More calculus with parametric A main motivation for looking at a different coordi- Given curves in parametric form we can carry out nate system is to find simpler ways to describe re- the same calculus problems we we have with curves gions. For instance the unit becomes r = 1. of the form y = f(x). For instance to find the area Other curves are more subtle. we can use the preced- from the down to the x-axis from time t = a to ing relationships to convert back and forth between t = b we have the two systems. This can help in identifying curves. b Curves in polar coordinates  Area = y(t) d x(t) . We can describe curves using polar coordinates. a Z These usually take the form r = f(θ). We can use This follows the same philosophy as before in split- these to find symmetry, for instance if f(−θ) = f(θ) ting things up into small rectangles. The height is then the curve will be symmetric around the x-axis. y(t) and the width is x0(t). Now we could use hori- If f(π − θ) = f(θ) then the curve will be symmetric zontal rectangles if we want to go to the x-axis, this around the y-axis. would be We can also take the curves in polar coordinates and b Area = x(t) dy(t). describe them in terms of parametric curves, where we treat the θ as the parameter. This gives Za One thing to be careful of is that we must pay atten- x(θ) = f(θ) cos(θ), tion to the orientation of the curve, orientation can effect the sign. y(θ) = f(θ) sin(θ). Similarly we can find the area of a of revolu- tion. This is a simple adaptation of our generic idea. From this we can, for example find the derivative Namely we had which gives slopes of tangent lines to the curve. In particular we have b Area = 2π (radius)(arc length) d∗ dy f0(θ) sin(θ) + f(θ) cos(θ) = . Za dx f0(θ) cos(θ) − f(θ) sin(θ) and we have already seen arc length. Therefore we have the following two possibilities. Around the x- We can also plot the curves. This can be done, for ex- axis ample, by putting in individual θ and plotting points. Another useful method is to understand the distance b q 2 2 of r; in particular identify which we have dis- Area = 2π y(t) x0(t) + y0(t) dt a tance 0, for which angles we have distance maxi- Z mized, and for which angles we have distance min- and around the y-axis imized. Once we know that we can identify “key b points” and then play our traditional game of con- q 2 2 Area = 2π x(t) x0(t) + y0(t) dt. necting the dots. a Z Area and length in polar coordinates Polar coordinates We can find the area bounded by parametric curves, A key idea in calculus is understanding the under- r = f(θ) by slicing things into little pieces. The basic lying which starts by describing where we idea is to carve up our region into little “pie slices”. 1 2 are. So far we have been using Cartesian coordinates Each such slice has area 2 f(θk) ∆θk. So that the to describe locations which work by indicating dis- area is approximately tance from two different axes. There is an alternative 1 2 method which is to describe location by a combina- Area ≈ f(θ ) ∆θ 2 k k tion of direction (θ) and distance (r). So we first start k at the origin (the “pole”), rotate as needed, then move X out as needed. which taking the limit as the slices get smaller and smaller this becomes

β 1 2 r Area = f(θ) dθ. y α 2 Z θ Note that this will generally require us to become x better at doing integration involving trigonometric integrals. y This also allows us to find area between two curves. x = r cos θ, y = r sin θ, x2 + y2 = r2, = tan θ. x In particular if f(θ) and g(θ) intersect at angles α and β then the area between is Quiz 13 problem bank

β 1 2 2 1. Find the area inside the region enclosed by the Area = f(θ) − f(θ) dθ. 2 3 2 parametric curves (1 − t , t − t ). (See below.) Zα We can also find the arc length of a polar curve start- ing at angle α and going to angle β. We can start with our parametrization in terms of θ and then note that

β q 2 2 Length = x0(θ) + y0(θ) dθ α 2. Find the surface area which results from Z β q 2 2 rotating the parametric curve = f(θ) + f0(θ) dθ. α x(t) = ln(sec t + tan t) − 2 sin t, y(t) = 2 cos t for Z π 0 6 t 6 4 around the x-axis. 3. Find the surface area which results from rotating the parametric curve x(t) = ln(sec t + tan t) − sin t, y(t) = cos t for π 0 6 t 6 4 around the x-axis. 4. Rewrite x2 + xy + y2 = 1 as an equation in polar coordinates. 5. Sketch the curve r = 1 − sin(θ).

6. Sketch the region described by sec θ 6 r 6 2. 7. Find the area inside the curve r = 2 sin(3θ).

8. Find the area inside the unit circle centered at (1, 0) and outside the unit circle centered at (0, 0). (See below.)

9. Find the length of the curve r = θ2 for 0 6 θ 6 π. (See below.)

10. [Mystery problem]