MIDTERM 1 FOR Multivariable Calculus, Math 251(22-24), FALL 2020, Dr. Z. NAME:Gillian Mulvoy RUID:202002783 EMAIL:gillian.mulvoy@rutgers.edu BELOW WRITE THE LIST OF THE ANSWERS Answer[ 1 ]=Dz/dy = (-x*z^2 - x*2*y*z - y) / 2*z (WRONG ANS. FORGOT TO PLUGIN NUMBERS, -10 POINTS) Answer[ 2 ]= decreasing Answer[ 3 ]=24 Answer[ 4 ]=Does not exist (WRONG ANS. STARTED CORRECTLY THEN MESSED UP, -5 POINTS) Answer[ 5 ]=10 Answer[ 6 ]= Does not exist (WRONG ANS., WRONG WAY -10 POINTS) Answer[ 7 ]=0 (WRONG ANS. MESSED UP CALCULATIONS -10 POINTS) Answer[ 8 ]= [2*pi, 0] (WRONG ANS., RIGHT WAY -5 POINTS) Answer[ 9 ]=23 Answer[ 10 ]= (1,7,2) ----------------------------------------------------------------- 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]=2 , a[2]= 1 , a[3]=2 , a[4]=1 , a[5]=1 , a[6]=2, a[7]=7 , a[8]=8 , 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 x^2+y^1+z^2+1*x*y*z^2 = 3+1 x^2+y^1+z^2+1*x*y*z^2 = 4 Here is how I do it (Explain everything) (x^2+y^1+z^2+1*x*y*z^2)' = (4)' take the derivative of each side with respect to y (x is a constant) 0+y+2z*dz/dy + 1*x(y*x^2)'= 0 derivative of x^2 becomes zero because in dz/dy x is just like a constant. 1*x can be pulled out of 1*x*y*z^2 because it is a constant y+2*z*dz/dy + 1*x(1*z^2+y*2*z)= 0 the product rule finds the derivative of of y*z^2 y+2*z*dz/dy + (x*z^2+ x*2*y*z)= 0 distributed the 1*x and removed the 1's because they do not mean anything 2*z*dz/dy = -(x*z^2+ x*2*y*z)- y brought anything not connected to the dz/dy onto the other side. Separating variable from derivative Dz/dy = -(x*z^2+ x*2*y*z)- y / 2*z divide both sides by 2*z to get dz/dy alone Dz/dy = (-x*z^2 - x*2*y*z - y) / 2*z made neater Ans.: Dz/dy = (-x*z^2 - x*2*y*z - y) / 2*z --------------------------------------------- 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 <2,2,-1>? Here is how I do it (Explain everything) |<2,2,-1>| = square root (2^2 + 2^2 + (-1)^2)= square root (9) = 3 Find the magnitude of the vector u= 1/3<2,2,-1> = <2/3,2/3,-1/3> Divide the vector by its magnitude to find the unit vector Gradient . Unit vector = <2,-1,9> . <2/3,2/3,-1/3> = 2*2/3 + -1*2/3 + 9*-1/3 = 4/3-2/3-9/3 = -7/3 The directional derivative is negative meaning the function is changing at a negative rate or decreasing. Ans.: 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 Find the directional derivative of the function f(x,y,z) x^3*2+y^3*2+z^3*8 2*x^3+2*y^3+8*z^3 At the point P=(1,-1,1) in the direction pointing to Q=(1,-1,3) Here is how I do it (Explain everything) fx=6*x^2 fy=6*y^2 fz= 24z^2 take the partial derivatives of the function Gradient= <6*x^2, 6*y^2, 24*z^2> Fill into gradient (1,-1,3) - (1,-1,1) = <0,0,2> Find vector PQ by subtracting P from Q |<0,0,2>| = square root (0^2 + 0^2 + 2^2)= square root (4)= 2 Find the magnitude of the vector u=1/2<0,0,2> = <0,0,1> Find the unit vector by divided the vector by its magnitude Gradient(xo,yo,zo) = <6*x^2, 6*y^2, 24*z^2> Gradient(1,-1,1) = <6*1^2, 6*-1^2, 24*1^2> = <6,6,24> Find the gradient at P the initial point Gradient . U = <6,6,24> . <0,0,1> = 6*0 + 6*0 +24*1 = 24 Take the dot product of the gradient and the unit vector 24 is the directional derivative of the function Ans.: 24 is the directional derivative of the function --------------------------------------------- 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-2) Here is how I do it (Explain everything) fx= (exp(x-1)-(x-1)*exp(y-2))' fx= exp(x-1) - (1*exp(y-2)+ (x-1)*exp(y-2)*0) = exp(x-1)-exp(y-2)-0 take the derivative of the function with respect to x fy= (exp(x-1)-(x-1)*exp(y-2))' fy= exp(x-1)*0 - ((x-1)*exp(y-2)+ 1*exp(y-2)) = 0-(x-1)*exp(y-2)+exp(y-2) take the derivative of the function with respect to y fxx= (exp(x-1)-exp(y-2))' fxx= exp(x-1)- exp(y-2)*0 = exp(x-1) take the second derivative of the function with respect to x fyy= (-(x-1)*exp(y-2)+exp(y-2))' fyy= -0*exp(y-2)+ (x-1)*exp(y-2) + exp(y-2)= (x-1)*exp(y-2)+exp(y-2) take the second derivative of the function with respect to x fxy= (exp(x-1)-exp(y-2))' fxy= exp(x-1)*0 - exp(y-2) = -exp(y-2) take the derivative of fx with respect to y exp(x-1)-exp(y-2) = 0 Set fx=0 -(x-1)*exp(y-2)+exp(y-2)= 0 Set fy= 0 -(x-1)*exp(y-2)= -exp(y-2) Solve for x -(x-1)= -exp(y-2)/exp(y-2) Divide exp(y-2) to the other side -x+1 = -1 right side equals -1 -x=-2 x=2 exp(x-1)-exp(y-2) = 0 Fill x=2 into fx to find y exp(2-1)-exp(y-2) = 0 exp(1) - exp(y-2) = 0 -exp(y-2)=-exp(1) Move the constants to the other side Exp(y-2)= exp(1) Remove the - sign on both sides ln(exp(y-2)) = ln(exp(1)) take the natural log of each side to remove the exp Y-2= 1 y= 3 Critical point at (2,3) fxx (2,3)= exp(x-1)= exp(2-1) = exp(1) Plug critical point into fxx fyy (2,3)= (x-1)*exp(y-2)+exp(y-2)=(2-1)*exp(3-2) + exp(3-2)= exp(1)+exp(1) Plug critical point into fyy fxy (2,3)= -exp(y-2)= -exp(3-2)= -exp(1) plug critical point into fxy D= fxx*fyy-fxy = exp(1)*(exp(1)+exp(1)) - (-exp(1)) = 17.49 approximately Solve for the determinate Since the determinate is positive, there is no saddle point Does not exist Ans.: Does not exist --------------------------------------------- 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 +7*y +1 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [2,1], B = [2,1], C = [1,2] Here is how I do it (Explain everything) fx= 1 fy= 7 Since fy and fx never go to 0, there are no critical points F(2,1) = 1*2 + 7*1 +1 = 10 F(2,1) = 10 F(1,2) = 1*1 +7*2 +1 = 15 Since there are no critical points the only options are the endpoints which as the vertices of the triangle. When they are plugged in it shows that (2,1) is the minimum and therefore the absolute minimum with the value of 10. Ans.: 10 --------------------------------------------- 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) (x^2-y^2)/(x-y) Simplified F(1,1)= (1^2-1^2) / (1-1) = 0/0 Undefined so the problem must continue (x^2-y^2)/(x-y) but y=cx so (x^2-(c*x)^2)/(x-c*x) Plug in y=cx so see if the limit is dependent on the slope (x^2-c^2*x^2)/(x-c*x) x=1 (1^2-c^2*1^2)/(1-c*1)= (1-c^2)/(1-c) Plug in x=1 to the equation to see if the limit would depend on c Since no normal number is found from plugging in for x, it is assumed that the limit of the function is dependent on the slope of the equation. This means that there are different limits for different lines and the list does not exist. Answer- The limit does not exist --------------------------------------------- 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) = [2, 1*t, 2*t^2] At the point (2,0,0) Here is how I do it (Explain everything) <2,1*t,2*t^2> = <2,0,0> Find t at the point given 2= 2 1*t = 0 2*t^2= 0 t=0 R'(t)= [0,t,4t] Find the first derivative of r(t) R"(t)= [0,1,4] Find the second derivative of r(t) R'(t) x R"(t) = i j k = i |t 4t| - j |0 4t| + k |0 t| = (4*t- 1*4*t)I - (0*4-4*t*0)j + (0*1- 0*t)k take the cross product of r'(t) and r"(t) 0 t 4t |1 4 | |0 4 | |0 1| 0 1 4 (4*0- 1*4*0)I - (0*4-4*0*0)j + (0*1- 0*0)k = 0i - 0j +0k Fill in t= 0 |R'(t) x R"(t)| = square root (0^2 + 0^2 +0^2) = 0 k(t) = 0 / 0 The curative is 0 Answer- the curvature is 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 [-2*sin(t), -1*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 Here is how I do it (Explain everything) v(t)= integral a(t) find v(t) by taking the integral of a(t) v(t)= [2*cos(t), -1*sint(t)] +Vo V(0)= [2,0] V(t)= [2*cos(t), -1*sint(t)] + [2,0] add velocity at time t=0 v(t)= [2(1+cos(t), -1*sin(t)] x(t)= integral v(t) find x(t) by taking the integral of v(t) x(t)= [2(t+sin(t)), 1*cos(t)] + xo x(0)= [0,1] x(t)= [2(t+sin(t), 1*cos(t)] + [0,1] add position at time t=0 x(t)= [2(t+sin(t)), 1*cos(t)+1] X(pi)= [2(pi+sin(pi), 1*cos(pi)+1] plug in t= pi x(pi)= [2(pi+0), -1+1] x(pi)= [2*pi,0] simplify The particle is located at [2*pi, 0] at time t=pi --------------------------------------------- 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, 7 Both x and y depend on time Right now the rate of change of x with respect to time is, 2 and the rate of change of y with respect to time is, 3 How fast is the function changing right now? Here is how I do it (Explain everything) Df/dx= 1 Df/dy= 7 Dx/dt= 2 Dy/dt= 3 Df/dt= df/dx * dx/dt + df/dy * dy/dt Df/dt= 1*2 + 7*3 = 2+ 21 = 23 The rate of change of the function with respect to time is 23. The function is changing at a rate of 23 right now. --------------------------------------------- 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 = 7, z = 2 Here is how I do it (Explain everything) This point will be when they all equal each other <1,0,0> <0,7,0> <0,0,2> x= 1 y=7 z=2 1a+0y+0z=d 0x+7b+0z= d 0x+0y+2c=d write out the equations of the planes 1a=d 7b=d 2c=d therefore they must meet only at (1,7,2)