MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME: Rahul Paleja RUID: 191003667 EMAIL: rpp109@scarletmail.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= -10/3 Answer[ 2 ]=This is the directional derivative which is negative((-8*sqrt(3))/3) at this point in this direction so f is decreasing at the direction <1,1,-1> Answer[ 3 ]= 60*sqrt(11)/11 (WRONG ANS., STARTED CORRECTLY, BUT CONFUSED DIRECTION, -5 POINTS) Answer[ 4 ]= (1,3) is a saddle point Answer[ 5 ]= (NO ANS. ,-10 POINTS) Answer[ 6 ]= limit does in fact exist and equal 0 (WRONG ANS., WRONG WAY, -10 POINTS) Answer[ 7 ]= 2/81 Answer[ 8 ]= pi*I = (WRONG ANS., RIGHT WAY, -5 POINTS) Answer[ 9 ]= (NO ANS. -10 POINTS) Answer[ 10 ]= (1,6,1) SCORE: 60 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]= 1, a[4]=1 , a[5]=1 , a[6]=3 , a[7]=6 , a[8]=6 , 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 find dz/dy at the point (1,1,1) if z(x,y) is given implicitly by the equation x+y^9+z+x*y*z^2 = 3+1 Here is how I do it (Explain everything) To begin, the equation is x+y^9+z+x*y*z^2 = 3+1 This simplifies to x+y^9+z+x*y*z^2-4 = 0 If we differentiate to find dz/dy which is z', the initial differentiation is 9*y^8+z'+x(y*z^2)' = 0 Using product rule, this further simplifies to 9*y^8+z'+x*(2*y*z*z'+z^2) = 0 This simplifies to: 9y^8+z'+2*x*y*z*z'+x*z^2 = 0 Subtract the portions of this equation without z' to the other side z' + 2*x*y*z*z' = -9*y^8-x*z^2 Factor out z' from the left side: z'(1+2*x*y*z) = -9*y^8-x*y^2 Divide (1+2*x*y*z) to the left side: z' = -9*y^8-x*z^2/(1+2*x*y*z) z' = dz/dy = dz/dy(1,1,1) = -9*(1)^8-(1)*(1)^2/(1+2*(1*1*1)) dz/dy at (1,1,1) = -10/3 Ans.: -10/3 --------------------------------------------- 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)=<1,-1,8>. Is f increasing or decreasing at the direction <1,1,-1>? Here is how I do it (Explain everything) To begin, we already have the gradient of f at a desired point p. Now we need the unit vector of direction The magnitude of the direction = sqrt(1^2+1^2+(-1)^2) = sqrt(3) So unit vector of direction = <1/sqrt(3),1/sqrt(3),-1/sqrt(3)> The dot product of the grad(f)(P) and the unit vector of direction = <1,-1,8>.<1/sqrt(3),1/sqrt(3),-1/sqrt(3)> = 1/sqrt(3) - 1/sqrt(3) -8/sqrt(3) = -8/sqrt(3) = (-8*sqrt(3))/3 This is the directional derivative which is negative at this point in this direction so f is decreasing at the direction <1,1,-1> Ans.: This is the directional derivative which is negative((-8*sqrt(3))/3) at this point in this direction so f is decreasing at the direction <1,1,-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*3+y^3*1+z^3*6 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 grad(f) = <9*x^2,3*y^2,18*z^2> at the point (1,-1,1), the grad(f) = <9,3,18> The unit vector of direction = <1/sqrt(11),-1/sqrt(11),3/sqrt(11)> where sqrt(11) = the magnitude of the direction vector = sqrt(1^2+(-1^2)+3^2) The dot product of the grad(f) at point p and the unit vector of direction = 60/sqrt(11) = 60*sqrt(11)/11 Ans.: 60*sqrt(11)/11 --------------------------------------------- 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-3) 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) e^(x-1) - (x-1)*e^(y-3) f_x = e^(x-1)-e^(y-3) f_xx = e^(x-1) f_y = (x-1)(e^(y-3)) f_yy = (x-1)(e^(y-3)) + e^(y-3) f_xy = -e(y-3) setting f_x = 0 and f_y = 0 we get values of x = y-2 and substituting this into f_y we get e^y-3(y-3) = 0 so y must be 3 because e to any exponent cannot equal 0. Thus x = 1. Using this point in the determinant we get: (e^0)(0)-(-e^(3-3))^2 = -1. Thus the determinant is less than 0 and (1,3) is a saddle point Ans.: (1,3) is a 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 + 6*y + 9 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [1, 9], B = [1, 1], C = [1, 3] 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) 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) To begin, take the lim as (x,y)->(1,1) of (x^2-y^2)/(x-y) Plugging in directly does not work so we must set y = cx We get lim(x->0) of (x^2-(cx)^2)/(x-cx) Factor out an x^2 in numerator and x in denominator we get: lim as x-> 0 of x(1-c^2)/(1-c) -> this could equal 0 but we do not know yet because c could vary depending on slope Convert to polar coordinates we get: lim as r->0 (r^2*cos^2(theta)-r^2*sin^2(theta))/(rcos(theta)-rsin(theta)) factoring out r lim as r-> 0 r(cos^(theta)-sin^2(theta))/(cos(theta)-sin(theta)) and substituting r = 0, we get that the limit does in fact exist and equal 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, 1*t^2] At the point (1,0,0) Here is how I do it (Explain everything) To begin t = 0 if we solve for t using point ` r'(t) = <0,9,2t> and r'(0) = <0,9,0> r''(t) = <0,0,2> r'(t)Xr''(t) = 18i the magnitude of this = 18 magnitude of r'(0)^3 = 9^3 = 729 Thus the curvature = 18/729 = 2/81 --------------------------------------------- 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 [-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) a(t) = int(-sin(t)I-9cos(t)j) = cos(t)I-9sin(t)*j + c a(0 = I = c v(t) = (cos(t)+1)I-9sin(t)*j integral of this and c = 9j we get: P(t) = (sin(t)+t)I-(-9cos(t)+9)j Substitute pi in for t we get pi*I = --------------------------------------------- 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, 6 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) --------------------------------------------- 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 = 6, z = 1 Here is how I do it (Explain everything)