9.5: Calculus and Polar Functions
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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(θ) into a set of parametric equations. Using the identities x=rcosθ and y=rsinθ, we can create the parametric equations x=f(θ)cosθ, y=f(θ)sinθ and apply the concepts of Section 9.3.
Polar Functions and dydx
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 dydx. Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ. Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.
Using Key Idea 37 we have dydx=dydθ/dxdθ.
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 dydx with Polar Functions
Let r=f(θ) be a polar function. With x=f(θ)cosθ and y=f(θ)sinθ,
dydx=f′(θ)sinθ+f(θ)cosθf′(θ)cosθ−f(θ)sinθ.
Example 9.5.1: Finding dydx with polar functions.
Consider the limacon r=1+2sinθ on [0,2π].
- Find the equations of the tangent and normal lines to the graph at θ=π/4.
- Find where the graph has vertical and horizontal tangent lines.
Solution
- We start by computing dydx. With f′(θ)=2cosθ, we have dydx=2cosθsinθ+cosθ(1+2sinθ)2cos2θ−sinθ(1+2sinθ)=cosθ(4sinθ+1)2(cos2θ−sin2θ)−sinθ.
When θ=π/4, dydx=−2√2−1 (this requires a bit of simplification). In rectangular coordinates, the point on the graph at θ=π/4 is (1+√2/2,1+√2/2). Thus the rectangular equation of the line tangent to the limacon at θ=π/4 is y=(−2√2−1)(x−(1+√2/2))+1+√2/2≈−3.83x+8.24. The limacon and the tangent line are graphed in Figure 9.47.
The normal line has the opposite--reciprocal slope as the tangent line, so its equation is
y≈13.83x+1.26.
- To find the horizontal lines of tangency, we find where dydx=0; thus we find where the numerator of our equation for dydx is 0.
cosθ(4sinθ+1)=0⇒cosθ=0or4sinθ+1=0.
On [0,2π], cosθ=0 when θ=π/2, 3π/2.
Setting 4sinθ+1=0 gives θ=sin−1(−1/4)≈−0.2527=−14.48∘. We want the results in [0,2π]; we also recognize there are two solutions, one in the 3rd quadrant and one in the 4th. Using reference angles, we have our two solutions as θ=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 black--filled dots.
To find the vertical lines of tangency, we set the denominator of dydx=0.
2(cos2θ−sin2θ)−sinθ=0.Convert the cos2θ term to 1−sin2θ:2(1−sin2θ−sin2θ)−sinθ=04sin2θ+sinθ−1=0.Recognize this as a quadratic in the variable sinθ. Using the quadratic formula, we havesinθ=−1±√338.
We solve sinθ=−1+√338 and sinθ=−1−√338:
sinθ=−1+√338sinθ=−1−√338θ=sin−1(−1+√338)θ=sin−1(−1−√338)θ=0.6399θ=−1.0030
In each of the solutions above, we only get one of the possible two solutions as sin−1x only returns solutions in [−π/2,π/2], the 4th and 1st quadrants. Again using reference angles, we have:
sinθ=−1+√338⇒θ=0.6399, 3.7815 radians
and
sinθ=−1−√338⇒θ=4.1446, 5.2802 radians.
These points are also shown in Figure 9.47 with white--filled dots.
When the graph of the polar function r=f(θ) intersects the pole, it means that f(α)=0 for some angle α. Thus the formula for dydx in such instances is very simple, reducing simply to
dydx=tanα.
This equation makes an interesting point. It tells us the slope of the tangent line at the pole is tanα; some of our previous work (see, for instance, Example 9.4.3) shows us that the line through the pole with slope tanα has polar equation θ=α. Thus when a polar graph touches the pole at θ=α, the equation of the tangent line at the pole is θ=α.
Example 9.5.2: Finding tangent lines at the pole.
Let r=1+2sinθ, a limacon. Find the equations of the lines tangent to the graph at the pole.
Solution
We need to know when r=0.
1+2sinθ=0sinθ=−1/2θ=7π6, 11π6.
Thus the equations of the tangent lines, in polar, are θ=7π/6 and θ=11π/6. In rectangular form, the tangent lines are y=tan(7π/6)x and y=tan(11π/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 θ is A=12θr2.
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 θ=α 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(θ) on [α,β] is given. (Note how the "sides'' of the region are the lines θ=α and θ=β, whereas in rectangular coordinates the "sides'' of regions were often the vertical lines x=a and x=b.)
Partition the interval [α,β] into n equally spaced subintervals as α=θ1<θ2<⋯<θn+1=β. The length of each subinterval is Δθ=(β−α)/n, representing a small change in angle. The area of the region defined by the ith subinterval [θi,θi+1] can be approximated with a sector of a circle with radius f(ci), for some ci in [θi,θi+1]. The area of this sector is 12f(ci)2Δθ. This is shown in part (b) of the figure, where [α,β] has been divided into 4 subintervals. We approximate the area of the whole region by summing the areas of all sectors:
Area≈n∑i=112f(ci)2Δθ.
This is a Riemann sum. By taking the limit of the sum as n→∞, 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 non-negative on [α,β], where 0≤β−α≤2π. The area A of the region bounded by the curve r=f(θ) and the lines θ=α and θ=β is
A = 12∫βαf(θ)2 dθ = 12∫βαr2 dθ
The theorem states that 0≤β−α≤2π. This ensures that region does not overlap itself, which would give a result that does not correspond directly to the area.
Example 9.5.3: Area of a polar region
Find the area of the circle defined by r=cosθ. (Recall this circle has radius 1/2.)
Solution
This is a direct application of Theorem 83. The circle is traced out on [0,π], leading to the integral
Area=12∫π0cos2θ dθ=12∫π01+cos(2θ)2 dθ=14(θ+12sin(2θ))|π0=14π.
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 ∫cos2θ dθ. 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 cos2θ and sin2θ is required often. We offer here these indefinite integrals as a time--saving measure.
∫cos2θ dθ=12θ+14sin(2θ)+C
∫sin2θ dθ=12θ−14sin(2θ)+C
Example 9.5.4: Area of a polar region
Find the area of the cardiod r=1+cosθ bound between θ=π/6 and θ=π/3, as shown in Figure 9.50.
Solution
This is again a direct application of Theorem 83.
Area=12∫π/3π/6(1+cosθ)2 dθ=12∫π/3π/6(1+2cosθ+cos2θ) dθ=12(θ+2sinθ+12θ+14sin(2θ))|π/3π/6=18(π+4√3−4)≈0.7587.
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 r2=f2(θ) and subtracting the area bounded by r1=f1(θ) on [α,β]. Thus
Area = 12∫βαr22 dθ−12∫βαr21 dθ=12∫βα(r22−r21) dθ.
KEY IDEA 42 area between polar curves
The area A of the region bounded by r1=f1(θ) and r2=f2(θ), θ=α and θ=β, where f1(θ)≤f2(θ) on [α,β], is
A=12∫βα(r22−r21) dθ.
Example 9.5.5: Area between polar curves
Find the area bounded between the curves r=1+cosθ and r=3cosθ, 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:
1+cosθ=3cosθcosθ=1/2θ=±π/3
Thus we integrate 12((3cosθ)2−(1+cosθ)2) on [−π/3,π/3].
Area=12∫π/3−π/3((3cosθ)2−(1+cosθ)2) dθ=12∫π/3−π/3(8cos2θ−2cosθ−1) dθ=(2sin(2θ)−2sinθ+3θ)|π/3−π/3=2π.
Amazingly enough, the area between these curves has a "nice'' value
Example 9.5.6: Area defined by polar curves
Find the area bounded between the polar curves r=1 and r=2cos(2θ), 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(2θ)=1⇒cos(2θ)=12⇒2θ=π/3⇒θ=π/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 θ=0. The dashed line breaks the region into its component parts. Below the dashed line, the region is defined by r=1, θ=0 and θ=π/6. (Note: the dashed line lies on the line θ=π/6.) Above the dashed line the region is bounded by r=2cos(2θ) and θ=π/6. Since we have two separate regions, we find the area using two separate integrals.
Call the area below the dashed line A1 and the area above the dashed line A2. They are determined by the following integrals:
A1=12∫π/60(1)2 dθA2=12∫π/4π/6(2cos(2θ))2 dθ.
(The upper bound of the integral computing A2 is π/4 as r=2cos(2θ) is at the pole when θ=π/4.)
We omit the integration details and let the reader verify that A1=π/12 and A2=π/12−√3/8; the total area is A=π/6−√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=∫ba√f′(t)2+g′(t)2 dt=∫ba√x′(t)2+y′(t)2 dt.
Now consider the polar function r=f(θ). We again use the identities x=f(θ)cosθ and y=f(θ)sinθ to create parametric equations based on the polar function. We compute x′(θ) and y′(θ) as done before when computing dydx, then apply Equation ???.
The expression x′(θ)2+y′(θ)2 can be simplified a great deal; we leave this as an exercise and state that x′(θ)2+y′(θ)2=f′(θ)2+f(θ)2.
This leads us to the arc length formula.
key idea 43 arc length of polar curves
Let r=f(θ) be a polar function with f′ continuous on an open interval I containing [α,β], on which the graph traces itself only once. The arc length L of the graph on [α,β] is
L=∫βα√f′(θ)2+f(θ)2 dθ=∫βα√(r′)2+r2 dθ.
Example 9.5.7: Arc length of a limacon
Find the arc length of the limacon r=1+2sint.
Solution
With r=1+2sint, we have r′=2cost. The limacon is traced out once on [0,2π], giving us our bounds of integration. Applying Key Idea 43, we have
L=∫2π0√(2cosθ)2+(1+2sinθ)2 dθ=∫2π0√4cos2θ+4sin2θ+4sinθ+1 dθ=∫2π0√4sinθ+5 dθ≈13.3649.
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(θ), where f′ is continuous on an open interval containing [α,β] on which the graph does not cross itself.
- The surface area of the solid formed by revolving the graph about the initial ray (θ=0) is: Surface Area=2π∫βαf(θ)sinθ√f′(θ)2+f(θ)2 dθ.
- The surface area of the solid formed by revolving the graph about the line θ=π/2 is: Surface Area=2π∫βαf(θ)cosθ√f′(θ)2+f(θ)2 dθ.
Example 9.5.8: Surface area determined by a polar curve
Find the surface area formed by revolving one petal of the rose curve r=cos(2θ) 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,π/4] about the initial ray. Using Key Idea ??? and the fact that f′(θ)=−2sin(2θ), we have
Surface Area=2π∫π/40cos(2θ)sin(θ)√(−2sin(2θ))2+(cos(2θ))2 dθ≈1.36707.
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.
Contributors and Attributions
Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/