MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Afsana Rahman RUID: 193007377 EMAIL: afsana.rahman@rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= dz/dy = -(9*y^8 + x*z^2)/(3*z^2 + x*y*2*z) (WRONG ANS., ANS> MUST BE A NUMBER, -10 POINTS) Answer[ 2 ]= decreasing Answer[ 3 ]= 42 (WRONG ANS., ESSENTIALLY RIGHT WAY, MISREAD THE COMPUTER NOTATION, -3 POINTS) Answer[ 4 ]= Does not exist (WRONG ANS., STARTED THE RIGHT WAY, -5 POINTS) Answer[ 5 ]= (NO ANS. -10 POINTS) Answer[ 6 ]= The limit exists and is 0. (WRONG ANS. WRONG WAY, -10 POINTS) Answer[ 7 ]= curvature = 6/81 Answer[ 8 ]= <0,-9> Answer[ 9 ]= The function is changing at 8 units per unit of time. (WRONG ANS., WRONG WAY, -10 POINTS) Answer[ 10 ]= intersection at point (1,3,3) SCORE: 52 POINTS (out of 100) ----------------------------------------------------------------- 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]= 1 , a[2]= 9 , a[3]= 3 , a[4]= 1 , a[5]= 1 , a[6]= 7 , a[7]= 3 , a[8]= 7 , a[9]= 7 -------------------------------------------- --------------------------------------------- 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 x^1 + y^9 + z^3 + 1*x*y*z^2 = 3+1 Here is how I do it (Explain everything) You cannot easily solve for z. Therefore, must solve implicitly. Since the problem is asking for dz/dy, we must derive in terms of y, and then when deriving z, [z]' = z'. Treat x as any other constant. x^1 + y^9 + z^3 + x*y*z^2 = 4 will become 9*y^8 + 3*z^2*z' + x*[y*z^2]' = 0 Using product rule for [y*z^2]' 9*y^8 + 3*z^2*z' + x*[y*z^2]' = 0 will become 9*y^8 + 3*z^2*z' + x*[z^2 + y*2*z*z'] = 0 Expand/simplify: 9*y^8 + 3*z^2*z' + x*[z^2 + y*2*z*z'] = 0 will become 9*y^8 + 3*z^2*z' + x*z^2 + x*y*2*z*z' = 0 Bring all z' to one side 9*y^8 + 3*z^2*z' + x*z^2 + x*y*2*z*z' = 0 will become 9*y^8 + x*z^2 = - 3*z^2*z' - x*y*2*z*z' Factor out z': 9*y^8 + x*z^2 = - 3*z^2*z' - x*y*2*z*z' will become z'(-3*z^2 - x*y*2*z) = 9*y^8 + x*z^2 Solve for z': z'(-3*z^2 - x*y*2*z) = 9*y^8 + x*z^2 will become z' = (9*y^8 + x*z^2)/(-3*z^2 - x*y*2*z) Ans.: dz/dy = -(9*y^8 + x*z^2)/(3*z^2 + x*y*2*z) --------------------------------------------- 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)=<1,-1,5>. Is f increasing or decreasing at the direction <1,3,-1>? Here is how I do it (Explain everything) Find directional derivative (if positive, is increasing, if negative, is decreasing) deriv = <1,-1,5> * (1/(sqrt(11))*<1,3,-1> = 1/(sqrt(11) - 3/(sqrt(11) - 5sqrt(11) 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*7+y^3*3+z^3*7 = x^21 + y^9 + z^21 At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) Here is how I do it (Explain everything) Find each partial derivative: f_x = 21*x^20 f_y = 9*y^8 f_z = 21*z^20 Solve for gradient: gradient = Plugging in point P: <21*x^20, 9*y^8, 21*z^20> will become <21*1^20, 9*-1^8, 21*1^20> Simplified, this is <21,9,21> Find unit vector (distance between points) direction = = <1-1, -1-(-1), 3-1> = <0,0,2> |u| = sqrt(0^2 + 0^2 + 2^2) = sqrt(4) = 2 u = (1/2)*<0,0,2> = <0,0,1> Directional derivative = gradient * unit vector directional derivative = <21,9,21> * <0,0,2> = 0 + 0 + 21*2 = 42 Ans.: 42 --------------------------------------------- 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-7) Here is how I do it (Explain everything) Find f_x, f_y, f_xx, f_xy and f_yy f_x = exp(x-1) - exp(y-7) f_y = (x-1)*exp(y-7) f_xx = exp(x-1) - exp(y-7) f_xy = exp(y-7) f_yy = (x-1)*exp(y-7) Set first derivatives = 0 f_x = 0 leads to exp(x-1) = exp(y-7) f_y = 0 leads to (x-1)*exp(y-7) Solve for a variable exp(x-1) = exp(y-7) becomes x-1 = y-7, y = x+6 Plug into other equation (x-1)*exp(y-7) becomes (x-1)*exp(x+6-7) = (x-1)*exp(x-1) = 0 (x-1)*exp(x-1) = 0 is only true when x-1 = 0, so x = 1 Plug x into first equation, solve for y: exp(1-1) = exp(y-7) becomes 0 = y-7, therefore, y = 7 CRITICAL POINT: (2,7) Plug critical point into second derivatives f_xx(2,7) = exp(2-1) - exp(7-7) = exp - 1 f_xy(2,7) = exp(7-7) = 1 f_yy(2,7) = (2-1)*exp(7-7) = 1 Plug into D formula: D = f_xx*f_yy - [f_xy]^2 = (exp-1)*1 - 1 = exp - 2 Since D > 0, there are no saddle points. Ans.: Does not exist --------------------------------------------- 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 1*x + 3*y + 9 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [1, 9], B = [3, 1], C = [1, 7] Here is how I do it (Explain everything) Ans.: --------------------------------------------- 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*1^2-y^2*1^2)/(x*1-y*1) = (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) First, try plugging in: (x^2-y^2)/(x-y) becomes (1-1)/(1-1) = 0/0. Try something else Plug in "cx" for y: (x^2-y^2)/(x-y) becomes (x^2 - c^2*x^2)/(x-cx) Simplify: x^2*(1-c^2)/x(1-c) = x*(1-c^2)/(1-c) = x*(1+c) Limit COULD exist - try converting to polar coordinates: ((r^2*cos^2) - (r^2*sin^2))/(r*cos - r*sin) Simplify: r*(cos^2-sin^2)/(cos-sin) = r*(cos+sin) Find limit as r approaches 0: lim(r->0)r*(cos+sin) = 0 --------------------------------------------- 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) = [1, 9*t, 3*t^2] At the point (1,0,0) Here is how I do it (Explain everything) Find t using point and equations given: 1 = 1, 9*t = 0, t = 0, 3*t^2 = 0, t= 0 Find first and second derivatives: r'(t) = <0,9,6*t>, r"(t) = <0,0,6> Plug in for t (t=0, solved before): r'(0) = <0,9,0>, r"(0) = <0,0,6> Curvature = |r'(t) x r"(t)|/|r'(t)|^3 Plug in what we know: |<0,9,0> x <0,0,6>| / |<0,9,0>|^3 Solve: |<54,0,0>|/(9^3) = 54/(9^3) = 6/81 --------------------------------------------- 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 [-1*sin(t), -9*cos(t)] At time t=0 its position is [0, 9] and its velocity is [1, 0] Where is it located at time , t = Pi? Here is how I do it (Explain everything) Integrate acceleration to get velocity function: integral(<-sin(t), -9*cos(t)>) = <-cos(t), 9*sin(t)> + C Solve for C at t = 0 and set equal to given velocity point <-cos(0), 9*sin(0)> + C = <-1,0> + C = <1,0> C = <-1,0> + <1,0> = <0,0>, therefore v = <-cos(t), 9*sin(t)> + <0,0> Integrate v to get position function: integral(<-cos(t), 9*sin(t)>) = + C Same steps as before, Solve for C at t = 0 and set equal to given position point + C = <0,9> + C = <0,9>, C = <0,9>-<0,9> = <0,0> Therefore the position function is + <0,0> To find location at time pi, plug in for pi position = = <0,-9> --------------------------------------------- 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 3 Both x and y depend on time Right now the rate of change of x with respect to time is 1 and the rate of change of y with respect to time is 7 How fast is the function changing right now? Here is how I do it (Explain everything) Create formulas: z = x + y, x = t, y = 7*t Plug in t for x and y variables: z = t + 7*t = 8*t Derive to find rate of change: z'(t) = 8 The function is changing at 8 units per unit of time. --------------------------------------------- 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 = 3, z = 3 Here is how I do it (Explain everything) Add the normals: <1,0,0> + <0,3,0> + <0,0,3> = (1,3,3)