MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Ashwin Haridas RUID: 184009626 EMAIL: ah1058@scarletmail.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= -3/2 Answer[ 2 ]= Increasing Answer[ 3 ]= 6 Answer[ 4 ]= Test is inconclusive (we dont know) Answer[ 5 ]= 8 Answer[ 6 ]= Does not exist Answer[ 7 ]= 1/8 Answer[ 8 ]= v(Pi) = <-1, 0>, r(Pi)= <0, 8> Answer[ 9 ]= sqrt(10) Answer[ 10 ]= (0, 6, 4) ----------------------------------------------------------------- ------------------------------- WRONG ANSWERS : #2 (BUT right way, -5 points) #4 (RIGHT WAY BUT WRONG CALCULATIONS, -5 points) #5 (RIGHT WAY [BUT WAY TOO LONG WAY, THERE IS A SHORTCUT (see answer key)], Wrong numbers, -5 points) #6 (YOU FORGOT TO CHANGE THE 0-s to 1's in your RUID, getting a nonsense question, -5 points) #8 (RIGHT WAY, SIGN ERROR, -5 POINTS) #9 (BIG CONCEPTUAL ERROR, MISUNDERSTANDING OF THE CHAIN RULE, -10 points) Note: In #10 YOU FORGOT TO CHANGE the 0 to 1 in your RUID, BUT OTHERWISE CORRECT. I DIT NOT Take points off SCORE: 65 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]= 8, a[3]= 4, a[4]=0 , a[5]= 0, a[6]= 9, a[7]= 6, a[8]= 2, a[9]= 6 -------------------------------------------- --------------------------------------------- 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^8+z^4+a^0*x*y*z^2 = 3+^0 Here is how I do it (Explain everything) We can first simplify our equation: x^1 + y^8 + z^4 + x * y * z^2 = 1 Let's use the short cut for implicit differentiation! dz/dy = (-f_y)/(f_z) Move everything to left x + y^8 + z^4 + x * y * z^2 - 1 = 0 f_y = 8*y^7 + x*z^2 f_z = 4*z^3 + 2*x*y*z Then: dz/dy = (-8*y^7 - x*z^2) / (4*z^3 + 2*x*y*z) Plugging in (1,1,1) at dz/dy gives us: -9/6 = -3/2 Ans.: -3/2 --------------------------------------------- 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)=<1,0,8>. Is f increasing or decreasing at the direction <1,4,0>? Here is how I do it (Explain everything) We can take the dot product of grad(f)(P) . direction vector So: <1,0,8> . <1,4,0> This equals: (1)(1) + (0)(4) + (8)(0) = 1 + 0 + 0 = 1 Since the dot product is greater than 0, we know that f is INCREASING at the direction. Ans.: f is INCREASING at the direction --------------------------------------------- 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*9+y^3*4+z^3*2 At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) Here is how I do it (Explain everything) So we first need to find the gradient, and to do this we need f_x, f_y, and f_z f_x = 27*x^2 f_y = 12*y^3 f_z = 6*z^2 So the gradient is: <27*x^2, 12*y^3, 6*z^2> Now, we also need the unit vector in the direction pointing to Q So, first find the vector PQ, which would be <0, 0, 2> Then find the unit vector by dividing by length: |PQ| = sqrt(0^2 + 0^2 + 2^2) = sqrt(4) = 2 Unit vector = <0,0,1> Plug in P into gradient: <27, -12, 6> Take dot product of what we found and unit vector: <27, -12, 6> . <0, 0, 1> This gives us: (27)*(0) + (-12)*(0) + (6)*(1) = 6 Thus the directional derivative at the Point P is 6. Ans.: 6 --------------------------------------------- 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-0)-(x-0)*exp(y-9) Here is how I do it (Explain everything) First find f_x, f_y, f_xx, f_xy, and f_yy f_x = exp(x) - exp(y-9) f_y = -x*exp(y-9) f_xx = exp(x) f_xy = -x*exp(y-9) f_yy = -x*exp(y-9) Now let f_x and f_y = 0 Solving this gets us x = 0, y = 9 which is our critical point! Now plug these points into f_xx, f_xy, f_yy f_xx = exp(0) = 1 f_xy = -(0)*exp(0) = 0 f_yy = -(0)*exp(0) = 0 Find discriminant: D = f_xx*f_yy - [f_xy]^2 D = 1*0 - [0]^2 The discriminant is equal to 0, so we don't know (the test is inconclusive) Ans.: 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 Let f(x,y) be the function 0*x + 6*y + 8 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [1, 8], B = [4, 0], C = [0, 9] Here is how I do it (Explain everything) f(x,y) = 6*y + 8 f_x = 0 f_y = 6 f_xx = 0, f_yy = 0, f_xy = 0 Let's set f_x = 0 and f_y = 0 0 = 0, 6 = 0 These are inconsistent!! So there are NO critical points! But, we still have to find absolute min in each boundary L1: f(x,y) = f(0,y) = 6*y + 8 Derivative = 6. So we can test endpoints only, (4,0) gives us 8 (1,8) gives us 72. (4,0) is a contender L2: f(x,y) = f(x, 0) = 6*0 + 8 = 8 Derivative = 0. So we can test endpoints only, (4,0) gives us 8 (9,0) gives us 8, so both are contenders L3: f(x,y) = Doing the same thing, and testing our endpoints we get contenders again. 8 is the minimum value! Ans.: The absolute minimum value is 8. --------------------------------------------- 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) 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) When simplifying the function, I end up getting f(x,y) = 0/0 This does not make sense, so the limit DOES NOT EXIST :) Ans: Limit DNE --------------------------------------------- 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, 8*t, 4*t^2] At the point (1,0,0) Here is how I do it (Explain everything) First, compute r'(t) and r''(t) r'(t) = [0, 8, 8*t] r''(t) = [0, 0, 8] At the point (1,0,0), t = 0, so let's plug in r'(0) and r''(0) to make things easier r'(0) = [0, 8, 0] r''(0) = [0, 0, 8] Again, lets find |r'(0)| which is sqrt(0^2 + 8^2 + 0^2) = 8 Find cross product of r'(0) x r''(0) = i * [(8)*(8)-0] - j[0-0] + k[0-0] = 64*i - <64, 0, 0> Find magnitude of cross product = sqrt(64^2 + 0^2 + 0^2) = 64 Now, the equation for curvature is: k(t) = |r'(t) x r''(t)| / (|r'(t)|)^3 Since t = 0, k(0) = 64 / (8)^3 k(0) = 64 / 512 = 1/8 Thus the curvature at the point (1,0,0) is 1/8 ANS: 1/8 --------------------------------------------- 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 A particle is moving in the plane with ACCELERATION given by [-1*sin(t), -8*cos(t)] At time t=0 its position is , [0, 8] and its velocity is , [1, 0] Where is it located at time , t = Pi Here is how I do it (Explain everything) The problem does not state what we need to do, So I assue we are getting v(Pi) and r(Pi) a(t) = [-sin(t), -8*cos(t)] To find v(t), take integral of a(t) v(t) = cos(t)*i - 8*sin(t)*i + C v(0) = cos(0)*i - 8*sin(0)*j + C = i v(0) = i + C = i Solving for C: C = 0 So v(t) = Now solve for r(t) by taking integral of v(t) r(t) = sin(t)*i + 8*cos(t)*j + C r(0) = sin(0)*i + 8*cos(0)*j + C = 8j r(0) = 8*j + C = 8*j Solving for C, C = 0 Therefore, r(t) = Now, plug in t = Pi v(Pi) = = <-1, 0> r(Pi) = = <0, 8> Ans: v(Pi) = = <-1, 0> r(Pi) = = <0, 8> --------------------------------------------- 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, 9 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, 6 How fast is the function changing right now? Here is how I do it (Explain everything) So rate of change initial is <0, 9> Rate of change final is <1, 6> I am not entirely sure how to do this problem, my attempt is find the difference and then take the magnitude. So: <1-0, 6-9> = <1, -3> Magnitude: sqrt(1^2 + (-3)^2) = sqrt(1 + 9) = sqrt(10) The function is changing sqrt(10) ANS: sqrt(10) --------------------------------------------- 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 = 6, z = 4 Here is how I do it (Explain everything) The equations are really simple, since it is just x = 0, y = 6, z = 4 So the point of intersection of these three planes is just (0, 6, 4) ANS: (0, 6, 4)