MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Irina Mukhametzhanova RUID: 190008204 EMAIL: iim15@scarletmail.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= DNE Answer[ 2 ]= Increasing Answer[ 3 ]= 0 Answer[ 4 ]= (0,8) Answer[ 5 ]= 9 Answer[ 6 ]= DNE Answer[ 7 ]= 0 Answer[ 8 ]= [0,-9] Answer[ 9 ]= 8 Answer[ 10 ]= (0,2,0) TOO MANY ERRORS ----------------------------------------------------------------- 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]= 0 , a[4]= 0 , a[5]= 0 , a[6]= 8 , a[7]= 2 , a[8]= 0 , 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^1+y^9+z^0+0*x*y*z^2 = 3+0 x+y^9+1 = 3 Here is how I do it (Explain everything) The equation that we got from the problem does not contain a z. So, the answer cannot be determined Ans.: DNE --------------------------------------------- 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)=<1,0,4>. Is f increasing or decreasing at the direction <1,0,0>? Here is how I do it (Explain everything) The formula for the directional derivative of function f is: grad(f) . u Where u is the unit vector in the desired direction. However, the question does not ask to find the numerical value of the directional derivative - we just need to see whether it is positive or negative. So, we can just use the direction vector itself: grad(f) . v = <1,0,4> . <1,0,0> = (1*1) + (0*0) + (4*0) = 1 + 0 + 0 = 1 Because 1>0, f is increasing at the desired direction Ans.: Increasing --------------------------------------------- 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*8+y^3*0+z^3*0 x^3*8 At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) Here is how I do it (Explain everything) The formula for the gradient vector at point (x0,y0,z0) is: grad(f)(x0,y0,z0) = So, first, we find all of the partial derivatives of f: f_x = (x^3*8)' = 24*x^2 f_y = (x^3*8)' = 0 f_z = (x^3*8)' = 0 Find all of the partial derivatives at (1,-1,1): f_x(1,-1,1) = 24*1^2 = 24 f_y(1,-1,1) = 0 f_z(1,-1,1) = 0 So, our gradient vector is: grad(f) = <24,0,0> To find the directional derivative, we also need the unit vector in the desired direction. Our direction vector is: v = <1-1, -1-(-1), 3-1> = <0,0,2> And the unit vector of this is: u = <0,0,2> / sqrt(0^2 + 0^2 + 2^2) = <0,0,2> / sqrt(4) = <0,0,2> / 2 = <0,0,1> Finally, calculate the directional derivative: graf(f)(1,-1,1) * u = <24,0,0> . <0,0,1> = (24*0) + (0*0) + (0*1) = 0 Ans.: 0 --------------------------------------------- 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)-(x)*exp(y-8) 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) If we want to find any critical points (points with a possible min, max, or saddle point), we first need to take the first partial derivatives of f: f_x = (exp(x))'-((x)*exp(y-8))' = exp(x) - exp(y-8) f_y = (exp(x))'-((x)*exp(y-8))' = 0 - x*exp(y-8) = (-1)*x*exp(y-8) Set both equations equal to 0 and solve for x and y: (-1)*x*exp(y-8) = 0 exp(y-8) cannot be equal to 0 at any value of y. So, the only way to make the equation be equal to 0 is: x = 0 Plug it back into the first equation: exp(x) - exp(y-8) = 0 exp(x) = exp(y-8) exp(0) = exp(y-8) y-8 = 0 y = 8 So, our critical point that we need to check is (0,8) To determine what kind of point it is (min, max, saddle point), we need to perform the Second Derivative test. For this, we need to use the formula for the determinant: D(x0,y0) = (f_xx(x0,y0))*(f_yy(x0,y0)) - (f_xy(x0,y0))^2 The point is a saddle point if the determinant is negative. So, first, we find all of the second partial derivatives of f: f_xx = (exp(x))' - (exp(y-8))' = exp(x) f_xy = (exp(x))' - (exp(y-8))' = (-1)*exp(y-8) f_yy = ((-1)*x*exp(y-8))' = (-1)*x*exp(y-8) Calculzte their numerical values at point (0,8): f_xx(0,8) = exp(0) = 1 f_xy(0,8) = (-1)*exp(8-8) = (-1)*exp(0) = (-1)*1 = -1 f_yy(0,8) = (-1)*0*exp(8-8) = 0 Plug the values into the determinant equation: D(0,8) = (1)*(0) - (-1)^2 = 0 - 1 = -1 The determinant at point (0,8) is negative, so, (0,8) is a saddle point Ans.: (0,8) --------------------------------------------- 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 0*x + 2*y + 9 2*y + 9 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [1, 9], B = [0, 0], C = [0, 8] Here is how I do it (Explain everything) First, we need to find the first partial derivatives of f: f_x = 0 f_y = 2 To find the critical points, we set both equations equal to 0: 0 = 0 2 = 0 The second equation does not make sense. So, there are no critical points Next, we check the vertices themselves by calculating f at those points: f(1,9) = 2*9 + 9 = 27 f(0,0) = 2*0 + 9 = 9 f(0,8) = 2*8 + 9 = 25 Next, we check the edges. We first need to find the linear equations they are represented by: Edge AB: y = 9*x Edge BC: x = 0 Edge AC: y = x + 8 Edge AB: f(x,9*x) = 2*9*x + 9 = 18*x + 9 Find the derivative of that: f' = 18 Set it to 0: 18 = 0 - doesn't work Edge BC: f(0,y) = 2*y+9 Find the derivative of that: f' = 2 Set it to 0: 2 = 0 - doesn't work Edge AC: f(x,x+8) = 2*(x+8) + 9 = 2*x + 16 + 9 = 2*x + 25 Find the derivative of that: f' = 2 Set it to 0: 2 = 0 - doesn't work So, the values we have are 27, 9 and 25 The absolute minimum value is 9 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*0^2-y^2*0^2)/(x*0-y*0) 0/0 Find the LIMIT of f(x,y) as (x,y) goes to the point [0,0], or show that it does not exist Here is how I do it (Explain everything) The answer does not exist because my function turns out to be 0/0 --------------------------------------------- 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, 0*t^2] r(t) = [1, 9*t, 0] At the point (1,0,0) Here is how I do it (Explain everything) The formula for the curvature of the curve is: k(t) = (|r'(t) x r''(t)|)/|r'(t)|^3 First, we need to find the value of t that the point is on: 1 = 1 9*t = 0 0 = 0 t = 0 Next, find the two derivatives of the curve at t=0: r'(t) = [1', (9*t)', 0'] = [0,9,0] r'(0) = [0,9,0] r''(t) = [0,0,0] r''(0) = [0,0,0] Next, calculate the magnitude of the cross product of r'(0) and r''(0): r'(0) x r''(0) = |i j k| |0 9 0| |0 0 0| = i*(0-0) - j*(0-0) + k*(0-0) = [0,0,0] The magnitude of that is 0. Next, we need to find the magnitude cubed of r'(0) |r'(0)|^3 = sqrt(0^2 + 9^2 + 0^2)^3 = sqrt(81)^3 = 9^3 = 729 Finally, find the curvature using the formula above: k(0) = 0/729 = 0 --------------------------------------------- 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) Acceleration is the derivative of velocity. So, to find velocity we integrate the acceleration vector: v = int(a) = [int(-1*sin(t)), int(-9*cos(t))] = [cos(t), -9*sin(t)] + C, where C is an arbitrary constant We know what velocity is equal to at time 0. So, to find C, we plug in t=0 and what velocity is equal to: [1,0] = [cos(0), -9*sin(0)] + C [1,0] = [1,0] + C C = [0,0] So, our velocity vector is: v = [cos(t), -9*sin(t)] Velocity is the derivative of position. So, to find the position vector, we integrate the velocity vector: r = int(v) = [int(cos(t)), int(-9*sin(t))] = [sin(t), 9*cos(t)] + C, where C is an arbitrary constant We know what position is equal to at time 0. So, to find C, we plug in t=0 and what position is equal to: [0,9] = [sin(0), 9*cos(0)] + C [0,9] = [0,9] + C C = [0,0] So, our position vector is: r(t) = [sin(t), 9*cos(t)] Plug in t = Pi: r(Pi) = [sin(Pi), 9*cos(Pi)] = [0, 9*(-1)] = [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, 0 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, 1 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) To find the rate of change of f with respect to time, we can use the Chain Rule: df/dt = (df/dx)*(dx/dt) + (df/dy)*(dy/dt) We know all of the needed rates of change, so, we plug them in: df/dt = 0*1 + 2*4 = 8 --------------------------------------------- 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 = 0, y = 2, z = 0 Here is how I do it (Explain everything) The point of intersection of the three planes is (0,2,0)