MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Yash Khangura RUID: 200007720 EMAIL: ask223@scarletmail.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= dz/dy at the point (1,1,1) is -2/3 Answer[ 2 ]= f is decreasing at the direction of <2,1,-1> because D_u f = -6/sqrt(6) < 0 Answer[ 3 ]= D_u f(1,-1,1) = 3 (WRONG ANS, BUT RIGHT WAY, CARELESS ERROR, -3 POINTS) Answer[ 4 ]= (1,7) is a saddle point because D<0 Answer[ 5 ]= This is technically a trick. All the points lie of the same plane z = 0. So the ABSOLUTE MAX AND MIN ARE 0. (WRONG ANS. WRONG WAY , -10 POINTS) Answer[ 6 ]= lim (x,y) -> (1,1) (x^2-y^2)/(x-y) = 2 Answer[ 7 ]= k(0) = |r'(0) x r"(0)| / |r'(0)|^3 = 3 (WRONG ANS., CARELESS ERROR, -2 POINTS) Answer[ 8 ]= r(Pi) = <2*sin(Pi), cos(Pi)> --> r(Pi) = (0, -1) Answer[ 9 ]= df/dt = df/dx * dx/dt + df/dy * dy/dt = 1*2 + 7*1 = 9 Answer[ 10 ]= The planes x = 1, y = 7, z = 1 intersect at the point (1,7,1). ----------------------------------------------------------------- SCORE: 85 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]= 2 , a[2]= 1 , a[3]= 1 , a[4]= 1 , a[5]= 1, a[6]= 7, a[7]= 7, a[8]= 2, a[9]= 1 -------------------------------------------- --------------------------------------------- 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^2+y^1+z^1+1*x*y*z^2 = 4 Here is how I do it (Explain everything) 1 + dz/dy + x*z^2 + 2*x*y*z*dz/dy = 0 here, I took the derivatives of all terms that contain y and z. But when I took the derivative of z, I would write dz/dy. dz/dy + 2*x*y*z*dz/dy = -1 - x*z^2 dz/dy = (-1 - x*z^2)/(1 + 2*x*y*z) Here I solved for dz/dy dz/dy = (-1 - 1*1^2)/(1 + 2*1*1*1) = -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,-1,9>. Is f increasing or decreasing at the direction <2,1,-1>? Here is how I do it (Explain everything) u = <2,1,-1> / |<2,1,-1>| = <2/sqrt(6),1/sqrt(6),-1/sqrt(6)> D_u f = grad(f)(P) . u = <2,-1,9> . <2/sqrt(6),1/sqrt(6),-1/sqrt(6)> D_u f = 4/sqrt(6) - 1/sqrt(6) - 9/sqrt(6) = -6/sqrt(6) Ans.: f is decreasing at the direction of <2,1,-1> because D_u f = -6/sqrt(6) < 0. I found the unit vector. I then found the directional derivative by doing grad(f)(P) . u which was less than 0, so f is 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 x^3*7+y^3*1+z^3*2 Here is how I do it (Explain everything) u = <0,0,2>/|<0,0,2>| = <0,0,1> here I found the unit vector f_x(x,y,z) = 21*x^2, f_y(x,y,z) = 3*y^2, f_z(x,y,z) = 3*z^2 f_x(1,-1,1) = 21, f_y(1,-1,1) = 3, f_z(1,-1,1) = 3 here I found the three partial derivatives, and then I found their values at the point D_u f(1,-1,1) = grad(f) * u = 21*0 + 3*0 + 3*1 = 3 Ans.: D_u f(1,-1,1) = 3 --------------------------------------------- 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-7) Here is how I do it (Explain everything) f_x(x,y) = exp(x-1) - exp(y-7) = 0 x+6 = y f_y(x,y) = -(x-1)*exp(y-7) = 0 -(x-1)*exp(x-1) = 0 -> x -1 = 0 -> x = 1 y = 7 critical point: (1,7) I found the partial derivatives and set them equal to 0. Then I solved the system of equations to get the critical point. f_xx(x,y) = exp(x-1), f_yy(x,y) = -(x-1)*exp(y-7), f_xy(x,y) = -exp(y-7) (1,7): D = 1 * 0 - (-1)^2 = -1 --> (1,7) is a saddle point Ans.: (1,7) is a saddle point because D<0 --------------------------------------------- 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 A = [2, 1], B = [1, 1], C = [1, 7] Here is how I do it (Explain everything) AB = [-1, 0], AC = [-1, 6], BC = [0, 6] Ans.: This is technically a trick. All the points lie of the same plane z = 0. So the ABSOLUTE MAX AND MIN ARE 0. --------------------------------------------- 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*1^2-y^2*1^2)/(x*1-y*1) = (x^2-y^2)/(x-y) , [1,1] Here is how I do it (Explain everything) lim (x,y) -> (1,1) (x^2-y^2)/(x-y) = (x+y)(x-y)/(x-y) = x+y = 2 I found the limit algebraically by expanding the function. Then I cancelled out (x-y). Leaving me with x+y. Then I plugged in (1,1) and got 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], (2,0,0) --> t = 0 Here is how I do it (Explain everything) r'(t) = <0,1,2t>, r"(t) = <0,0,2> r'(0) = <0,1,0>, r"(0) = <0,0,2> r'(t) x r"(t) = <3, 0, 0> k(0) = |r'(0) x r"(0)| / |r'(0)|^3 k(0) = |<3, 0, 0>| / |1|^3 = 3 here I first realized that t = 0, for at the given point. Then I found r'(t) and r"(t). I plugged in t=0 into them. I then found the cross product of r'(0) and r"(0). I then used the curvature equation k(0) = |r'(0) x r"(0)| / |r'(0)|^3 and got that the curvature at t=0 is k(0) = 3. --------------------------------------------- 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, 1] and its velocity is , [2, 0] Where is it located at time , t = Pi With my RUID data the question is [-2*sin(t), -1*cos(t)] Here is how I do it (Explain everything) a(t) = [-2*sin(t), -1*cos(t)] v(t) = <2*cos(t) + c, -sin(t) +c> v(0) = <2*cos(0) + c, -sin(0) +c> = <2,0> v(t) = <2*cos(t), -sin(t)> r(t) = <2*sin(t)+c, cos(t)+c> r(0) = <2*sin(0)+c, cos(0)+c> = <0,1> r(t) = <2*sin(t), cos(t)> r(Pi) = <2*sin(Pi), cos(Pi)> --> r(Pi) = (0, -1) Here I integrated a(t) to get v(t). I then plugged in t=0 to get the values of c, here c was 0 for both components. I then integrated v(t) to get r(t). I then plugged in t=0 to get the values of c, here c was also 0 for both components. I then plugged in Pi into r(t) and got that the postion of the particle at t = Pi was (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] = 1 and the rate of change of the function with respect to y is, a[7] = 7 Both x and y depend on time Right now the rate of change of x with respect to time is, a[1] = 2 and the rate of change of y with respect to time is, a[9] = 1 How fast is the function changing right now? With my RUID data the question is Right now the rate of change of the function with respect to x is, a[5] = 1 and the rate of change of the function with respect to y is, a[7] = 7 Both x and y depend on time Right now the rate of change of x with respect to time is, a[1] = 2 and the rate of change of y with respect to time is, a[9] = 1 Here is how I do it (Explain everything) df/dt = df/dx * dx/dt + df/dy * dy/dt = 1*2 + 7*1 = 9 The function is changing at a rate of 9. I did this by using the chain rule and then plugging in the given values for each part: df/dx = 1, dx/dt = 2, df/dy = 7, dy/dt =1. --------------------------------------------- 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 = 7, z = 1 Here is how I do it (Explain everything) The planes x = 1, y = 7, z = 1 intersect at the point (1,7,1). I know this because all three equations are just flat planes, and do not do anything special. x and y are both vertical planes and z is the horizantal plane. They will all converge at the point (1,7,1).