Math 53 Worksheet Solutions - Polar Coordinates and VectorsNote: The solutions have not been double-checked for accuracy, so there may be somenumerical errors.1. Find the points on r = 1 − sin θ where the tangent line is horizontal or vertical.Solution. Using x = r cos θ = (1 −sin θ) cos θ, y = r sin θ = (1 − sin θ) sin θ, we havedxdθ= −cos2θ − sin θ(1 −sin θ),dydθ= −cos θ sin θ + cos θ(1 − sin θ).As a result,dydx=cos θ(1 − 2 sin θ)sin2θ − cos2θ − sin θ.To have a horizontal tangent, the numerator must be zero and the denominator mustbe nonzero. For a vertical tangent, the reverse must occur. And if both are zero, wemust employ L’Hopital. Solving cos θ(1 − 2 sin θ) = 0, we have either cos θ = 0 orsin θ =12. The solutions are θ =π6,π2,5π6,3π2. Solving sin2θ − cos2θ − sin θ = 0, werewrite the equation as2 sin2θ − sin θ − 1 = 0,and the quadratic formula offerssin θ =1 ± 34,so sin θ = 1 or sin θ = −12. Thus θ =π2,7π6,11π6. The only overlap between these two isπ2, so we must apply L’Hospital. We havelimθ→π2cos θ(1 − 2 sin θ)sin2θ − cos2θ − sin θ= limθ→π2−sin θ(1 − 2 sin θ) − 2 cos2θ4 sin θ cos θ − cos θ.Though this limit is undefined, it’s numerator’s limit is nonzero and denominator’slimit is zero. Thus, in terms of slope, it corresponds to a vertical tangent line. Thusθ =π6,5π6,3π2all correspond to points with horizontal tangent lines, whereasθ =π2,7π6,11π6all correspond to points with vertical tangent lines. From this, you cancalculate the r values and write the points as ordered pairs.2. Show that the curves r = a sin θ and r = a cos θ intersect at right angles.Solution. The curves intersect when tan θ = 1, or when θ =π4or θ =5π4. For the firstcurve, which we’ll call r1, we calculatedy1dx1=ddθa sin2θddθ(a sin θ cos θ)=2 sin θ cos θcos 2θ,sody1dx1θ=π4= ∞,dy1dx1θ=5π4= ∞,1where we are liberally interpreting an undefined derivative as infinite. (This is onlyappropriate in the context of slopes of “nice” functions.) We also havedy2dx2=cos 2θ−2 cos θ sin θ,by similar reasoning, anddy2dx2θ=π4= 0,dy2dx2θ=5π4= 0.As zero slope is horizontal and infinite slope is vertical, we’re done.3. Find the exact length of the curve r = e2θ, 0 ≤ θ ≤ 2π.Solution.L =Z2π0sr2+drdθ2dθ=Z2π0pe4θ+ 4e4θdθ=√5Z2π0e2θdθ=√512e2θ2π0=√52e4π− 1.4. Find the area between a large loop and the enclosed small loop of the curver = 1 + 2 cos 3θ.Solution. Draw a picture and use symmetry, so you need only evaluate half of theloops. The large loop has areaA1= 2Z2π/9012(1 + 2 cos 3θ)2dθ = ··· =2π3+√32.The smaller loop has areaA2= 2Zπ8π/912(1 + 2 cos 3θ)2dθ = ··· = −√32+π3.The area between them is thusA1− A2=π3+√3.25. Consider the points P such that the distance from P to A(−1, 5, 3) is twice thedistance from P to B(6, 2, −2). Show that the set of all such points is a sphere, andfind its center and radius.Solution. The distance equation isp(x + 1)2+ (y − 5)2+ (z − 3)2= 2p(x − 6)2+ (y − 2)2+ (z + 2)2.Square both sides, complete the square, and obtainx −2533+ (y − 1)2+ (z + 5)2= 2√1093!2.6. Describe in words the region of R3represented by the equation x2+ y2+ z2> 2z.Solution. Complete the square with z. Rewrite asx2+ y2+ (z − 1)2> 1.Region is that outside a sphere of radius 1 centered at (0, 0, 1).7. If r = hx, yi, r1= hx1, y1i, and r2= hx2, y2i, describe the set of all points (x, y) suchthat |r − r1| + |r − r2| = k, where k > |r1− r2|.Solution. By definition, an ellipse.8. Find the angle between a diagonal of a cube and one of its faces.Solution. Use the point r1= (1, 1, 0) as the diagonal of one of its faces andr2= (1, 1, 1) as the diagonal of the cube. Thencos θ =r1· r2|r1| |r2|=2√2√3=r23.Thenθ ≈ 0.61 = 35.3◦.9. If c = |a|b + |b|a, where a, b, and c are all nonzero vectors, show that c bisects theangle between a and b.Solution. Draw the parallelogram. The key is that ||a|b| = |a| |b| and ||b|a| = |a| |b|.310. Use Theorem 3 from Section 12.3 to prove the Cauchy-Schwarz Inequality|a · b| ≤ |a| |b|.Solution.|a · b| = |a| |b| |cos θ| ≤ |a| |b|,since |cos θ| ≤ 1.Challenge ProblemsC1. The Triangle Inequality for vectors is|a + b| ≤ |a| + |b|.Give a geometric interpretation of the Triangle Inequality, and then use theCauchy-Schwarz inequality from problem 10 to prove it.Solution.|a + b|2= (a + b) ·(a + b)= a · a + 2a · b + b · b= |a|2+ 2a ·b + |b|2≤ |a|2+ 2|a| |b| + |b|2= (|a| + |b|)2Now take square roots.C2. The Parallelogram Law states that|a + b|2+ |a −b|2= 2|a|2+ 2|b|2.Give a geometric interpretation of the Parallelogram law and then prove it.Solution.|a + b|2+ |a −b|2= (a · a + 2a · b + b · b) + (a · a − 2a ·b + b · b)= 2a · a + 2b · b= 2|a|2+ 2|b|2.The law states that the sum of the squared lengths of the diagonals of a parallelogramare equal to twice the sum of the squared side