A: Derivatives of Circular Functions. ( x) The central angle measures one radian. Arc Length of r

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1 4: Derivatives of Circular Functions an Relate Rates Before we begin, remember tat we will (almost) always work in raians. Raians on't ivie te circle into parts; tey measure te size of te central angle in a sector of a unit circle wit a certain arc lengt (wew!). Tus, raians really measure te lengt of an arc. Te central angle measures one raian r r Arc Lengt of r Figure - Raian Measure In matematics, te raian is te preferre (an in many cases, only) option. In applie situations, egrees are preferre altoug certain applications require tat raians be use; wic forces conversions at te beginning an en of te problem! You soul be able to convert between egrees an raians. Since raians eal wit arc lengt, ten a full circle angle (360º) is equal to a number of raians tat is te same as te circumference (πr). Te angle part of tat is just π just remember tat 80º π (raians), an you'll always be able to set up a proportion to convert between egrees an raians. A: Derivatives of Circular Functions Let's jump in from first principles! Start wit y sin sin( + ) sin sin cos + cos sin sin y lim lim 0 0 (use te sum/ifference ientities you've learne ) sin cos sin cos sin lim + 0 sin cos sin cos sin lim + lim 0 0 cos sin (factor out te "constants") sin lim + cos lim 0. Once again, we've 0 reuce te problem to te point were we ave some limits tat nee investigating. So look at te graps! Te grap of Te grap of cos aroun 0 : sin aroun 0 : HOLLOMAN S IB HIGHER LEVEL YEAR HH NOTES 4, PAGE OF 5

2 sin an cos It appears tat lim 0 lim ; tus, 0 0 cos sin sin lim + cos lim cos 0 sin cos 0. I'll leave it to you to try tis wit te cosine function! Te result will be cos sin. For tangent, use te previous two results, te ientity tan tan sec. HOLLOMAN S IB HIGHER LEVEL YEAR HH NOTES 4, PAGE OF 5 B: Derivatives of Reciprocal Circular Functions sin an te quotient rule! cos You can use eiter te quotient rule, or te general power rule to fin te erivatives of sec cos, csc sin, an cot tan! sec sec tan csc csc cot cot csc C: Derivatives of Inverse Circular Functions Te Inverse Circular Functions In orer for a function to ave an inverse, te function must be one to one. Unfortunately, te circular functions on't ave tat property! Tis is a problem but it can be fie by restricting te omain of te function. Once we o tat, ten we can talk about te arc functions te inverse functions of te circular functions. If f sin on te interval, ten f sin ( ) arcsin ( ) on te interval [-, ]. Note tat tere are still a few people out tere wo migt talk about Arcsin () along wit arcsin () te capitalize version isn't a real inverse function; it completely cancels te sine function wen compose. Tis is a any feature, but a ar one to legitimize. Here are te oter inverse functions: Table - Inverse Circular Functions Inverse Function Domain Range arcsine [-, ]

3 arccosine [-, ] [0, π] arctangent R No one argues about tese everyone recognizes tem. Sometimes you may encounter inverse functions for te oter tree functions you will not ever fin tem on a calculator, owever. Derivatives of Inverse Circular Functions Rewrite te functions, ifferentiate implicitly, ten solve! y y arcsin sin( y). Differentiate: cos( y). Tis will require an ientity to get y y y sin y. Use a similar process/ientity to o cosine an tangent! In summary: arcsin arccos arctan + ri of te y: D: Maima/Minima wit Trigonometry Er wor problems. 'Nuff sai. E: Relate Rates We know all about relating two variables in equations wat we aven't one is investigate ow te rate of cange of one variable is relate to te rate of cange of anoter variable (or several oter variables), wit respect to a anoter variable (time). Di I mention tat we're talking about wor problems applications of calculus? Te Tecnique You're going to nee to write an equation tat relates te two variables. Ten, take te implicit erivative wit respect to t (time). Now you soul be able to plug in te given values an solve for te requeste item. Peraps tis soul be illustrate wit some eamples Eamples [.] A rock is roppe into a pon, wic causes a circular ripple. Te ripple moves away from te center (were te rock was roppe) at m/s. How fast is te area of te circle canging wen te raius is 3m? HOLLOMAN S IB HIGHER LEVEL YEAR HH NOTES 4, PAGE 3 OF 5

4 Let's inventory wat was given in te problem. Te ripple moves away from te center at m/s: tis is te rate of cange of te raius; r t. How fast is te area of te circle canging : tis is asking for te rate of cange in te area; A t. wen te raius is 3m: r 3. So, te variables are raius, area, an time. We nee to write an equation tat relates te area an te raius. Any ieas? Of course! Aπ r. Now tat we've got an equation, we nee to ifferentiate wit respect to t. A r π r. t t Now, we're trying to solve for A : tat means we nee to plug in value for everyting else t r an r t. r Joy! Tey were given as an r 3! t A ( 3) 6 t π π. Wen te raius is 3m, te area is canging at 6πm²/s. [.] Gravel is being poure out of a container onto te groun at a rate of y³/min. Te gravel forms a cone as it lans, were te eigt of te cone is about alf of te iameter. How fast is te eigt of te cone rising wen te iameter of te pile is 3y? V, r.5, ½ r, an we're aske about. We nee an equation tat relates V, r t t an V V 3 π r, so r π 3 r r t + t t. Plug in! r π t t. Oops wat are we going to o about r an? Well, r, so.5. Wat about r t t? Well, if r, ten r! So we get π t t + t t. Now we just nee to solve for t HOLLOMAN S IB HIGHER LEVEL YEAR HH NOTES 4, PAGE 4 OF 5

5 Factor! π + means, t π (.5)(.5) + (.5) t 6 6,,. Tat gives an approimate rate of π (.5) + (.5) t 3π (.5) t π(.5) t cange of 0.89y/min. [3.] Strane on Tatooine, R-D an C-3PO ave ecie to split up in te esert. R-D moves ue nort at 7km/r, an C-3PO moves ue east at 5km/r. How fast is te istance between tem increasing after ours? Let's let n istance nort (of R-D), e istance east (of C-3PO), an D istance n between te two. So we're given 7 t, e t 5, an t. Te only equation we can write tat relates tese is n² + e² D². Of course, tat means n e D n + e D. Wat are we going to plug in for n, e an D? How will we use te value t t t t? Of course! At t, n 4, e 0, an ( 4)( 7) + ( 0)( 5) 96 D, so t D D t 96 48, or approimately 8.60km/r HOLLOMAN S IB HIGHER LEVEL YEAR HH NOTES 4, PAGE 5 OF 5