MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Orion Kress Sanfilippo RUID: 204006944 EMAIL: omk23@scarletmail.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= -1/3 Answer[ 2 ]= f is decreasing Answer[ 3 ]= 12 Answer[ 4 ]= P = (1,6) Answer[ 5 ]= 12 Answer[ 6 ]= 1 Answer[ 7 ]= 8 Answer[ 8 ]= P = (0,1) Answer[ 9 ]= 38 Answer[ 10 ]= (1,9,4) ----------------------------------------------------------------- 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]= 6, a[7]= 9, a[8]= 4, a[9]= 4 -------------------------------------------- --------------------------------------------- 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^4+1*x*y*z^2 = 3+1 Here is how I do it (Explain everything) Have to find derivative of z with respect to y, start by using the prime (x^2+y^1+z^4+1*x*y*z^2 = 3+1)’ (w.r.t. y) = 0 + 1 + 4*z^3*z’ + x(z^2 + 2*y*z*z’) = 0 Isolate terms with z’ on one side: z’(4*z^3 + 2*x*y*z) = -x*z^2 - 1 Then divide: dz/dy = To find derivative at a point, plug in values: @ P = (1,1,1), dz/dy = Ans.: dz/dy = --------------------------------------------- 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,-1,11>. Is f increasing or decreasing at the direction <2,4,-1>? Here is how I do it (Explain everything) In order to determine whether f(x,y,z) is increasing or decreasing in a certain direction, we have to use dot product with the direction we are given: (grad(f)(P)) ૦ <2,4,-1> = (2*2 + -1*4 + -1*11) = 4 - 4 - 11 = -11 Since the dot product is negative, f is not increasing in this direction. Ans.: f is decreasing in the direction (2,4,-1) --------------------------------------------- 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*6+y^3*4+z^3*4 At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) Here is how I do it (Explain everything) First, find gradient of f: grad(f(x,y,z)) = < 18x^2, 12y^2, 12z^2 > Plug in x_0, y_0 and z_0 grad(f(1,-1,1)) = < 18, 12, 12 > To find directional derivative, take dot product of grad(f(P)) and provided direction, BUT we need to find and normalize direction vector first: PQ = (1 - 1, (-1) - (-1), 3 - 1)/|PQ| = PQ = ( 0 , 0 , 1) Now take dot prod: <18,12,12> ૦ <0,0,1> = 12 Ans.: The directional derivative of f at point P in the direction towards Q is 12. --------------------------------------------- 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-6) 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) First find the partial derivatives of f: df/dx = exp(x-1) - exp(y-6) df/dy = -(x-1)*exp(y-6) Then create a system of equations, set partial derivatives equal to 0 and solve: df/dx = exp(x-1) - exp(y-6) = 0 df/dy = -(x-1)*exp(y-6) = 0 In df/dy, (exp(y-6)) can never equal 0, therefore 1-x = 0, x = 1 Plugging into df/dx, exp(1-1) = exp(y-6) 1 = exp(y-6) ln(1) = y - 6, y = 6 To test point categorization, find second derivatives: d^2f/dx^2 = exp(x-1) d^2f/dy^2 = -(x-1)exp(y-6) d^2f/dx*dy = -exp(y-6) Next, plug in P = (1, 6) found in previous part and find the discriminant: D = (d^2f/dx^2)*(d^2f/dy^2) - (d^2f/dx*dy)^2 = (exp(1-1)*-(1-1)*exp(6-6)) - = 0 - 1 = -1 Since D< 0, P = (1,6) is this function’s only saddle point Ans.: P = (1,6) is the only saddle point. --------------------------------------------- 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 + 9*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, 6] Here is how I do it (Explain everything) Must find extreme values and all critical points of f within the triangular interval: df/dx = 1 df/dx = 9 Since the partial derivatives are constants, all extreme values can be found by plugging in the bounds of the area, then checking between these values f(A) = 1*2 + 9*1 + 1 = 12 f(B) = 1*4 + 9*1 + 1 = 14 f(C) = 1*1 + 9*6 + 1 = 56 Since f is linear along each side of the triangle, no need to check intermediate values, the absolute min of f on this interval exists at f(A), and is equal to 12 Ans.: The absolute minimum value of f on this interval is 12, and occurs at point A. --------------------------------------------- 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) 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) Plugging in values here is trivial and unnecessary, clearly yields 0/0. Must test what happens when we approach 1,1 along the line y = cx: Since this limit is not dependent on the slope of the line on which we approach, the limit of f as x and y both approach 1 is 1 Ans: The limit of f as x and y both approach 1 is 1. --------------------------------------------- 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, 1*t, 4*t^2] At the point (2,0,0) Here is how I do it (Explain everything) First find the prime and double prime of r: r’ = [0, 1, 8t] r’’ = [0,0,8] The formula for curvature is k = |r’ x r’’| = |i(8 - 0) - j(0) - k(0) | = 8 |r’(t)|^3 @ P = (2,0,0) = Ans: The curvature at this point is k = 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 Problem 8: 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) In order to go from the acceleration to the velocity and then position, take 2 integrals: Ans: The position at this time is r(pi) = (0,1) --------------------------------------------- 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, 9 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, 4 How fast is the function changing right now? Here is how I do it (Explain everything) Since f is dependent on x and y, AND s and y are both dependent on t, we must use the chain rule to determine the derivative of f with respect to t: df/dt = df/dx*dx/dt + df/dy*dy/dt df/dx = 1, dx/dt = 2, df/dy = 9, dy/dt = 4 df/dt = 1*2 + 9*4 = 38 Ans: The speed of the function at the given point with respect to t is 38. --------------------------------------------- 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 = 9, z = 4 Here is how I do it (Explain everything) Since each of these shapes are flat and must intersect at at least one point, the point corresponds to the value of each plane: P = (1,9,4) Ans: (1,9,4)