}{dx}). Given (r=f( heta)), we are generally not concerned with (r^prime =f^prime ( heta)); that describes how fast (r) changes with respect to ( heta). Instead, we will use (x=f( heta)cos heta), (y=f( heta)sin heta) to compute (frac{dy}{dx}).
Using Key Idea 37 we have [frac{dy}{dx} = frac{dy}{d heta}Big/frac{dx}{d heta}.]
Each of the two derivatives on the right hand side of the equality requires the use of the Product Rule. We state the important result as a Key Idea.
key idea 41 Finding (frac{dy}{dx}) with Polar Functions
Let (r=f( heta)) be a polar function. With (x=f( heta)cos heta) and (y=f( heta)sin heta),
[frac{dy}{dx} = frac{f^prime ( heta)sin heta+f( heta)cos heta}{f^prime ( heta)cos hetaf( heta)sin heta}.]
Example (PageIndex{1}): Finding (frac{dy}{dx}) with polar functions.
Consider the limacon (r=1+2sin heta) on ([0,2pi]).
 Find the equations of the tangent and normal lines to the graph at ( heta=pi/4).
 Find where the graph has vertical and horizontal tangent lines.
Solution
 We start by computing (frac{dy}{dx}). With (f^prime ( heta) = 2cos heta), we have [egin{align*}frac{dy}{dx} &= frac{2cos hetasin heta + cos heta(1+2sin heta)}{2cos^2 hetasin heta(1+2sin heta)}&= frac{cos heta(4sin heta+1)}{2(cos^2 hetasin^2 heta)sin heta}.end{align*}]
When ( heta=pi/4), (frac{dy}{dx}=2sqrt{2}1) (this requires a bit of simplification). In rectangular coordinates, the point on the graph at ( heta=pi/4) is ((1+sqrt{2}/2,1+sqrt{2}/2)). Thus the rectangular equation of the line tangent to the limacon at ( heta=pi/4) is [y=(2sqrt{2}1)ig(x(1+sqrt{2}/2)ig)+1+sqrt{2}/2 approx 3.83 x+8.24.] The limacon and the tangent line are graphed in Figure 9.47.
The normal line has the oppositereciprocal slope as the tangent line, so its equation is
[y approx frac{1}{3.83}x+1.26.]  To find the horizontal lines of tangency, we find where (frac{dy}{dx}=0); thus we find where the numerator of our equation for (frac{dy}{dx}) is 0.
[cos heta(4sin heta+1)=0quad Rightarrow quad cos heta=0 quad ext{or}quad 4sin heta+1=0.]
On ([0,2pi]), (cos heta=0) when ( heta=pi/2, 3pi/2).Setting (4sin heta+1=0) gives ( heta=sin^{1}(1/4)approx 0.2527 = 14.48^circ). We want the results in ([0,2pi]); we also recognize there are two solutions, one in the 3(^ ext{rd}) quadrant and one in the 4(^ ext{th}). Using reference angles, we have our two solutions as ( heta =3.39) and (6.03) radians. The four points we obtained where the limacon has a horizontal tangent line are given in Figure 9.47 with blackfilled dots.
To find the vertical lines of tangency, we set the denominator of (frac{dy}{dx}=0).
[egin{align*}2(cos^2 heta sin^2 heta)sin heta &= 0 . ext{Convert the (cos^2 heta) term to (1sin^2 heta):}& 2(1sin^2 hetasin^2 heta)sin heta &= 04sin^2 heta + sin heta 1 &= 0. ext{Recognize this as a quadratic in the variable (sin heta).}& ext{ Using the quadratic formula, we have} sin heta &= frac{1pmsqrt{33}}{8}.end{align*}]We solve (sin heta = frac{1+sqrt{33}}8) and (sin heta = frac{1sqrt{33}}8):
[egin{align*}sin heta &=frac{1+sqrt{33}}8 & sin heta &= frac{1sqrt{33}}{8} heta &= sin^{1}left(frac{1+sqrt{33}}8 ight) & heta &= sin^{1}left(frac{1sqrt{33}}8 ight) heta &= 0.6399 & heta &= 1.0030end{align*}]In each of the solutions above, we only get one of the possible two solutions as (sin^{1}x) only returns solutions in ([pi/2,pi/2]), the 4(^ ext{th}) and (1^ ext{st}) quadrants. Again using reference angles, we have:
[sin heta = frac{1+sqrt{33}}8 quad Rightarrow quad heta = 0.6399, 3.7815 ext{ radians}]
and
[sin heta = frac{1sqrt{33}}8 quad Rightarrow quad heta = 4.1446, 5.2802 ext{ radians.}]
These points are also shown in Figure 9.47 with whitefilled dots.
When the graph of the polar function (r=f( heta)) intersects the pole, it means that (f(alpha)=0) for some angle (alpha). Thus the formula for (frac{dy}{dx}) in such instances is very simple, reducing simply to
[frac{dy}{dx} = an alpha.]
This equation makes an interesting point. It tells us the slope of the tangent line at the pole is ( an alpha); some of our previous work (see, for instance, Example 9.4.3) shows us that the line through the pole with slope ( an alpha) has polar equation ( heta=alpha). Thus when a polar graph touches the pole at ( heta=alpha), the equation of the tangent line at the pole is ( heta=alpha).
Example (PageIndex{2}): Finding tangent lines at the pole.
Let (r=1+2sin heta), a limacon. Find the equations of the lines tangent to the graph at the pole.
Solution
We need to know when (r=0).
[egin{align*}
1+2sin heta &= 0
sin heta &= 1/2
heta &= frac{7pi}{6}, frac{11pi}6.
end{align*}]
Thus the equations of the tangent lines, in polar, are ( heta = 7pi/6) and ( heta = 11pi/6). In rectangular form, the tangent lines are (y= an(7pi/6)x) and (y= an(11pi/6)x). The full limacon con can be seen in Figure 9.47; we zoom in on the tangent lines in Figure 9.48.
Note: Recall that the area of a sector of a circle with radius r subtended by an angle ( heta) is (A=frac{1}{2} heta r^2).
Area
When using rectangular coordinates, the equations (x=h) and (y=k) defined vertical and horizontal lines, respectively, and combinations of these lines create rectangles (hence the name "rectangular coordinates''). It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral.
When using polar coordinates, the equations ( heta=alpha) and (r=c) form lines through the origin and circles centered at the origin, respectively, and combinations of these curves form sectors of circles. It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles.
Consider Figure 9.49 (a) where a region defined by (r=f( heta)) on ([alpha,eta]) is given. (Note how the "sides'' of the region are the lines ( heta=alpha) and ( heta=eta), whereas in rectangular coordinates the "sides'' of regions were often the vertical lines (x=a) and (x=b).)
Partition the interval ([alpha,eta]) into (n) equally spaced subintervals as (alpha = heta_1 < heta_2
[ ext{Area} approx sum_{i=1}^n frac12f(c_i)^2Delta heta.]
This is a Riemann sum. By taking the limit of the sum as (n oinfty), we find the exact area of the region in the form of a definite integral.
THEOREM 83 AREA OF A POLAR REGION
Let (f) be continuous and nonnegative on ([alpha,eta]), where (0leq etaalphaleq 2pi). The area (A) of the region bounded by the curve (r=f( heta)) and the lines ( heta=alpha) and ( heta=eta) is
[A = frac12int_alpha^eta f( heta)^2 d heta = frac12int_alpha^eta r^{,2} d heta]
The theorem states that (0leq etaalphaleq 2pi). This ensures that region does not overlap itself, which would give a result that does not correspond directly to the area.
Example (PageIndex{3}): Area of a polar region
Find the area of the circle defined by (r=cos heta). (Recall this circle has radius (1/2).)
Solution
This is a direct application of Theorem 83. The circle is traced out on ([0,pi]), leading to the integral
[egin{align*}
ext{Area} &= frac12int_0^pi cos^2 heta d heta
&= frac12int_0^pi frac{1+cos(2 heta)}{2} d heta
&= frac14ig( heta +frac12sin(2 heta)ig)Bigg_0^pi
&= frac14pi.
end{align*}]
Of course, we already knew the area of a circle with radius (1/2). We did this example to demonstrate that the area formula is correct.
Note: Example 9.5.3 requires the use of the integral (int cos^2 heta d heta). This is handled well by using the power reducing formula as found in the back of this text. Due to the nature of the area formula, integrating (cos^2 heta) and (sin^2 heta) is required often. We offer here these indefinite integrals as a timesaving measure.
[intcos^2 heta d heta = frac12 heta+frac14sin(2 heta)+C]
[intsin^2 heta d heta = frac12 hetafrac14sin(2 heta)+C]
Example (PageIndex{4}): Area of a polar region
Find the area of the cardiod (r=1+cos heta) bound between ( heta=pi/6) and ( heta=pi/3), as shown in Figure 9.50.
Solution
This is again a direct application of Theorem 83.
[egin{align*}
ext{Area} &= frac12int_{pi/6}^{pi/3} (1+cos heta)^2 d heta
&= frac12int_{pi/6}^{pi/3} (1+2cos heta+cos^2 heta) d heta
&= frac12left( heta+2sin heta+frac12 heta+frac14sin(2 heta)
ight)Bigg_{pi/6}^{pi/3}
&= frac18ig(pi+4sqrt{3}4ig) approx 0.7587.
end{align*}]
Area Between Curves
Our study of area in the context of rectangular functions led naturally to finding area bounded between curves. We consider the same in the context of polar functions. index{polar!functions!area between curves}
Consider the shaded region shown in Figure 9.51. We can find the area of this region by computing the area bounded by (r_2=f_2( heta)) and subtracting the area bounded by (r_1=f_1( heta)) on ([alpha,eta]). Thus
[ ext{Area} = frac12int_alpha^eta r_2^{,2} d heta  frac12int_alpha^eta r_1^{,2} d heta = frac12int_alpha^eta ig(r_2^{,2}r_1^{,2}ig) d heta.]
KEY IDEA 42 area between polar curves
The area (A) of the region bounded by (r_1=f_1( heta)) and (r_2=f_2( heta)), ( heta=alpha) and ( heta=eta), where (f_1( heta)leq f_2( heta)) on ([alpha,eta]), is
[A = frac12int_alpha^eta ig(r_2^{,2}r_1^{,2}ig) d heta.]
Example (PageIndex{5}): Area between polar curves
Find the area bounded between the curves (r=1+cos heta) and (r=3cos heta), as shown in Figure 9.52.
Solution
We need to find the points of intersection between these two functions. Setting them equal to each other, we find:
[egin{align*}
1+cos heta &= 3cos heta
cos heta &=1/2
heta &= pm pi/3
end{align*}]
Thus we integrate (frac12ig((3cos heta)^2(1+cos heta)^2ig)) on ([pi/3,pi/3]).
[egin{align*}
ext{Area} &= frac12int_{pi/3}^{pi/3} ig((3cos heta)^2(1+cos heta)^2ig) d heta
&= frac12int_{pi/3}^{pi/3} ig( 8cos^2 heta2cos heta1ig) d heta
&= ig(2sin(2 heta)  2sin heta+3 hetaig)Bigg_{pi/3}^{pi/3}
&= 2pi.
end{align*}]
Amazingly enough, the area between these curves has a "nice'' value
Example (PageIndex{6}): Area defined by polar curves
Find the area bounded between the polar curves (r=1) and (r=2cos(2 heta)), as shown in Figure 9.53 (a).
Solution
We need to find the point of intersection between the two curves. Setting the two functions equal to each other, we have
[2cos(2 heta) = 1 quad Rightarrow quad cos(2 heta) = frac12 quad Rightarrow quad 2 heta = pi/3quad Rightarrow quad heta=pi/6.]
In part (b) of the figure, we zoom in on the region and note that it is not really bounded between two polar curves, but rather by two polar curves, along with ( heta=0). The dashed line breaks the region into its component parts. Below the dashed line, the region is defined by (r=1), ( heta=0) and ( heta = pi/6). (Note: the dashed line lies on the line ( heta=pi/6).) Above the dashed line the region is bounded by (r=2cos(2 heta)) and ( heta =pi/6). Since we have two separate regions, we find the area using two separate integrals.
Call the area below the dashed line (A_1) and the area above the dashed line (A_2). They are determined by the following integrals:
[A_1 = frac12int_0^{pi/6} (1)^2 d hetaqquad A_2 = frac12int_{pi/6}^{pi/4} ig(2cos(2 heta)ig)^2 d heta.]
(The upper bound of the integral computing (A_2) is (pi/4) as (r=2cos(2 heta)) is at the pole when ( heta=pi/4).)
We omit the integration details and let the reader verify that (A_1 = pi/12) and (A_2 = pi/12sqrt{3}/8); the total area is (A = pi/6sqrt{3}/8).
Arc Length
As we have already considered the arc length of curves defined by rectangular and parametric equations, we now consider it in the context of polar equations. Recall that the arc length (L) of the graph defined by the parametric equations (x=f(t)), (y=g(t)) on ([a,b]) is
[L = int_a^b sqrt{f^prime (t)^2 + g^prime(t)^2} dt = int_a^b sqrt{x^prime(t)^2+y^prime (t)^2} dt.label{eq:polar_arclength}]
Now consider the polar function (r=f( heta)). We again use the identities (x=f( heta)cos heta) and (y=f( heta)sin heta) to create parametric equations based on the polar function. We compute (x^prime( heta)) and (y^prime ( heta)) as done before when computing (frac{dy}{dx}), then apply Equation ef{eq:polar_arclength}.
The expression (x^prime( heta)^2+y^prime ( heta)^2) can be simplified a great deal; we leave this as an exercise and state that [x^prime( heta)^2+y^prime ( heta)^2 = f^prime ( heta)^2+f( heta)^2.]
This leads us to the arc length formula.
key idea 43 arc length of polar curves
Let (r=f( heta)) be a polar function with (f^prime ) continuous on an open interval (I) containing ([alpha,eta]), on which the graph traces itself only once. The arc length (L) of the graph on ([alpha,eta]) is
[L = int_alpha^eta sqrt{f^prime ( heta)^2+f( heta)^2} d heta = int_alpha^etasqrt{(r^prime )^2+ r^2} d heta.]
Example (PageIndex{7}): Arc length of a limacon
Find the arc length of the limacon (r=1+2sin t).
Solution
With (r=1+2sin t), we have (r^prime = 2cos t). The limacon is traced out once on ([0,2pi]), giving us our bounds of integration. Applying Key Idea 43, we have
[egin{align*}
L &= int_0^{2pi} sqrt{(2cos heta)^2+(1+2sin heta)^2} d heta
&= int_0^{2pi} sqrt{4cos^2 heta+4sin^2 heta +4sin heta+1} d heta
&= int_0^{2pi} sqrt{4sin heta+5} d heta
&approx 13.3649.
end{align*}]
Figure 9.54: The limacon in Example 9.5.7 whose arc length is measured.
The final integral cannot be solved in terms of elementary functions, so we resorted to a numerical approximation. (Simpson's Rule, with (n=4), approximates the value with (13.0608). Using (n=22) gives the value above, which is accurate to 4 places after the decimal.)
Surface Area
The formula for arc length leads us to a formula for surface area. The following Key Idea is based on Key Idea 39.
KEY IDEA 44 SURFACE AREA OF A SOLID OF REVOLUTION
Consider the graph of the polar equation (r=f( heta)), where (f^prime ) is continuous on an open interval containing ([alpha,eta]) on which the graph does not cross itself.
 The surface area of the solid formed by revolving the graph about the initial ray (( heta=0)) is: [ ext{Surface Area} = 2piint_alpha^eta f( heta)sin hetasqrt{f^prime ( heta)^2+f( heta)^2} d heta.]
 The surface area of the solid formed by revolving the graph about the line ( heta=pi/2) is: [ ext{Surface Area} = 2piint_alpha^eta f( heta)cos hetasqrt{f^prime ( heta)^2+f( heta)^2} d heta.]
Example (PageIndex{8}): Surface area determined by a polar curve
Find the surface area formed by revolving one petal of the rose curve (r=cos(2 heta)) about its central axis (see Figure 9.55.
Solution
We choose, as implied by the figure, to revolve the portion of the curve that lies on ([0,pi/4]) about the initial ray. Using Key Idea ef{idea:surface_area_polar} and the fact that (f^prime ( heta) = 2sin(2 heta)), we have
[egin{align*}
ext{Surface Area} &= 2piint_0^{pi/4} cos(2 heta)sin( heta)sqrt{ig(2sin(2 heta)ig)^2+ig(cos(2 heta)ig)^2} d heta
&approx 1.36707.
end{align*}]
The integral is another that cannot be evaluated in terms of elementary functions. Simpson's Rule, with (n=4), approximates the value at (1.36751).%; with (n=10), the value is accurate to 4 decimal places.
This chapter has been about curves in the plane. While there is great mathematics to be discovered in the two dimensions of a plane, we live in a three dimensional world and hence we should also look to do mathematics in 3D  that is, in space. The next chapter begins our exploration into space by introducing the topic of vectors, which are incredibly useful and powerful mathematical objects.
Top 5 Hardest Calculus Problems In The World
I’m currently midway through my Calculus 2 course and already on the verge of mental exhaustion. Yet, I desperately want to know what the most difficult Calculus problems are and what it might take to solve them. Also, I want to know what realworld applications they might have if they were solved.
I did some research and what I found out was really exciting. There are actually several unsolved Calculus problems which, if solved, could have some revolutionary realworld applications across several fields.
Additionally, two of the problems that made this list could earn a person $1,000,000, awarded by the Clay Mathematics Institute if a solution is found. These 5 unsolved problems are among the hardest in the world that fall into the realm of Calculus.
Reviews
Reviewed by Andy Rich, Professor of Mathematics, PALNI, Manchester University on 12/19/19
Has all the usual topics and then some. I liked the development of differential forms towards the end and having chapter 11 as a teaser for higher level stuff. The development was clear enough that I hope most students at this level could get. read more
Reviewed by Andy Rich, Professor of Mathematics, PALNI, Manchester University on 12/19/19
Comprehensiveness rating: 5 see less
Has all the usual topics and then some. I liked the development of differential forms towards the end and having chapter 11 as a teaser for higher level stuff. The development was clear enough that I hope most students at this level could get it. Some of my favorite examples were missing: e.g., the cycloid and deriving Kepler's laws from Newton's laws, but everybody has their own favorites so I am okay with that. In discussing multivariable continuity, it would have been nice to pull out the two path discussion which appears in the text and highlight it as a theorem, but these are all minor points. I still give this text a 5 for comprehensiveness.
Content Accuracy rating: 5
My only complaint here is in the discussion of torsion. He describes torsion as "wobbling" which to me gives the wrong idea. Wobbling means going back and forth which can happen in the plane. Torsion is "twisting" out of the plane. The mathematics is all correct and the author is honest about where things are being swept under the rug, e.g., in the proof of Green's Theorem.
Relevance/Longevity rating: 5
This is not much of an issue for math texts.
Excellent! Good clear explanations of the ideas behind the theory, e.g., Lagrange multipliers which are sometimes presented as magic, but here are motivated with geometrical ideas. Nice treatment of multidimensional chain rule via matrix multiplication with one section on the conceptual picture and one on computations. Good diagrams throughout, present whenever needed to help with understanding, e.g., showing the relationship of two different parametrizations when proving that the value of the integral is independent of the parametrization.
The division into parts, chapters, and sections made sense and the sections were not too long.
Organization/Structure/Flow rating: 5
Good. The preface and table of contents outline the organization and make it easy to find topics. It might be helpful to introduce polar coordinates earlier.
No problems that I saw. It easy to navigate jumping around with the bookmarks pane in Adobe. Good color illustrations showing geometric pictures when helpful.
Grammatical Errors rating: 5
Cultural Relevance rating: 5
I really liked this text. It is more theoretical, more proofbased than what I usually teach, and I might skip some of that if I were teaching from this text, but I think it is good to include the proofs in the text. I liked the presentation of differential forms towards the end. I think one addition to chapter 11 as a further teaser and to try to show the relationship of the various theorems, would be to set up the deRham sequence in R^3 and show how gradient, curl, and divergence come into that single unifying sequence. (This tiptoes up to topology of the underlying space which the author has already alluded to at various places.) The conversational style was great: "trying to find the maximum of a function like . is silly" for example. Humor is scattered throughout, for example, including a picture of a porcupine as an example of a mammal with an orientation. There seemed to be plentiful exercises. Overall, I think this is an excellent text for multivariable calculus.
Problems on Converting Rectangualar Equations to polar form
Problem 1
 Let us rewrite the equations as follows:
2 ( x 2 + y 2 )  x + y = 0  We now use the formulas giving the relationship between polar and rectangular coordinates: R 2 = x 2 + y 2 , y = R sin t and x = R cos t:
2 ( R 2 )  R cos t + R sin t = 0  Factor out R
R ( 2 R  cos t + sin t ) = 0  The above equation gives:
R = 0
or
2 R  cos t + sin t = 0  The equation R = 0 is the pole. But the pole is included in the graph of the second equation 2 R  cos t + sin t = 0 (check that for t = π / 4 , R = 0). We therefore can keep only the second equation.
2 R  cos t + sin t = 0
or
R = (1 / 2)(cos t  sin t)
Problem 2
 Use y = R sin t and x = R cos t into the given equation:
x + y = 0
R cos t + R sin t = 0
Math 181: Calculus II
Math 181 is the second semester of the standard threesemester calculus sequence. As such, its goal is to continue the study of calculus on the real line, started in Math 180 (Calculus I), with a focus on integration, the basics of sequences and series, as well as parametric descriptions for sets in the plane.
Credit Awarded
Right
Course Materials
Textbook
Calculus: Early Transcendentals by William Briggs and Lyle Cochran, 3rd edition, published by AddisonWesley.
Note that a MyLabMath code is required for this course while the printed textbook is optional.
The ISBN for one semester access is 9780135329221, the ISBN for multiple semester access is 9780135329276. The site will become available in week 0, which is the week before the classes begin. Note that only these ISBNs will work with the MyLabMath course. A book bought with MyLabMath access from Amazon or other sources most likely won’t work with the MyLabMath access code.
Chapters 610 are covered in Math 181.
MyMathLab Access Code
A MyLabMath code can be purchased online after registering for MyMathLab through Blackboard, or at the UIC bookstore, with or without the textbook. Make sure your MyMathLab code is linked to the Blackboard course. The MyMathLab code includes an electronic version of the textbook, buying a physical copy is optional.
Students coming from Math 180 at UIC can use the same access code for Math 181, if it has not expired.
Worksheet Bundle
The worksheet bundle will be used in the problem solving sessions on Tuesdays and Thursdays. An electronic, printable copy of it can be found on the Blackboard site or by contacting any course instructor or TA. A physical copy can be bought directly at the UIC bookstore.
Check Blackboard for updated schedule and semesterspecific information.
Calculus Animations with Mathcad
The statementcan be intuitively interpreted as follows:
the number L is approached by the function values f(x) corresponding to x values that approach c.
Two examples are illustrated:
Note that the positive rise/run values are indicated in green negative values are indicated in red.
This animation can be viewed for the following functions:
Currently, animations corresponding to the following pseries are available:
This definition is illustrated for two ellipses with different eccentricities
curve  the rays start at  
a focus  other point(s)  
Ellipse x 2 /9 + y 2 /4 = 1  ( sqrt(5), 0 )  ( 1, 1 ), ( 0, 0 ), ( 0,  sqrt(2) ), (  2, 0 ) 
Parabola y 2 = x  ( 0.25, 0 )  ( 1, 0 ) 
The area inside a parametric curve (x(t),y(t)) can be calculated by the following formulas
We illustrate the use of these formulas for calculation of area in the two examples below. Definite integrals are approximated with the corresponding Riemann sums, and each animation frame corresponds to adding one term to the sum. The area corresponding to the current term is enclosed in a rectangle  if the term is positive, then the rectangle is green, otherwise it is red.
The partial sum of the terms included already is depicted using a dotted pattern  the dots are green within region whose area is accounted for with a factor +1 the red dots correspond to regions accounted for with a factor 1.
 

Area inside the curve  Formula used  
counterclockwise  AVI file size: 134K (or try a shorter animation with 83K)  AVI file size: 166K (or try a shorter animation with 102K) 
clockwise  AVI file size: 75K  AVI file size: 73K 
During the animation of each polar plot, the corresponding values of and will be tabulated. Animations of the following polar curves are available:
9.5: Calculus and Polar Functions  Mathematics
We know that, if a smooth curve is given by the parametric equations
provided that f ′(t) ≠ 0.
To find the slope of a polar curve r = f (θ), we must first express the curve in parametric form. Since
If f (θ) is differentiable, so are x and y then
Also, if then
In doing an exercise, it is often easier simply to express the polar equation parametrically, then find dy/dx, rather than to memorize the formula.
(a) Find the slope of the cardioid r = 2(1 + cos θ) at See Figure N4–24.
(b) Where is the tangent to the curve horizontal?
FIGURE N4–24
(a) Use r = 2(1 + cos θ), x = r cos θ, y = r sin θ, and r ′ = −2 sin θ then
At
(b) Since the cardioid is symmetric to θ = 0 we need consider only the upper half of the curve for part (b). The tangent is horizontal where (provided ). Since factors into 2(2 cos θ − 1)(cos θ + 1), which equals 0 for cos or −1, or π. From part (a), does equal 0 at π. Therefore, the tangent is horizontal only at (and, by symmetry, at ).
It is obvious from Figure N4–24 that r ′(θ) does not give the slope of the cardioid. As θ varies from 0 to the slope varies from −∞ to 0 to +∞ (with the tangent rotating counterclockwise), taking on every real value. However, r ′(θ) equals −2 sin θ, which takes on values only between −2 and 2!
Chapter Summary
In this chapter we reviewed many applications of derivatives. We’ve seen how to find slopes of curves and used that skill to write equations of lines tangent to a curve. Those lines often provide very good approximations for values of functions. We have looked at ways derivatives can help us understand the behavior of a function. The first derivative can tell us whether a function is increasing or decreasing and locate maximum and minimum points. The second derivative can tell us whether the graph of the function is concave upward or concave downward and locate points of inflection. We’ve reviewed how to use derivatives to determine the velocity and acceleration of an object in motion along a line and to describe relationships among rates of change.
For BC Calculus students, this chapter reviewed finding slopes of curves defined parametrically or in polar form. We have also reviewed the use of vectors to describe the position, velocity, and acceleration of objects in motion along curves.
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Math 2210  Calculus III
These lecture videos are organized in an order that corresponds with the current book we are using for our Math2210, Calculus 3, courses (Calculus, with Differential Equations, by Varberg, Purcell and Rigdon, 9th edition published by Pearson). We have numbered the videos for quick reference so it's reasonably obvious that each subsequent video presumes knowledge of the previous videos' material. Along with the video lecture for each topic, we have included the "prenotes" and "postnotes" which are the notes of the lecture before we did the problems and after we worked everything out during the lecture, respectively. You may want to download the notes to use as a reference while watching the lecture video, or for later reference.
If you find an error in the lecture or a problem with the video, or if you would like to give feedback to us about these lectures, please email [email protected] to do so.
NOTE: These videos were not recorded in stereo sound. If you are listening to these videos on headphones, you may want to consider setting your sound channels to come through both sides. This document will give you an idea of how to accomplish that.
 1 Parametric Equations lecture video
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The Polar Form of a Complex Number
are points on the circle of radius one centered at the origin.
Think of the point moving counterclockwise around the circle as the real number moves from left to right. Similarly, the point moves clockwise if decreases. And whether increases or decreases, the point returns to the same position on the circle whenever changes by or by or by where k is any integer.
Exercise: Prove de Moivre's formula
Now picture a fixed complex number on the unit circle
Consider multiples of z by a real, positive number r .
As r grows from 1, our point moves out along the ray whose tail is at the origin and which passes through the point z . As r shrinks from 1 toward zero, our point moves inward along the same ray toward the origin. The modulus of the point is r . We call the angle which this ray makes with the xaxis, the argument of the number z . All the numbers rz have the same argument. We write
Just as a point in the plane is completely determined by its polar coordinates , a complex number is completely determined by its modulus and its argument.
Notice that the argument is not defined when r =0 and in any case is only determined up to an integer multiple of .
Why not just use polar coordinates? What's new about this way of thinking about points in the plane?
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Watch the video: polar points (December 2021).