MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME:Yongshan Li RUID:204000235 EMAIL:yl1524@roseprogram.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]=-1/3 Answer[ 2 ]=decreasing Answer[ 3 ]=9/sqrt(11) Answer[ 4 ]=(1,1) Answer[ 5 ]=4 Answer[ 6 ]=does not exist Answer[ 7 ]=8 Answer[ 8 ]= Answer[ 9 ]=12 Answer[ 10 ]=(1,2,4) ----------------------------------------------------------------- Instructions: Download this file with its original name, mt1.txt, then rename it, in your computer mt1FirstLast.txt Edit it with your answers and solutions USING COMPUTEREZE: e.g.: x times y IS x*y, x to the power y is x^y and Email DrZcalc3@gmail.com, 80 minutes (or sooner) after starting (for most people 10:00am, Oct. 15) Subject: mt1 with an attachment. YOU MUST NAME IT EXACTLY mt1FirstLast.txt --------------------------------------------------------------------------- For each of the questions you MUST first figure, YOUR version, with the following convention For i=1,2,3,4,5,6,7,8,9 , a[i]:= The i-th digit of your RUID, BUT of it is zero make it 1 Example: RUID=413200125; a[1] = 4, a[2] = 1, a[3] = 3, a[4] = 2, a[5] = 1, a[6] = 1, a[7] = 1, a[8] = 2, a[9] = 5 ----------------------------------------------------------------------------------------------------------------------------------------------- HERE WRITE THE ACTUAL a[i] a[1]=2 , a[2]=1 , a[3]=4 , a[4]=1 , a[5]=1 , a[6]=1 , a[7]=2 , a[8]=3 , a[9]=5 -------------------------------------------- --------------------------------------------- Problem 1: Find dz/dy at the point (1,1,1) if z(x,y) is given implicitly by the equation x^a[1]+y^a[2]+z^a[3]+a[5]*x*y*z^2 = 3+a[5] With my RUID data the question is Find dz/dy at the point (1,1,1) if z(x,y) is given implicitly by the equation x^2+y+z^4+x*y*z^2 = 3+1 Here is how I do it (Explain everything) Differentiate with respect to y the given relation gives: 1+4z^3*z'+x*z^2+x*y*2z*z'=0 (4z^3+x*y*2z)*z'+x*z^2+1=0 (4z^3+x*y*2z)*z'=-(x*z^2+1) z'=-(x*z^2+1)/(4z^3+x*y*2z) plug in (1,1,1) dz/dy at the point (1,1,1)=-(1*1^2+1)/(4*1^3+1*1*2*1)=-1/3 Ans.:-1/3 --------------------------------------------- Problem 2: Suppose that grad(f)(P)=. Is f increasing or decreasing at the direction ? With my RUID data the question is Suppose that grad(f)(P)=<2,-1,2+2>. Is f increasing or decreasing at the direction <2,4,-1>? Here is how I do it (Explain everything) |<2,4,-1>|=sqrt(4+16+1)=sqrt(21) u=<2/sqrt(21), 4/sqrt(21), -1/sqrt(21)> grad(f)·u=<2,-1,2+2>·<2/sqrt(21), 4/sqrt(21), -1/sqrt(21)>=4/sqrt(21)-4/sqrt(21)-4/sqrt(21)=-4/sqrt(21)<0 Therefore, f is decreasing at the direction <2,4,-1> Ans.: decreasing --------------------------------------------- Problem 3: Find the directional derivative of the function f(x,y,z) x^3*a[6]+y^3*a[3]+z^3*a[8] At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) With my RUID data the question is Find the directional derivative of the function f(x,y,z) x^3+y^3*4+z^3*2 At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) Here is how I do it (Explain everything) fx=3x^2 fy=12y^2 fz=6z^2 grad(f)=<3x^2,12y^2,6z^2> grad(f)(1,-1,1)=<3,12,6> |<1,-1,3>|=sqrt(1+1+9)=sqrt(11) u=<1/sqrt(11),-1/sqrt(11),3/sqrt(11)> grad(f)·u=<3,12,6>·<1/sqrt(11),-1/sqrt(11),3/sqrt(11)>=3/sqrt(11)-12/sqrt(11)+18/sqrt(11)=9/sqrt(11) the directional derivative is 9/sqrt(11). Ans.: 9/sqrt(11) --------------------------------------------- Problem 4: Find a saddle point of the function f(x,y)= exp(x-a[4])-(x-a[4])*exp(y-a[6]) If there is no saddle point, write in the Answers: "Does Not exist". Explain what you are doing With my RUID data the question is Find a saddle point of the function f(x,y)= exp(x-1)-(x-1)*exp(y-1) If there is no saddle point, write in the Answers: "Does Not exist". Explain what you are doing Here is how I do it (Explain everything) fx=e^(x-1)-e^(y-1) fy=-e^(y-1)*(x-1) {e^(x-1)-e^(y-1)=0,-e^(y-1)*(x-1)=0}{x,y} {1,1} fxx=e^(x-1) fxy=-e^(y-1) fyy=-e^(y-1) fxx*fyy-fxy^2=-1-1=-2<0 Therefore,{1,1}is saddle point Ans.:(1,1) --------------------------------------------- Problem 5: Let f(x,y) be the function a[4]*x + a[7]*y + a[2] Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [a[1], a[2]], B = [a[3], a[4]], C = [a[5], a[6]] With my RUID data the question is Let f(x,y) be the function x + 2*y + 1 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [2, 1], B = [4,1], C = [1, 1] Here is how I do it (Explain everything) fx=1 fy=2 The critical points does not exist. f(1,1)=4 f(4,1)=7 Therefore, the absolute minimum value is f(1,1)=4 Ans.:4 --------------------------------------------- Problem 6: Let f(x,y) be the function (x^2*a[4]^2-y^2*a[5]^2)/(x*a[4]-y*a[5]) Find the LIMIT of f(x,y) as (x,y) goes to the point [a[5],a[4]], or show that it does not exist With my RUID data the question is Let f(x,y) be the function (x^2-y^2)/(x-y) Find the LIMIT of f(x,y) as (x,y) goes to the point [1,1], or show that it does not exist Here is how I do it (Explain everything) plug in [1,1] both bottom and up will be 0 so go on plug in y=cx lim x->1 (x^2-(cx)^2)/(x-(cx)) =(1-c^2)x^2/(1-c)x =(1-c^2)/(1-c) The limit does not exist since you get different limits when you approach the point (1,1) on different lines. --------------------------------------------- Problem 7: Find the curvature of the curve r(t) = [a[1], a[2]*t, a[3]*t^2] At the point (a[1],0,0) With my RUID data the question is Find the curvature of the curve r(t) = [2, t, 4*t^2] At the point (2,0,0) Here is how I do it (Explain everything) r'(t)=1+8t r"(t)=8 r'(t)×r"(t)=8 |r'(t)×r"(t)|=8 |r'(t)|=sqrt(1+64t^2) k(t)=8/(sqrt(1+64t^2))^3 At the point (2,0,0), t=0 k(2,0,0)=8 --------------------------------------------- Problem 8: A particle is moving in the plane with ACCELERATION given by [-a[1]*sin(t), -a[2]*cos(t)] At time t=0 its position is , [0, a[2]] and its velocity is , [a[1], 0] Where is it located at time , t = Pi With my RUID data the question is A particle is moving in the plane with ACCELERATION given by [-2*sin(t), -1*cos(t)] At time t=0 its position is , [0, 1] and its velocity is , [2, 0] Where is it located at time , t = Pi Here is how I do it (Explain everything) integral(-2*sin(t)) from 0 to Pi is -4 integral(-1*cos(t)) from 0 to Pi is 0 Velocity:[-2,2] --------------------------------------------- Problem 9: A certain function depends on variables x and y Right now the rate of change of the function with respect to x is, a[5] and the rate of change of the function with respect to y is, a[7] Both x and y depend on time Right now the rate of change of x with respect to time is, a[1] and the rate of change of y with respect to time is, a[9] How fast is the function changing right now? With my RUID data the question is A certain function depends on variables x and y Right now the rate of change of the function with respect to x is, 1 and the rate of change of the function with respect to y is, 2 Both x and y depend on time Right now the rate of change of x with respect to time is, 2 and the rate of change of y with respect to time is, 5 How fast is the function changing right now? Here is how I do it (Explain everything) df/fx=1, df/dy=2, dx/dt=2, dy/dt=5 df/dt=(df/dx)*(dx/dt)+(df/dy)*(dy/dt) =1*2+2*5 =12 --------------------------------------------- Problem 10: Find the point of intersection of the three planes x = a[5], y = a[7], z = a[3] With my RUID data the question is Find the point of intersection of the three planes x = 1, y = 2, z = 4 Here is how I do it (Explain everything) Since the three plane contain points that are: p(x)=(1,y,z) p(y)=(x,2,z) p(z)=(x,y,4) the point of intersection of the three planes is (1,2,4)