Math 0230 Calculus 2 Lectures

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1 Mth Clculus Lectures Chpter 9 Prmetric Equtions nd Polr Coordintes Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition Section 91 Prmetric Curves Prmetric equtions: x = ft, y = gt, where ft nd gt re continuous functions They define curve in the xy-plne when t runs long n intervl [, b]: x, y = ft, gt, when t b Initil point of the curve is f, g nd terminl point is fb, gb In the prmetric form y is not necessrily must be function of x Exmple 1 Sketch the curve x = t 1, y = t, when t 4 It is prbol: t x y Sketch its grph Algebric justifiction tht it i prbol: t = x + 1, y = x + 1 = x + x Exmple Sketch the curve x = cos t, y = sin t, when t π It is the unit circle: x + y = 1 b Sketch the curve x = h + r cos t, y = k + r sin t, when t π It is circle centerd t h, k with rdius r: x h + y k = r cos t + r sin t = r 1

2 Section 9 Clculus with Prmetric Curves Tngents x = ft, y = gt Here we ssume tht ft nd gt re differentible functions y cn be expressed s y = xt Using the Chin Rule we get = or = / / = y x The derivtive evluted t point x, y, such tht x = xt, y = yt for some t, is slope of the tngent line to the prmetric curve t tht point If =, then the tngent is horizontl If =, then the tngent is verticl For the second derivtive, d d y = d = Exmple 1 The curve x = cos t, y = sin t, t π is n ellipse Find n eqution of the tngent line to the ellipse t point where t = π/ b Find ll points on the ellipse t which the tngent line is verticl Slope m = / when t = π/ / = cos t = cosπ/ = / t=π/ t=π/ = sin t = sinπ/ = t=π/ t=π/ So, m = = The point on the curve is cos π/, sin π/ = The point slope form of the tngent line is y = b =, sin t =, t =, t = π 1, x 1 or y = x +

3 Ares The re under the curve given by prmetric equtions x = ft, y = gt, α t β is β A = gtf t or α α A = gtf t if fβ, gβ is the most left endpoint β Exmple Find the re enclosed by the ellipse x = cos t, y = sin t, t π A = π sin t sin t = 1 sin t = 6 π 1 cos t [ = 6 t 1 ] π sin t π = 6 [π ] = 6π Arc Length The length L of the curve x = ft, y = gt, α t β is β α + Exmple Find the length L of the unit circle x = cos t, y = sin t, t π = sin t, = cos t Then, π π sin t + cos t = = π

4 Section 9 Polr Coordintes Polr vs Crtesin coordintes x = r cos θ, y = r sin θ, x + y = r, tn θ = y x Exmple 1 Convert the point 4, from polr to Crtesin coordintes r = 4, θ =, x = 4 cos = 4 =, y = 4 sin = 4 1 =,, Exmple Convert the points 1, b 1, c 1, d 1, from Crtesin to polr coordintes x = ±1, y = ±, r = 1 + = tn θ =, θ = π/ The point is, π b tn θ =, θ = π/ The point lies in the 4th qunt: c tn θ =, θ = π/ The point lies in the nd qunt: b tn θ =, θ = 4π/ The point lies in the rd qunt:, π =, 5π, π, 4π Polr Curves Generl eqution either r = rθ or F r, θ = Exmple Sketch the curve r = 1 Exmple 4 x + y = 1, it is the unit circle y x Sketch the curve θ = π/4 = tnπ/4 = 1, y = x, it is line Exmple 5 Sketch the curve with the polr eqution r = 4 sin θ sin θ = r r, y = r sin θ = 4 4, 4y = r = x +y, x +y 4y =, x +y 4y +4 = 4, x + y =, it is circle with center t, nd rdius 4

5 In generl Consider the polr eqution r = cos θ, where is positive constnt Multiplying both sides by r, we obtin r = r cos θ, which in rectngulr coordintes becomes x + y = x or x x + y = Next we complete the squre on the terms in prentheses by dding /4 to both sides, obtining x / + y = / Now we recognize this s the eqution of circle centered t the point /, with rdius / Similrly, the eqution r = sin θ represents circle centered t the point, / with rdius / Note tht both of these circles re trversed twice s θ vries from to π Tngents to Polr Curves r = rθ, x = r cos θ = rθ cos θ, y = r sin θ = rθ sin θ = cos θ r sin θ, = sin θ + r cos θ, = / / = If Exmple 6 θ = π/4 = nd, then the tngent is verticl sin θ + r cos θ cos θ r sin θ Find the slope of the tngent to the curve r = cos θ t the point where m = = / when θ = π/4 / = sin θ = sinπ/4 = θ=π/4 θ=π/4, rπ/4 = = θ=π/4 cosπ/4 sinπ/4 = = 1 = = θ=π/4 sinπ/4 + cosπ/4 = + = m = = + + = =

6 Section 94 Ares nd Lengths in Polr Coordintes Ares Let R be region bounded by the polr curves r = rθ, θ =, nd θ = b Its re is A = 1 b r θ If R is region bounded by the polr curves r = r 1 θ, r = r θ, θ =, nd θ = b, where r 1 θ r θ, then its re is A = 1 b r θ r 1θ Exmple 1 Find the re enclosed by one loop of the three-leved rose r = cos θ θ Hence, A = 1 cos θ = cos θ = cos 6θ = 1 [θ + 16 ] sin 6θ = 1 [ π ] 6 + = π 1 Exmple r = 1 + cos θ Find the re tht lies inside the curve r = cos θ nd outside the curve The first curve is circle centered t the point /, with rdius / The second curve is crdioid Their points of intersections: cos θ = 1 + cos θ, cos θ = 1/, θ = pi/ nd θ = π/ Hence, A = 1 π/ π/ 9 cos θ 1 + cos θ = 1 π/ π/ 8 cos θ 1 cos θ = π/ 8 cos θ 1 cos θ = π/ cos θ 1 cos θ = π/ 4 cos θ + cos θ = [ ] π/ sin θ + θ sin θ = π = π 6

7 Arc Length Let r = rθ, θ b be polr curve r = rθ, θ =, nd θ = b We regrd θ s prmeter Then x = rθ cos θ, y = rθ sin θ, = cos θ r sin θ, = sin θ + r cos θ nd + = + r And the length of the curve is b + b = r + Exmple Find the lenght of the curve r = cos θ, θ = sin θ Hence, cos θ + sin θ = = π Exmple 4 [Hrd] Find the lenght of the crdioid r = 1 cos θ θ π Hence, π 1 cos θ + sin θ = π cos θ = 8 Exmple 5 [Hrd] Find the lenght of one loop of the three-leved rose r = cos θ θ Hence, cos θ + sin θ = cos θ + 9 sin θ 7

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