MultivariateCalculus;Fall2013 S.Jamshidi 4.3 Arc Length and Curvature Objectives ⇤ Iknowhowtocalculatearclength. ⇤ Iknowhowtofindtheunittangentvector. ⇤ Iknowtwodi↵erentthree-dimensionalequationsforcurvatureandIknowonetwo- dimensional equation for curvature. ⇤ I know how to calculate the normal vector and the binormal vector. A fundamental question in physics is, “How far did we go?” Since the paths travelled are described by displacement functions, which are vector functions, we want to come up with some kind of method of measuring this length. Looking at the picture left, how might you accomplish this? Let’s begin with what we know. We know how to measure the distance of a straight line. If we approximate a complicated curved surface like the one in the picture with line segments, we could measure those line segments and approximate the distance. Consider the curve below. We can approximate it by a straight line connecting the end points (in green). That’s clearly an under- estimate. We could also approximate the curve with three line segments (in gray), which is much better than the first one. Notice, that we can do better by using more line segments (like that in blue). Let us recall the distance formula. (∆x)2 +(∆y)2 +(∆z)2 p 75 of 134 MultivariateCalculus;Fall2013 S.Jamshidi We want to sum this over each line segment. n 2 2 2 (∆xi) +(∆yi) +(∆zi) i=0 X p If you recall the definition of an integral, then you’ll notice that letting these line segments get infinitesimally small will produce an integral and derivatives! That is, b dx 2 dy 2 dz 2 Arc Length from a to b = + + dt a s dt dt dt Z ✓ ◆ ✓ ◆ ✓ ◆ Put another way, b 2 2 2 Arc Length from a to b = (f 0(t)) +(g0(t)) +(h0(t)) dt Za q or b Arc Length from a to b = ~r 0(t) dt | | Za These equations aren’t mathematically di↵erent. They are just di↵erent ways of writing the same thing. 4.3.1 Examples Example 4.3.1.1 Find the length of the curve ~r (t)= 3cos(t), 3sin(t),t h i when 5 t 5. − However you choose to think about calculating arc length, you will get the formula 5 L = ( 3sin(t))2 +(3cos(t))2 +(1)2dt 5 − Z− We can simplify this integral withp the equation sin2 t +cos2 t =1 76 of 134 MultivariateCalculus;Fall2013 S.Jamshidi That gives us 5 L = 9[sin2(t)+cos2(t)] + 1 dt 5 − Z 5 q = p9+1dt 5 − Z 5 = p10 dt 5 Z− =5p10 ( 5)p10 = 10p10 − − Example 4.3.1.2 Find the length of the curve t ~r (t)= p2t, e ,e−t D E when 0 t 1. The formula tells us 1 L = (p2)2 +(et)2 +( e t)2dt − − Z0 q This integral looks very complicated. How can we solve it? We need to use some clever algebra. 1 2 t 2 t 2 L = (p2) +(e ) +( e− ) dt 0 − Z 1 q 2t 2t = p2+e + e− dt 0 Z 1 t 2t 4t 2t = e− p2e + e +1dt we factor out e− ,thentakeitoutofthesquareroot 0 Z 1 t 2t 2 = e− (e +1) dt Z0 1 p t 2t = e− (e +1)dt 0 Z 1 t t = e + e− dt Z0 1 = e + e 77 of 134 MultivariateCalculus;Fall2013 S.Jamshidi In addition to length, we’d like to have some idea of the curvature of a path. For example, when probes are sent in outer space, engineers care a great deal about how many turns it must take since this impacts fuel consumption. But how do we measure curvature? We want to measure the amount a curve changes direction. Since it is a measurement, it will be a scalar value. When we imagine a curvy line like the one below, we see that the vector ~r 0(t)inrelationtothe path created by ~r (t)changesdirectionwildly. Notice, however, that not only do these vectors change direction, they also change in size! The blue vectors are longer than the green vector, which is longer than the red vectors. We don’t want to measure changes in magnitude (which would correspond to speed), we only want to look at the direction. Enter the unit tangent vector.Thisisavectorwhoistangenttothecurvebutlength1.It is defined to be ~r (t) T~(t)= 0 ~r (t) | 0 | When we graph the curve with the unit tangent vectors for the same values of t as above, we get a set of vectors that only change in direction. We still have not addressed curvature. Curvature will be the amount T~(t)changesaswetravel along a segment. How can we measure this? If we think about it, this is the change of T~(t)with respect to the changes in arc length up to t.Thissecondfunctionisrepresentedas t s(t)= ~r 0(u) du | | Za It is the same equation we had for arc length earlier except our end point is the variable t. 78 of 134 MultivariateCalculus;Fall2013 S.Jamshidi Therefore, curvature, (t), is dT~ (t)= ds Remember, measurements are scalar values, so we need to take the magnitude. Looking at this equation, we can make two observations: 1. It makes perfect sense as a definition for what is happening. 2. It’s stupidly unusable. Let’s be honest, finding s(t)isapainandwhatdoesdT/ds~ really mean? We need to do a little more math to get a more usable equation. First, remember that di↵erentials work like fractions. That’s one of the reasons they are written that way. Next, make the following observation: dT~ dT~ dT~ dt = = dt ds dt ds ds dt The fundamental theorem of calculus tells us t if s(t)= ~r 0(u) du, then s0(t)= ~r 0(t) | | | | Za Therefore, we get the following equation: dT~ dT~ T~ (t) = dt = 0 ds ds ~r (t) | 0 | dt Since curvature is the magnitude of the above vector, we get dT~ T~ 0(t) (t)= = | | ds ~r 0(t) | | 79 of 134 MultivariateCalculus;Fall2013 S.Jamshidi 4.3.2 Examples Example 4.3.2.1 Let ~r (t)= t, 3cost, 3sint . Find the curvature of the path at time t. h i To calculate this, we’re going to follow these steps 1. Find ~r 0(t). 2. Find ~r 0(t) .Simplifywhenpossible. | | 3. Find T~(t). 4. Find T~ 0(t). 5. Find T~ 0(t) .Simplifywhenpossible. | | 6. Divide (5) by (2) to get (t) We’ll often get messy answers, but that’s ok. Let’s do each step. 1. Find ~r 0(t). ~r 0(t)= 1, 3sint, 3cost h − i 2. Find ~r 0(t) .Simplifywhenpossible. | | 2 2 2 ~r 0(t) = 1 +( 3sint) +(3cost) | | − = p12 +9 = p10 3. Find T~(t). ~r (t) T~(t)= 0 ~r (t) | 0 | 1, 3sint, 3cost = h − i p10 1 3sint 3cost = , − , p10 p10 p10 ⌧ 80 of 134 MultivariateCalculus;Fall2013 S.Jamshidi 4. Find T~ 0(t). 3cost 3sint T~ 0(t)= 0, − , p10 p10 ⌧ 5. Find T~ 0(t) .Simplifywhenpossible. | | 9cos2 t 9sin2 t 3 T~ 0(t) = 0+ + = | | r 10 10 p10 6. Divide (5) by (2) to get (t) 3 3 1 3 (t)= p10 = = p10 p10 · p10 10 So curvature for this equation is a nonzero constant. This means that at every time t,we’re turning in the same way as we travel. The graph shows exactly this kind of movement As you might guess, doing donuts with your car would also result in constant nonzero curvature. Now, let’s look at a messier example. 81 of 134 MultivariateCalculus;Fall2013 S.Jamshidi 1 2 2 Example 4.3.2.2 Let ~r (t)= t, 2 t ,t . Find the curvature of the path at time t. We will go through the steps.⌦ As you↵ might expect, they will not simplify nicely. 1. Find ~r 0(t). ~r 0(t)= 1,t,2t h i 2. Find ~r 0(t) .Simplifywhenpossible. | | 2 2 2 ~r 0(t) = p1+t +4t = p1+5t | | 3. Find T~(t). 1 t 2t T~(t)= , , p1+5t2 p1+5t2 p1+5t2 ⌧ 4. Find T~ 0(t). This is where is starts to get ugly. For the derivatives, I used product rule; however, you could use quotient rule as well. This might be a better approach since it keeps everything in one fraction. Also, keep in mind that z(t)isjust2y(t). I will use this to do less work. 10t 10t2 1 10t2 1 T~ 0(t)= − , − + , 2 − + 2(p1+5t2)3 2(p1+5t2)3 p1+5t2 2(p1+5t2)3 p1+5t2 ⌧ ✓ ◆ 5t 5t2 (p1+5t2)2 5t2 (p1+5t2)2 = − , − + , 2 − + *(p1+5t2)3 (p1+5t2)3 (p1+5t2)3 (p1+5t2)3 (p1+5t2)3 !+ 5t 5t2 +1+5t2 5t2 +1+5t2 = − , − , 2 − (p1+5t2)3 (p1+5t2)3 (p1+5t2)3 ⌧ ✓ ◆ 5t 1 1 = − , , 2 (p1+5t2)3 (p1+5t2)3 (p1+5t2)3 ⌧ ✓ ◆ 1 = 5t, 1, 2 (p1+5t2)3 h i 82 of 134 MultivariateCalculus;Fall2013 S.Jamshidi 5. Find T~ 0(t) .Simplifywhenpossible. | | 1 T~ 0(t) = 5t, 1, 2 | | (p1+5t2)3 |h− i| 1 = p25t2 +1+4 (p1+5t2)3 p25t2 +5 = (p1+5t2)3 p5p5t2 +1 = (p1+5t2)3 p5 = (p1+5t2)2 p5 = 1+5t2 Notice that I did one trick that made all my calculations easy. I factored out a constant and then took the magnitude of the simplified vector. If every entry has a constant in common, when we take the magnitude, the positive version of that constant comes out. Since our coefficient is always positive (t2 0), we can do this without any problem.