Answer Key for Sample Exam for Multivariable Calculus but with material from Calc1 and Calc2 for someone whose RUID is, 201001195 NAME: RUID: , 201001195 EMAIL: BELOW WRITE THE LIST OF THE ANSWERS `Answer[`, 1, ` ]= `, -2/3 `Answer[`, 2, ` ]= `, ` NEITHER ` `Answer[`, 3, ` ]= `, 27 `Answer[`, 4, ` ]= `, [1, 1] `Answer[`, 5, ` ]= `, 3 `Answer[`, 6, ` ]= `, 2 `Answer[`, 7, ` ]= `, 2 `Answer[`, 8, ` ]= `, [0, -1] `Answer[`, 9, ` ]= `, 7 `Answer[`, 10, ` ]= `, [1, 1, 1] --------------------------------------------- With my RUID data the question is --------------- Problem 1: Find dz/dy at the point (1,1,1) if z(x,y) is given implicitly by the equation x*y*z^2+x^2+y+z = 4 --------------- Here is how I do it (Explain everything) First we make sure that the point (1,1,1) lies on the surface, it does! Differentiating with respect to y, we get, let zy(x,y)=dz(x,y)/dy 1+zy(x,y)+x*z(x,y)^2+2*x*y*z*zy(x,y) = 0 Plugging-in x=1,y=1,z=1, set get , let z1=z_y(1,1) 3*z1+2 = 0 Solving for zy(1,1) , aka z1, we get -2/3 --------------------------------------------- With my RUID data the question is --------------- Problem 2: Suppose that grad(f)(P) is [2, -1, 3] If f increasing or decreasint at the direction [2, 1, -1] Here is how I do it (Explain everything) Since we only care about the SIGN of the directional derivative, we don't have to normalize the direction The dot product is 0 Since it is ZERO it is STATIONARY (NEITHER INCREASING NOR DECREASING) NEITHER --------------------------------------------- With my RUID data the question is Problem 3: Find the directional derivative of the function f(x,y,z) x^3+y^3+9*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) The gradient is 2 2 2 [3 x , 3 y , 27 z ] Pluging in x=1, y=-1, z=1, we get [3, 3, 27] The vector PQ is <0,0,2> dividing by its magnitude we have the unit vector <0,0,1> Taking the dot prodcut of , [3, 3, 27], and <0,0,1> we get 27 Ans.:, 27 --------------------------------------------- With my RUID data the question is Problem 4: Find a saddle point of the function f(x,y)= exp(x - 1) - (x - 1) exp(y - 1) If there is no saddle point, write in the Answer "Done Not exist". Explain what you are doing Here is how I do it (Explain everything) We first take the first derivatives f_x, f_y f_x is exp(x - 1) - exp(y - 1) f_y is -(x - 1) exp(y - 1) For future reference we take f_xx,f_xy, f_yy f_xx is exp(x - 1) f_xy is -exp(y - 1) f_yy is -(x - 1) exp(y - 1) We have to solve the system of two equations {-(x - 1) exp(y - 1) = 0, exp(x - 1) - exp(y - 1) = 0} In the two unknowns , {x, y} There is only one solutin {x = 1, y = 1} Plugging in (x,y)=, [1, 1] That fxx is exp(0) That fxy is -1 That fyy is 0 The discrimiant is -1 Since this is negative it follows that the critical point, [1, 1] is a saddle point Ans.: , [1, 1] --------------------------------------------- With my RUID data the question is Problem 5: Let f(x,y) be the function x + y + 1 Find the ABSOLUTE MINIMUM VALUE of f(x,y) INSIDE the TRIANGLE whose VERTICES ARE A = [2, 1], B = [1, 1], C = [1, 1] Here is how I do it (Explain everything) Since the function f(x,y) is LINEAR, there are no local max or min points, and not even on the edges Hence the FINALISTS are the points A = [2, 1], B = [1, 1], C = [1, 1] The VALUE of f(A) is , 4 The VALUE of f(B) is , 3 The VALUE of f(C) is , 3 picking the SMALLEST among, 4, 3, 3 We get that the ABSOLUTE (aks GLOBAL) MINIMUM VALUE is 3 Ans.: , 3 --------------------------------------------- With my RUID data the question is Problem 6: Let f(x,y) be the function 2 2 x - y ------- x - y 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 Both top and bottom are 0 at the designated point, [1, 1], BUT SIMPLIFYING gives x + y NOW IT IS CONTINUOUS everywhere, so plugging-in, x = 1, y = 1 we get, 2 Ans.:, 2 --------------------------------------------- --------------------------------------------- With my RUID data the question is Here is how I do it (Explain everything) Problem 7: Find the curvatute of the curve 2 r(t) = [2, t, t ] At the point (a[1],0,0) Here is how I do it We first find the value of t. It is t=0 r'(t) is [0, 1, 2 t] r''(t) is [0, 0, 2] But we care about t=0 so r'(0) is [0, 1, 0] = j r''(0) is [0, 0, 2] = 2 k r'(0) x r''(0) is, 2 i whose magnitude is, 2 |r'(0)|^3 is, 1 Since the cruvatute at t=0 is |r'(0)x r''(0)|/|r'(0)|^3, we get that the answer is 2 --------------------------------------------- --------------------------------------------- With my RUID data the question is Problem 8: A particle is moving in the plane with ACCELERATION given by [-2*sin(t), -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) The velocity v(t) is obtained by INTEGRATING a(t) with respect to t, getting [2*cos(t), -sin(t)]+C Where C is a constant vector. But since the velocity is, [2, 0], at t=0, we get that C=[0,0] So the velocity is [2 cos(t), -sin(t)] The POSITION, r(t), is obtained by INTEGRATING the velocicty w.r.t. t, getting that it equals [2 sin(t), cos(t)] + C where C is another constant vector Since the position is, [0, 1], at t=0 , we get that C=[0,0], so the position in general is [2 sin(t), cos(t)] [2 sin(t), cos(t)] Finally, plugging-in t=Pi , we get [0, -1] --------------------------------------------- --------------------------------------------- With my RUID data the question is --------------------------------------------- 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, 1 and the rate of change of the function with respect to y is, 1 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, 5 How fast is the function changing right now? We use the CHAIN rule (df/dt)=(df/dx)*(dx/dt)+ (df/dy)*(dy/dt) We have (RIGHT NOW) df/dx=, 1 df/dy=, 1 dx/dt=, 2 dy/dt=, 5 Hence the rate of change of the function is 7 Ans.:, 7 --------------------------------------------- With my RUID data the question is --------------------------------------------- Problem 10: Find the point of intersection of the three planes x = 1, y = 1, z = 1 Here is how I do it (Explain everything) Obviously it is the point, [1, 1, 1]