Math 53 Worksheet Solutions- Vector Functions and Derivatives1. At what points does the helix r(t) = hsin t, cos t, ti intersect the spherex2+ y2+ z2= 5?Solution. We havesin2t + cos2t + t2= 5,so t2= 4 and t = ±2 are where we have intersection. The points are thenP1= r(2) = hsin 2, cos 2, 2i, P2= r(−2) = h−sin 2, cos 2, −2i.2. Find a vector function that represents the curve of intersection of the paraboloidz = 4x2+ y2and the parabolic cylinder y = x2.Solution. We have r(t) = hx(t), y(t), z(t)i. Set x(t) = t. Then y(t) = x(t)2= t2. Andz(t) = 4x(t)2+ y(t)2= 4t2+ t4. Thenr(t) = ht, t2, 4t2+ t4i.3. Find parametric equations for the tangent line to the curvex = ln t, y = 2√t, z = t2,at the point (0, 2, 1).Solution. The position vector is r(t) = hln t, 2√t, t2i, so the tangent vector isr0(t) = h1t,1√t, 2ti. The point (0, 2, 1) corresponds to t = 1, since x(t) = ln t = 0 impliesthat t = 1. Then the tangent vector at the point (0, 2, 1) isv = r0(1) = h1, 1, 2i.So the parametric equation of the line isr(t) = h0, 2, 1i + th1, 1, 2i = ht, 2 + t, 1 + 2ti.4. If u(t) = r(t) ·[r0(t) × r00(t)], show thatu0(t) = r(t) · [r0(t) × r000(t)].Solution. Laws of derivatives for cross products and dot products. We haveddt(r(t) · [r0(t) × r00(t)]) = r0(t) · [r0(t) × r00(t)] + r(t) ·ddt[r0(t) × r00(t)]= 0 + r(t) · (r00(t) × r00(t) + r0(t) × r000(t))= r(t) · (r00(t) × r00(t)) + r(t) · (r0(t) × r000(t))= r(t) · (r0(t) × r000(t))15. Reparametrize the curver(t) = e2tcos 2ti + 2j + e2tsin 2tkwith respect to arc length measured from the point where t = 0 in the direction ofincreasing t.Solution. First we find s(t), the arc length of the curve between t = 0 and t = t. Wecalculater0(t) = h2e2tcos 2t − 2e2tsin 2t, 0, 2e2tsin 2t + 2e2tcos 2ti.Thens(t) =Zt0|r0(u)|du=Zt0q(2e2u cos 2u −2e2usin 2u)2+ (2e2usin 2u + 2e2ucos 2u)2du= 2Zt0e2uq(cos 2u − sin 2u)2+ (sin 2u + cos 2u)2du= 2Zt0e2upcos22u + sin22u − 2 cos 2u sin 2u + sin22u + cos22u + 2 sin 2u cos 2u du= 2Zt0e2u√2 du=√2e2u u=tu=0=√2e2t− 1.Now we want to invert this to find t = t(s). So we solve for t. Thuss√2+ 1 = e2t,and thent =12lns√2+ 1.Substitute this in to r(t) and call the new parametrization r0(s). We haver0(s) = r(t(s)) = hs√2+ 1cos lns√2+ 1, 2,s√2+ 1sin lns√2+ 1i,after reducing with basic logarithm rules.6. What force is required so that a particle of mass m has the position functionr(t) = t3i + t2j + t3k?Solution. We know F = ma.a(t) = r00(t) = h6t, 2, 6ti,soF(t) = ma(t) = h6mt, 2m, 6mti.27. (Challenge Problem) A cable has radius r and length L and is wound around a spoolwith radius R without overlapping. What is the shortest length along the spool that iscovered by the cable?Solution. We look at points in the center of the bottom of the cable. These pointsform a helix. In fact, the parametrization can readily be determined from a simplepicture. It isx = (R + r) cos t, y = (R + r) sin t, z = ht2π,where R + r comes from the distance from the center of the cable to the point at thecenter of the bottom of the cable, which is one large radius and one small radius. Hereh is the vertical distance between consecutive loops; as the cable makes one loop in 2πunits of t, h is the distance traveled upwards in one loop. As soon as we find h, we canfind the arc length.Draw a picture. The width of the small cable is 2r. The height change of one fullrotation in h. That same height change can be achieved by moving 2π(r + R) unitsahead on the cable. Drawing these distances as straight lines that create triangles, asimilar triangle argument produces2r2π(r + R)=√h2− 4r2h,and so cross-multiplying and solving for h gives ush =2π(r + R)rpπ2(r + R)2− r2.Now, the length of one cycle, `, around the main axis is` =Z2π0px0(t)2+ y0(t)2+ z0(t)2dt=Z2π0s(R + r)2+h2π2dt= 2πs(R + r)2+(r + R)2r2π2(r + R)2− r2= 2π(R + r)sπ2(r + R)2− r2+ r2π2(r + R)2− r2=2π2(R + r)2pπ2(r + R)2− r2.3The number of complete cycles is n = floor(L/`), where the floor function gives thelargest integer less than the given number, in this case L/`. The shortest length isthen along the length before the final cycle closes, and this length istotal length = hn = 2π(r + R)rpπ2(r + R)2− r2!· floor Lpπ2(r + R)2− r22π2(R +