MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Aditya Sivakumar RUID: 201001195 EMAIL: aditya.sivakumar@rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= -2/3 Answer[ 2 ]= Neither as DuF = 0 Answer[ 3 ]= 27 Answer[ 4 ]= Does Not exist Answer[ 5 ]= The points given don't form a triangle and there are no critical points so there is on absolute minimum Answer[ 6 ]= 2 Answer[ 7 ]= 2 Answer[ 8 ]= <0,-1> Answer[ 9 ]= 7 Answer[ 10 ]= (1,1,1) WRONG ANSWERS: #4 (BUT RIGHT WAY!) Note: #5 is a DEGENERATE TRIANGLE (so there is an answer, but I did not take points off) SCORE: 95 (OUT OF 100) ----------------------------------------------------------------- Instructions: Download this file with its original name, mt1.txt, then rename it, in your computer mt1AdityaSivakumar.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 mt1AdityaSivakumar.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]=1 , a[4]=1 , a[5]=1 , a[6]=1 , a[7]=1 , a[8]=9 , 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^1+z^1+1*x*y*z^2 = 3+1 Here is how I do it (Explain everything) x^2 + y + z + xyz^2 = 4 1 + dz/dy + 2xyz(dz/dy) + xz^2 = 0 took derivative w.r.t y dz/dy + 2xyz(dz/dy) = -xz^2 - 1 put all dz/dy terms on one side dz/dy = (-xz^2 - 1)/(1 + 2xyz) solved for dz/dy Dz/dy @ (1,1,1) = (-1(1)^2 - 1)/(1 + 2(1)1(1)). Plugged in (1,1,1) for (x,y,z) = -2/3 Ans.: dz/dy at the point (1,1,1) is -2/3 --------------------------------------------- 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,-1,3>. Is f increasing or decreasing at the direction <2,1,-1>? Here is how I do it (Explain everything) Magnitude of <2,1,-1> = sqrt(4 + 1 + 1) = sqrt(6) Magnitude of vector of direction U = <2,1,-1> / sqrt(6) = <2/sqrt(6),1/sqrt(6),-1/sqrt(6)> Found unit vector of direction grad(f)(P) . U = <2,-1,3> . <2/sqrt(6),1/sqrt(6),-1/sqrt(6) = 4/sqrt(6) - 1/sqrt(6) - 3/sqrt(6) = 0 Solved for the directional derivative Ans.: f is neither increasing not decreasing in the direction <2,1,-1> as the directional derivative is 0 --------------------------------------------- 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 x^3*1+y^3*1+z^3*9 Here is how I do it (Explain everything) QP = <1,-1,3> - <1,-1,1> = <0, 0, 2> Found directional vector Unit vector QP = <0, 0, 2>/sqrt(4) = <0,0,1> Found unit directional vector grad(f)(P) = <3x^2, 3y^2, 27z^2> = <3,3,27> Found gradient vector grad(f)(P) . Unit vector QP = <3,3,27> = <0,0,1> = 27 Found directional derivative by taking the dot product of the two vectors above Ans.: The directional derivative of x^3*1+y^3*1+z^3*9 at the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) is 27. --------------------------------------------- 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 exp(x-1)-(x-1)*exp(y-1) Here is how I do it (Explain everything) F_x = exp(x-1)-*exp(y-1) Found F_x, f_y, F_xx, f_yy, F_xy f_xx = exp(x-1) F_y = -(x-1)*exp(y-1) f_yy = -(x-1)*exp(y-1) f_xy = 0 F_x = 0 = exp(x-1)-*exp(y-1 Set f_x and F_y = o to find critical points x = y f_y = 0 = -(x-1)*exp(y-1) = -(y-1)*exp(y-1) Y = 1 x=1 Critical Point at (1,1) D =(f_xx)(f_yy) - (f_xy)^2 Using the critical point I found the discriminant = (exp(1-1))*(-(1-1)*exp(1-1) = 0 D = 0, inconclusive, Saddle point does not exist D = 0, so the test is inconclusive, so there is no saddle point Ans.: Saddle point does not exist as D = 0, so the test is inconclusive --------------------------------------------- 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 f(x,y) = 1*x + 1*y + 1 f(x,y) = x + y + 1 A = [2, 1], B = [1, 1], C = [1, 1] Here is how I do it (Explain everything) f_x = 1 F_y = 1 f_xx = 0 f_yy = 0 f_xy = 0 The points given do not form a triangle There are also no critical points so there is no absolute minimum Ans.: The points given don't form a triangle and there are no critical points so there is on absolute minimum., --------------------------------------------- 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 (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) the LIMIT of f(x,y) as (x,y) goes to the point [1,1] = (x^2-y^2)/(x-y]) Difference of squares = (x-y)*(x+y)/(x-y) = x + y = 2 Ans.: The limit of (x^2-y^2)/(x-y]) as it approaches the point (1,1) is 2. --------------------------------------------- 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 r(t) = [2, 1*t, 1*t^2]. At the point (2,0,0) Here is how I do it (Explain everything) r(t) = [2, t, t^2] Found first and second derivatives of r(t) R'(t) = <0,1,2t> R''(t) = <0,0,2> 2 = 2 Found the value of t at the specified point T = 0 t^2 = 0 (2,0,0) occurs at t= 0 K = |r'(t) x r''(t)|/|r'(t)|^3 R'(0) = <0,1,0> Found r'(0) and r''(0) R''(0) = <0,0,2> R'(0) x R''(0) = <2,0,0> Found cross product as it is part of curvature formula |<2,0,0>| = 2 |R'(0)| = 1 Found |R'(0)"| as it is also part of curvature formula K = 2/1^3 = 2 Ans.: The curvature is 2. --------------------------------------------- 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 [-2*sin(t), -cos(t)], time t=0 its position is , [0, 1] & velocity is , [2, 0] Here is how I do it (Explain everything) v(t) = integral([-2*sin(t), -*cos(t)] dt) + C. Integral to find v(t), also found C v(t) = <2cos(t), -sin(t)> + C <2,0> = <2cos(0), -sin(0)> + C C = 0 v(t) = <2cos(t), -sin(t)> s(t) = integral(<2cos(t), -sin(t)>) + C Integral to find s(t), also found C s(t) = <2sin(t), cos(t)> + C <0,1> = <2sin(0), cos(0)> + C C = 0 s(t) = <2sin(0), cos(0)> s(pi) = <2sin(pi), cos(pi)> = <0,-1> Found s(pi) s(pi) = <0,-1> Ans.: The particle is located at <0,-1> when t = Pi --------------------------------------------- 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 rate of change of the function with respect to x is, 1 rate of change of the function with respect to y is, 1 the rate of change of x with respect to time is, 2 rate of change of y with respect to time is, 5 Here is how I do it (Explain everything) Df/dt = df/dx*dx/dt + df/dy*dy/dt. Chain rule Df/dx = 1 Solving for values Df/dy = 1 Dx/dt = 2 Dy/dt = 5 Df/dt = 1*2 + 1*5 = 7 Plugging in Ans.: The function is changing at a rate of 7 right now --------------------------------------------- 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 x = 1, y = 1, z = 1 Here is how I do it (Explain everything) x = 1, y = 1, z = 1 is the point of intersection Found point of intersection as all three planes don't share a variable (1,1,1) Ans.: The three planes intersect at the point (1,1,1)