The values n 0 1 2 will suffice to generate all cases

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Calculus: Early Transcendental Functions
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Chapter 13 / Exercise 53
Calculus: Early Transcendental Functions
Edwards/Larson
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, the valuesn= 0,-1,-2will suffice to generate all cases. Whenn=-1, we getr= 0, theorigin again.The second case gives-sin(θ+ 2+π) = sin(2θ)exactly as in the first case.The points of intersection are(x, y) = (0,0),θ=π3,r= sin(π3),(x, y) = (34,34),θ=5π3,r= sin(5π3),(x, y) = (-34,34),Figure 49. Polar curvesr= sin(θ)in blue andr= sin(2θ)in red.The figure shows what is happening. The only thing to note is that forθ=5π3,both values ofrare negative and the point is in the second quadrant.91
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Calculus: Early Transcendental Functions
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Chapter 13 / Exercise 53
Calculus: Early Transcendental Functions
Edwards/Larson
Expert Verified
We look at a moving particle in polar coordinates. We use a pair of vectorsk(θ) =(cos(θ),sin(θ))andl(θ) = (-sin(θ),cos(θ)). Note thatdk(θ) =l(θ)anddl(θ) =-k(θ)Now the position vector of a point given in polars isrk(θ). Therefore withtdenotingtime, the velocity isddtrk(θ)=drdtk(θ) +rdk(θ)dt=drdtk(θ) +rdtl(θ).This gives the velocity of the particle in the radial and transverse to radial directions.Differentiating again, we getd2dt2rk(θ)=ddtdrdtk(θ) +rdtl(θ)=d2rdt2k(θ) +drdtdk(θ)dt+drdtdtl(θ) +rd2θdt2l(θ) +rdtdl(θ)dt=d2rdt2-rdt2k(θ) +rd2θdt2+ 2drdtdtl(θ)This gives the acceleration of the particle in the radial and transverse to radial direc-tions.1.25Arclength of polar curves - March 7Givenr=r(θ)a polar equation we rewrite it as a parametric equationx(θ) =r(θ) cos(θ),y(θ) =r(θ) sin(θ)and computedx=drcos(θ)-rsin(θ),dy=drsin(θ) +rcos(θ)Then, we getdx2+dy2=dr2cos(θ)2-2rdrsin(θ) cos(θ) +r2sin(θ)2+dr2sin(θ)2+ 2rdrsin(θ) cos(θ) +r2cos(θ)2=dr2+r2.92
Thereforeds=dx2+dy2=dr2+r2dθ.This is the formula for computing arclengths in polars.Another way of thinking about this is thats)2r)2+ (rΔθ)2, givingΔsΔθΔrΔθ2+r2ExampleFind the arclength of the cardioid curver= 1-cos(θ)The range ofθto go around the curve once is0θ2π. We haveds=dr2+r2=sin(θ)2+ (1-cos(θ))2=2-2 cos(θ)=4sinθ22= 2 sinθ2The arclength is then2π0ds=2π02 sinθ2=2π02 sinθ2=-4 cosθ22π0= 8ExampleFind the arclength of the spiral curver=θ2for0θ2π.We haveds=dr2+r2=(2θ)2+θ4=|θ|4 +θ2The desired length is2π0ds=2π0|θ|4 +θ2=4π2+44u12·12du=83(π2+ 1)32-1sinceθ|geq0on the range of integration and after the substitutionu=θ2+ 4,du=2θdθ.ExampleFind the arclength of the curver= cosθ44.First, we need to discuss the range ofθ. We haveθ= 0,r= 1.93
θ=π,r=14.θ= 2π,r= 0.θ= 3π,r=14.θ= 4π,r= 1.withrdecreasing on0θ2πand increasing on2πθ4π. We are back towhere we started whenθ= 4π.Figure 50. The polar curver= cosθ44.We finddr=-sinθ4cosθ43and it follows thatds= cosθ43. Thearclength is now given by4π0cosθ43= 22π0cosθ4394
by symmetry. This is a trig integral and we solve by substituting

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