Dr. Z’s Math251 Handout #13.2 [Derivates and Integrals of Vector Functions]By Doron ZeilbergerProblem Type 13.2a: Find a parametric equation for the tangent line to the curve with thegiven parametric equation at the specified pointx = f1(t) , y = f2(t) , z = f3(t) ; P (p1, p2, p3)Example Problem 13.2a: Find a parametric equation for the tangent line to the curve with thegiven parametric equation at the specified pointx = t2− 1 , y = t2+ 1 , z = t + 1 ; (−1, 1, 1)Steps Example1. Find the relevant t at the designatedpoint. Let’s call it t0. Solve the equationsf1(t0) = p1, f2(t0) = p2, f3(t0) = p3. Ifthere is no solution refuse to do the prob-lem.1. −1 = t20− 1 , 1 = t20+ 1 , 1 = t0+ 1means t0= 0.2. Putting r(t) = hx(t), y(t), z(t)i, takethe derivative r0(t) by doing it component-by-componentr0(t) = hx0(t), y0(t), z0(t)i .2.r(t) = h t2− 1 , t2+ 1 , t + 1 ir0(t) = h (t2−1)0, (t2+1)0, (t+1)0i = h 2t , 2t , 1 i .3. Plug-in the specific value (t = t0), thatyou found in step 1, to get r0(t0), which isthe direction vector, let’s call it D. andusing the given point P as the startingpoint, the parametric equation of theline is hx, y, zi = P + tD. Finally, spellout the expressions for x, y, z.3. The direction vector is h 2t , 2t , 1 i plugged-in at t = 0, it is h0 , 0 , 1 i, and since thepoint is h −1 , 1 , 1 i the equation of thetangent line, in vector-form ish −1 , 1 , 1 i+th 0 , 0 , 1 i = h −1 , 1 , 1+t i ,and spelling it outx = −1 , y = 1 , z = 1 + t .This is the Ans..1Problem Type 13.2b: Find r(t) ifr0(t) = f1(t) i + f2(t) j + f3(t) kandr(t0) = a i + b j + c kExample Problem 13.2b: Find r(t) ifr0(t) = 2t i + 3t2j + 4t3kandr(1) = i + 4 j + 5 kSteps Example1. Find the indefinite integral by inte-grating every component and not forget-ting to add an arbitrary constant thatnow is a vector.1.r(t) =Z(2t i + 3t2j + 4t3k) dt= t2i + t3j + t4k + C .2. Plug-in t = t0and solve for C. 2.r(1) = 12i + 13j + 14k + C =i + j + k + C = i + 4 j + 5 k .Solving for C gives C = 3 j + 4 k3. Go back to step 1 and incorporate thespecific C that you’ve found in step 2.3.r(t) = t2i + t3j + t4k + 3 j + 4 k= t2i + (t3+ 3) j + (t4+ 4) k .This is the