1 Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider the circle exmple bove. We hve so = sin(t), = cos(t), = cos(t) sin(t) = x y. Wht kind of tngency do we hve on prmetric curve? 1. If / = nd /, then the curve should hve horizontl tngency.. If / = nd /, then the curve should hve verticl tngency. 3. If both derivtives re zero, then ll kinds of things cn hppen: we cn hve verticl or horizontl tngencies, we cn hve cusp, we cn even hve regulr point tht looks perfectly fine. Although we don t sy much bout it here (you will see much more in MATH 41!), we cn lso think of the tngent vector to curve s the vector: (, ) Its slope gives / in the computtions bove. Note tht it lso hs nother prmeter ssocited to it, its length: ( ) ( ) +, which we cn think of s speed t which we trverse the curve. (More on this below.) 1
2 1. Second (nd higher) derivtives Wht if we re interesting in computing where prmetric curve is concve up or down, for exmple? Here we would need the second derivtive d y/. How do we compute this? Recll the chin rule. If is function of x nd x is function of t, then or =, (1) = This is the formul we used bove, but with = y. To get the second derivtive d y/, let us choose = /, giving ( ) d / d y = d ( ) / = =. d(/) It is complicted formul. I wouldn t suggest memorizing it, but do recll how to derive it, becuse this nd relted formuls re useful! Cycloid, revisited Recll the equtions for the cycloid: x = r(t sin t), y = r(1 cos t). () Let us first compute the derivtive t ech point on the cycloid. Note tht we hve = / / = r sin t r(1 cos t) = sin t 1 cos t. We see tht the slope is zero when sin t = nd 1 cos t, which occurs t every odd multiple of π. Notice tht whenever the denomintor is zero, so is the numertor, so we hve to be clever to compute the derivtive. For exmple, let us consider t =. We formlly hve / indeterminte form here. Expnding in Tylor series t t =, we obtin = t t3 /6 + O(t 5 ) t / + O(t 4 = 1 ) t + O(t). We see tht the left nd right limits re different! So in fct we hve lim t + =, lim t =. This mkes sense from the picture: the slope is pointing down s we pproch ech cusp from the left, nd pointing up s we pproch ech cusp from the right.
3 3 Other exmples with indeterminte derivtives We hve lre seen from the cycloid tht point with n indeterminte derivtive cn hve verticl tngency in cusp-like mnner. We cn lso see tht horizontl nd verticl tngencies re possible. For exmple, consider the prmetric curve x = t 3, y = t 5, t [ 1, 1]. We plot this: Notice tht x (t) = 3t, so x () =, nd y (t) = 5t 4, so y () =. We cn see from the picture tht the curve seems to hve horizontl tngency. In fct, note tht x, y stisfy y = x 5/3, nd therefore = 5 3 x/3, so t x = this slope is, in fct, zero. We cn lso obtin verticl tngency by flipping the role of x, y in the previous exmple: x = t 5, y = t 3, t [ 1, 1]. 3
4 Also, note tht we cn ctully obtin ny slope we wnt t n indeterminte point. For exmple, consider x = t 3, y = αt 3. Note tht t t =, both of these functions hve zero derivtive. But we cn see tht y = αx, so the curve is just line with slope α. We cn choose α to be ny number we wnt to get ny slope we wnt. Bsiclly, when both functions hve zero derivtive, nything cn hppen! 4 Are We might now try to compute the re underneth prmetric curve, or more specificlly, the (signed) re between curve nd the x-xis. Now, if we hve curve y = f(x) where x [, b], then the re under the curve is given by Now consider the prmetric curve f(x) = y. x = f(t), y = g(t), t [c, d], where f(c) = nd f(d) = b. Writing = f (t), we then obtin y = d c g(t)f (t). In the cse where we hve x = t, y = g(t), notice tht this recovers the stndrd integrl, but this llows us to consider the generl prmetric cse. 4
5 For exmple, wht is the re under one lef of the cycloid? We hve r(1 cos t) r(1 cos t) = r (1 cos t) = r 1 cos t + cos t = r 1 cos t + 1 (1 + cos t). The two trig functions will hve zero verge, so the finl nswer we obtin here is 5 Arc Length r 3 = 3πr. Recll our derivtion of rc length erlier in the semester. If we brek up curve into mny smll line segments, notice tht we hve ds = +, nd thus our integrnd is d c (/) + (/) For exmple, if we wnt the length of one lef of the cycloid, we compute r (1 cos t) + r sin t = r (1 cos t). To do this integrl, let us recll the trig formul Solving gives This mens we cn lso write So Then our integrl is r cos t = 1 (1 cos t). cos t = cos t 1 = 1 sin t. 1 cos t = 1 (1 sin (t/)) = sin (t/). (1 cos t) = 4 sin (t/) = sin(t/). sin(t/) = 4r cos(t/) t=π t= = 4r( 1 1) = 8r. We cn lso consider the rc length of circle, which is much simpler clcultion. A circle of rdius one is prmeterized by x = cos t, y = sin t, t [, π]. sin t + cos t = = π. Of course we lre knew tht. 5