Polar Coordinates. July 30, 2014

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1 Polr Coordintes July 3, 4 Sometimes it is more helpful to look t point in the xy-plne not in terms of how fr it is horizontlly nd verticlly (this would men looking t the Crtesin, or rectngulr, coordintes x nd y), but rther how fr wy it is from the origin, nd wht ngle it mkes with the positive x-xis; these re the coordintes r nd θ used in sonr nd mny other pplictions. The equtions relting polr coordintes to x nd y re: r = x + y, θ = tn (y/x) The equtions relting Crtesin coordintes (x nd y) to polr ones re: x = r cos θ, y = r sin θ For exmple, the point (, π/) in polr coordintes becomes (, ) in rectngulr coordintes becuse cos(π/) = nd sin(π/) =. The points (, π/), (, 5π/), (, 3π/), (, π/), etc. ll represent the sme point. The point (3, π/3) in polr coordintes becomes (3/, 3 3/) in Crtesin coordintes. The point (, 3) in rectngulr coordintes becomes ( 3, tn (3/)) in polr coordintes. The point (3, 3) in rectngulr coordintes becomes (3, π/4) in polr coordintes. Exmple. Now let s look t some equtions in polr coordintes. The eqution r = 3 is the circle centered t the origin with rdius 3. In generl, for ny rel number r, the eqution r = r is the circle centered t the origin with rdius r (if r = then the circle with rdius just consists of the origin). Exmple. The eqution θ = π/4 is the line through the origin with slope. In generl, for ny rel number θ, the eqution θ = θ

2 is the line through the origin which mkes n ngle of θ with the positive x-xis. If θ is n odd multiple of π/, then it is simply the y-xis. If θ is not n odd multiple of π/, then the slope of this line is tn(θ ). Exmple 3. To grph the eqution r = cos θ we convert it into rectngulr coordintes by first substituting x = r cos θ, or rther x/r = cos θ, into the eqution: r = x/r, r = x Now r = x + y so the eqution becomes x + y = x x x + y = x x + + y = (x ) + y = Therefore, this is the circle of rdius centered t (, ). In generl, sketching curves given s n eqution in polr coordintes cn be difficult. You cn t lwys convert them into n eqution in Crtesin coordintes tht you know how to drw. For exmple, r = + sin θ is wht s known s crdioid becuse it is shped like hert. To convert this eqution into rectngulr coordintes, let us substitute: r = + y/r r = r + y x + y y = x + y (x + y y) = x + y This is not n eqution tht we know how to drw, but using polr coordintes we cn get pretty good ide of wht it looks like (this will be done in clss). The bsic ide is tht you cn plug in few points for θ, find r = + sin θ, convert these points into Crtesin coordintes, nd grph

3 them. Then connect the dots smoothly, using the knowledge of whether r is incresing or decresing. To be ble to sketch polr curves correctly nd efficiently, it is relly mtter of prctice, so be sure to get lots of prctice! It cn lso help to memorize the shpe of certin curves. The slope of the tngent line to the grph of r = f(θ) t some vlue of θ cn be clculted using the following formul: dy dx = dy/ dx/ d(r sin θ)/ = d(r cos θ)/ (f(θ) sin θ) = (f(θ) cos θ) = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ I would recommend not memorizing this formul, but rther remembering how to derive it. Exmple 4. To clculte the slope of t θ =, first compute r = + sin θ f(θ) = + sin θ, f (θ) = cos θ Now the slope is f() =, f () = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ = + = To find the re enclosed by the curve r = f(θ) for θ b, we use the formul b f(θ) This little formul is more complicted thn it looks; good del of thought needs to go into the problem before we cn ctully write down the integrl. Exmple 5. Find the re enclosed by the curve r = + sin θ. 3

4 Solution. By sketching the curve r = + sin θ, we see tht it tkes the intervl [, π] to sketch the whole curve, so we need to integrte from to π: π π ( + sin θ) = ( + sin θ + sin θ) = π ( + sin θ + cos(θ) ) = ( 3 θ cos θ ) π 4 sin(θ) ( ) 6π = 4 ( ) = 3π Exmple 6. Find the re enclosed by the curve r = cos θ. Solution. By sketching the curve, we see tht it tkes the intervl [, π] to sketch the curve, so we integrte from to π: π π (cos θ) = + cos(θ) = ( θ + ) π 4 sin(θ) = π 4 In this cse, we cn verify our result: the curve is circle with rdius, so the re is π ( ) = π. 4 Exmple 7. Find the re enclosed between the two curves r = cos θ nd r = sin θ. Solution. The curve r = cos θ is the circle with rdius / with center (/, ); the curve r = sin θ is the circle with rdius / with center (, /). Drwing the two circles, we see tht the re lies in the first qudrnt, nd no prt of the re ctully lies between the two curves in the sense of polr coordintes, i.e. direction wy from the origin. In other words, if we drw line from the origin through the re tht we re interested in, then it first touches curve tht encloses the re from the outside, nd then the other curve (which hs no relevnce to the re). If we drw line tht mkes n 4

5 ngle between nd π/4 with the positive x-xis, then the line first touches r = sin θ. If we drw line tht mkes n ngle between π/4 nd π/ with the positive x-xis, then the line first touches r = cos θ. Thus, we cn find the re s follows: π/4 (sin θ) + π/ π/4 (cos θ) But by symmetry, we know tht the res represented by the two integrls re the sme, so we relly need to compute only one nd just double it: ( ) π/4 π/4 (sin θ) cos(θ) = = θ 4 sin(θ) π/4 = π 8 4 Exmple 8. () Find the re in the first qudrnt enclosed between r = + sin θ nd r = cos θ. (b) Find the re in the fourth qudrnt enclosed between r = + sin θ nd r = cos θ. Solution. () In the first qudrnt, r = + sin θ is the outermost curve nd r = cos θ is the innermost curve, so we wnt to find the re enclosed by the curve r = + sin θ nd subtrct off the re enclosed by r = cos θ. We set up the integrl s follows: π/ (( + sin θ) (cos(θ)) ) = = π/ π/ ( + sin θ + sin θ cos θ) ( + sin θ cos(θ)) = θ cos θ π/ 4 sin(θ) ( π ) = ( ) 4 = + π 4 5

6 (b) In the fourth qudrnt, the sitution is reversed: the curve r = +sin θ is innermost nd the curve r = cos θ is outermost, so we set up the integrl s follows: π/((cos(θ)) ( + sin θ) ) = π/ (( + sin θ) (cos(θ)) ) = θ + cos θ + 4 sin(θ) π/ = π 4 Note tht we hd lredy computed the negtive of the integrl, so we just used the computtion of the originl integrl (if you cn sve yourself time by not redoing computtion, do so!) Length of Curve. To find the length of curve r = f(θ) from θ = to θ = b, use the following formul: b r + ( ) dr If you re curious bout how this formul is derived, we use the clcultions of dx dy nd bove: ( ) ( ) dx dy + = +(f (θ) cos θ f(θ) sin θ) + (f (θ) sin θ + f(θ) cos θ) = f (θ) cos θ f (θ)f(θ) sin θ cos θ + f(θ) sin θ + f (θ) sin θ + f (θ)f(θ) sin θ cos θ + f(θ) cos θ = f (θ) + f(θ) ( ) dr = r + Now the curve whose length we wnt to compute is just the curve prmetrized by x = r cos θ = f(θ) cos θ, y = r sin θ = f(θ) sin θ so the length is b (dx ) + ( dy ) b = r + ( ) dr 6