Name:TEST 2No Calculators1:2:3:4:5:6:7:8:1 (16 points) A particle’s acceleration at time t is given byd2rdt2(t) = < −sin t − cos t, cos t − sin t, 0 >,initial velocitydrdt(0) = < 1, 1, 1 >, initial position r(0) = < 0, 0, 0 > .Find the particle’s position r(t) at time t and the arc length of its trajectory from timet = 0 to t = 1.Integration gives velocity r0= < cos t − sin t, sin t + cos t, 0 > +c and c needs to besuch that the initial veloc ity is correct. Hence r0= < cos t − sin t, sin t + cos t, 1 >.Integration gives p osition r = < sin t + cos t, −cos t + sin t, t > +c1and c1needs tobe such that the initial position is correct. Hencer(t) = < sin t + cos t − 1, 1 − cos t + sin t, t > .Since |r0|2= (cos t−sin t)2+(sin t+cos t)2+1 = 3 we have that the arc length equalsZ10|r0(t)|dt =√3.2 (14 points) Does f(x, y) = xy/(x−y) have a limit as (x, y) approaches (0, 0)? Justifyyour answer.No.Solving f(x, y ) = 1 gives y = x/(1 + x), hence f(x, x/(1 + x)) = 1 as x → 0.Solving f(x, y ) = −1 gives y = x/(1 − x), hence f(x, x/(1 − x)) = −1 as x → 0.By the Two-Path Test the limit does not exist.3 (14 points) Find the value of ∂z/∂x at the point (1, 1, 1) if the equationxy + z3x − 2yz = 0defines z as a function of the two independent variables x and y and the partial derivativeexists.Differentiation gives y + 3z2zxx + z3− 2yzx= 0 hence zx= −2.4 (14 points) Let w = f(r, φ), where r and φ are the polar coordinates, i.e. x = r cos φand y = r sin φ. Express wxas a function of r and φ.r =px2+ y2, φ = tan−1(y/x)rx= x/r = cos φ, φx= −y/(x2+ y2) = −(sin φ)/rwx= frrx+ fφφx= frcos φ − fφ(sin φ)/r.5 (14 points) Find the equations of tangent plane and of the normal line to the surfacez = x2− y2at the point (2, −1, 3).f = x2− y2− z = 0fx= 2x = 4, fy= −2y = 2, fz= −1, ∇f = < 4, 2, −1 >Tangent plane: 4(x − 2) + 2(y + 1) − (z − 3) = 0Normal line: x = 2 + 4t, y = −1 + 2t, z = 3 − t6 (14 points) Let f(x, y , z) = x/y −yz. Give a good estimate of the maximum increaseof f as we move a distance 0.01 from the point (1, 1, −1).fx= 1/y = 1, fy= −x/y2− z = 0, fz= −1∇f = < 1, 0, −1 >, |∇f| =√2f increases the most in the direction u = ∇f/|∇f|dfdsu= ∇f · u = |∇f|df = |∇f|ds = 0.01√27 (14 points) Let f(x, y) = ex2y−1cos(x2− y3). Find the linearization of f at the point(1, 1).fx=2xyex2y−1cos(x2− y3) − ex2y−1sin(x2− y3) (2x) = 2fy= −x2y2ex2y−1cos(x2− y3) − ex2y−1sin(x2− y3) (−3y2) = −1Linearization:L(x, y) = f(1, 1) + fx(1, 1)(x − 1) + fy(1, 1)(y − 1) = 1 + 2(x −1) −(y − 1) = 2x −y8 (10 points) (extra credit points only if you get 90 or more points on problems 1-7)The difference r between the positions of two bodies moving in their gravitational fieldcan be rescaled to satisfyr00= −r|r|3.Show that r0× r is a constant.(r0× r)0= r00× r + r0× r0= r00× r = −|r|−3r × r =