MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME:Rachel Balji RUID:199001403 EMAIL:rjb356@scarletmail.rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]= -10/11 Answer[ 2 ]= Answer[ 3 ]= < 0, 0, 3> Answer[ 4 ]= Answer[ 5 ]= abs min is 14 and the abs max of the function is 54. Answer[ 6 ]= lim(x^2 - cx)/(x-cx) = positive infinity Answer[ 7 ]= Answer[ 8 ]= Answer[ 9 ]= Answer[ 10 ]= The intersection of the three planes is (1,4,9). ----------------------------------------------------------------- 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]= 9 , a[4]=0 , a[5]= 0, a[6]=1 , a[7]=4 , a[8]= 0, a[9]=3 -------------------------------------------- --------------------------------------------- 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^9+1*x*y*z^2 = 3+1 Here is how I do it (Explain everything) First I found the derivative of the function with respect to y implicity. 9*y^8 + 9*z^8*z' + x*z^2 + 2*z*z'*y*x = 0 Then I seperated the variable z' (which is also known as dz/dy) z' = (-9*y^8 - x*z^2) / ( 9*z^8 - 2*z*y*x) Then I subsitued in the given values of (x,y,z) which is (1,1,1) into the final equation: dz/dy (1,1,1) = (-9*(1)^8 - (1)*(1)^2) / ( 9*(1)^8 - 2*(1)*(1)*(1)) Ans.: dz/dy(1,1,1) = -10/11 --------------------------------------------- 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,-1,6>. Is f increasing or decreasing at the direction <1,9,-1>? Here is how I do it (Explain everything) Ans.: --------------------------------------------- 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*1+y^3*9+z^3*1 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 I determined the gradient: gradf = < f_x , f_y , f_z > f_x = 3*x^2 f_y = 27*y^26 f_x = 3*z^2 Therefore, gradf = <3*x^2 , 27*y^26 , 3*z^2 > Then I subsituted in the given point P =(1, -1, 1) into the x, y, and z values of gradf: gradf (1,-1,1) = < 3*(1)^2 , 27*(-1)^26 , 3*(1)^2 > gradf (1,-1,1) = < 3, 27, 3 > Then I found the vector from P to Q by subtracting the two points: Vector PQ = < q1-p1 , q2-p2, q3-p3> Vector OQ = <1-1, -1-(-1) , 3-1 > => <0,0,2> Then I found the value of |u| by determining the magnitude of PQ and then dividing all values of the vector by the magnitude: |PQ| = sqrt(0^2 + 0^2 + 2^2) = 2 ||u|| = <0/2, 0/2, 2/2> => <0,0,1> Then I mulitpled the gradf by ||u|| in order to find the directional derivative: gradf * ||u|| = <3, 27, 3> * <0,0,1> => <0,0,3> Ans.: <0,0,3> --------------------------------------------- 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 Here is how I do it (Explain everything) Ans.: --------------------------------------------- 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 + 4*y + 9 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [1, 9], B = [9, 1], C = [1, 1] Here is how I do it (Explain everything) First I found the range and domain of the function: If the vertices are A = [1, 9], B = [9, 1], C = [1, 1] then the range is 1<=y<=9 and the domain is 1<=x<=9 Then I found the critical point of the function: f_x = 1 f_y = 4 critical point = (1,4) f(1,4) = 26 The for each part of the boundary, I found the absolute minimum and maxium Fr the left side : F(1,y) = 10 +4y F'(1,4) = 4 F(4) = 26 F(1) = 14 F(9) = 46 For the right side: F(9, y) = 18 + 4y F'(9,4) = 4 F(4) = 34 F(1) = 22 F(9) = 54 For the down side: F(x,1) = x +13 F'(x,1) = 1 F(1)1 = 14 F(1) = 14 F(9) = 22 For the up side: F(x, 9) = x +45 F(1) = 46 F(9) = 54 The final abs min contenders are 14, 22, 14, 46 and 26. The final abs max contenders are 46, 22, 26, 54, 54. Ans.: The abs min is 14 and the abs max of the function is 54. --------------------------------------------- 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) I first determined if the top and bottom both vanish. In this case, they both do and the limit turns to 0/0 which is undefined. Then, I proved that the limit DNE by finding the limit along a straght line y=cx in f(x,y). lim(x^2 - cx)/(x-cx) as x approaches 1 In this case, the limit would approach infinity (positive) and the value of cx would not matter. Hence, the limit DOES exist and is inifnity since you will only get infinity even if you approach the point (1,1) on different lines. lim(x^2 - cx)/(x-cx) = positive infinity --------------------------------------------- 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, 9*t^2] At the point (1,0,0) Here is how I do it (Explain everything) First I determined the dereivative of r(t) and then found the magnitude r'(t) = <0, 9, 18t> |r'(t)| = sqrt(0^2 + 9^2 + (18t)^2) = --------------------------------------------- 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) First I integrated the acceleration vector in order to find velocity at the point (1,0). The integral of --------------------------------------------- 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 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 = 4, z = 9 Here is how I do it (Explain everything) Since each plane is given by the linear equation of the form ax + by + cz = d, one can say that the above equations gives us the three unknowns for the point (x,y,z). The intersection of the three planes is (1,4,9).