Solution 2. ((3)(1) (2)(1), (4 3), (4)(2) (3)(3)) = (1, 1, 1) D u (f) = (6x + 2yz, 2y + 2xz, 2xy) (0,1,1) = = 4 14

Transcription

1 Vector and Multivariable Calculus L Marizza A Bailey Practice Trimester Final Exam Name: Problem 1. To prepare for true/false and multiple choice: Compute the following (a) (4, 3) ( 3, 2) Solution 1. (4)( 3) + (3)(2) = = 6 (b) (4, 3, 1) (3, 2, 1) Solution 2. ((3)(1) (2)(1), (4 3), (4)(2) (3)(3)) = (1, 1, 1) (c) Let f(x, y, z) = 3x 2 y 2 + 2xyz. Find the derivative of f in the direction of (3, 1, 2) at the point (0, 1, 1). Solution 3. (3, 1, 2) D u (f) = (6x + 2yz, 2y + 2xz, 2xy) (0,1,1) = (2, 2, 0) (3, 1, 2) 1 = = 4 (d) Find the area of the parallelogram spanned by the vectors (2, 3) and (1, 1). Solution 4. To find the area of the parallelogram, you need only take the determinant of the matrix ( 2 1 ) 3 1 which has determinant (2)( 1) (3)(1) = 2 3 = 5 so the area is 5. (e) Find the equation of the plane through the point (1, 2, 0) normal to the vector (1, 2, 1). Solution 5. The plane equation (1, 2, 1) (x, y, z) = (1, 2, 0) (1, 2, 1) can be simplified to x + 2y + z = 5. (f) Find the equation of the line through the points (1, 2, 0) and ( 1, 3, 2). Solution 6. The equation of a line requires a direction vector (not necessarily a unit vector) and a point. The direction vector is (1, 2, 0) ( 1, 3, 2) = (2, 1, 2) and any of the two points will do. So the line equation (2, 1, 2)t + (1, 2, 0).

2 (g) Find the equation of the tangent line to the curve r(t) = (3t 2 1, 2t + 1, t 2 + 2) at the point t = 2. Solution 7. The direction vector for the tangent line is the velocity vector of the curve at time t = 2. r (t) = (6t, 2, 2t) at t = 2 is r (2) = (12, 2, 4). A point on the curve can be achieved by finding the point of tangency on the curve, r(2) = (11, 5, 6). Therefore, the equation of the line is (12, 2, 4)t + (11, 5, 6). (h) Find the tangent and normal vector, T and N, of the space curve r(t) = (2 + t, 6 + ln(sec(t)), 2). Solution 8. The tangent vector is the unit vector in the direction of the velocity vector. Since, r sec(t) tan(t) (t) = (1,, 0) = (1, tan(t), 0). sec(t) Since, r (t) = 1 + tan 2 (t) = sec(t), then T (t) = (cos(t), sin(t), 0) and N(t) = ( sin(t), cos(t), 0) (i) Find the first order partial derivatives of the function f(x, y) = sin(x) cos(y) + x 2 y 3. Solution 9. f xx = cos(x) cos(y) + 2xy 3 and f yy = sin(x) sin(y) + 3x 2 y 2 (j) Find the direction of greatest increase of the function f(x, y, z) = 3xyz + 2 cos(xy) at the point (0, 1, 2). Solution 10. The direction of greatest increase is the unit vector in the direction of the gradient of f. f = (3yz 2y sin(xy), 3xz 2x sin(xy), 3xy) (0,1,2) = (6, 0, 0). The unit vector in this direction is (1, 0, 0). (k) Find the equation of the tangent plane to the surface x 2 2xy + y 2 z = 1 at the point (1, 0, 0). Solution 11. If I allow f(x, y, z) = x 2 2xy + y 2 z, then the surface above is the level surface of f at 1. Therefore, f (1,0,0) is normal to the surface at the point (1, 0, 0), and, therefore, normal to the tangent plane. f = (2x 2y, 2x + 2y, 1) (1,0,0) = (2, 2, 1), and (1, 0, 0) is a point on the plane. So the tangent plane equation is 2x 2y z = 2. Problem 2. Answer the following questions to the best of your ability. (A) How are vectors added and subtracted geometrically? When two vectors, u, v are added, their sum is the diagonal of the parallelogram generated by u and v, starting at the initial point of u. When two vector u and v are subtracted, u v is the vector starting at the terminal point of v, and ending at the terminal point of u. (B) What is the geometric interpretation of dot product? The dot product of vectors, u and v is the projection of u onto v times the length of v.

3 (C) What is the geometric or physical interpretation of cross product? The cross product of u and v is the vector orthogonal to u and v, satisfying the right hand rule whose length is the area of the parallelogram spanned by u and v. (D) What are the necessary pieces of information required to find the equation for a line? A direction vector, v, and a point, p, will yield L(t) = vt + p (E) What are the necessary pieces of information required to find the equation of a plane? A normal vector, N, and a point p are required for the equation of a plane N (x, y, z) = n p (F) How do you find the distance from a point to a line? to a plane? If p is a point and N (x, y, z) = N p 0 is a plane, then the distance from p to the plane is the vector projection of p p o onto N. N This is achieved by (p p o ). N If p is a point and L(t) = vt + p o is a line, then the distance from p to L(t) is (p p o ) sin(θ) where θ is the angle between p p 0 and v. This can be achieved by v (p p o) v = v p p o sin(θ). v (G) Can you prove that a curve whose position vector has constant length is orthogonal to its derivative? Hopefully, you can. If not, look at your notes and a previous quiz. (H) Describe how to parameterize by arc length. You find the arclength function s(t) = t t o r(τ)dτ and solve for t. This is unsually complicated and I don t recommend it for the faint of heart. (I) What is the geometric or physical interpretation of (i) arc length The distance traveled by the imaginary particle that travels along the path of the curve at parameterized by time. (ii) Tangent vector The unit vector in the direction the particle is moving. (iii) Normal Vector The unit vector in the direction of the center of curvature of the curve. (iv) curvature The length of the change in the tangent vector with respect to parametrization by arc length. (v) Binormal Vector The cross-product of the tangent and normal vector, it is the unit vector normal to the osculating plane on which the curve sits at a point p.

4 (vi) torsion The length of change in binormal vector with respect to time. (vii) Normal component of acceleration The curvature times the speed squared is the force of the acceleration pointing in the direction of the center of curvature. (viii) Tangential component of acceleration The rate of the change in the speed of the particle and the force of the acceleration pointing in the direction the particle is moving. (ix) Osculating Plane The plane of the curve at the point, p. (J) Is it important to assume a curve is parameterized by arc length when computing the Tangent, Normal and Binormal vector? Why or why not? It s not important for computing the T NB frame, but it does help. It is however, necessary to deduce that T and N are orthogonal (K) What does it mean for a set in R 2 or R 3 to be open? Why does it matter? A set in R n is open if for every point in the set has a ball of radius r that is completely contained in the set. Multivariable functions which are differentiable on open sets are able to satisfy the hypotheses of theorems. (L) Why is it more difficult to compute the limit of a multi-variable function than the limit of a function in single variable calculus? In order for a limit of a function to exist at a point in its domain, there are many more directions which need to be tested than the right and left side for a single variable function. (M) Why is there no true derivative for a multi-variable function? A true derivative measures the change in the value of the function with respect to one unit. A multivariable function has many units from which to choose. One may consider the change in value of a function in any direction, but to ask for the change in the value of the function with respect to an area unit would not be well defined. (N) What are the conditions necessary to assume that f xy = f yx? Both first order partials and second order partials must be continuous. (O) What is the chain rule for f r if f : R 3 R is a function in three variables and r(t) : R R 3 is a space curve? f(x, y, z) r(t) dt = f dx x dt + f dy y dt + f dz z dt (P) What are the conditions for (a, b) to be a local maxima, local minima, or saddlepoint of f(x, y)? For any of these to exist, we must have f xy = f yx in an open set containing (a, b) and

5 (a, b) must be a critical point. For a local max f xx < 0 and f xx f yy (f xy ) 2 > 0. For a local min f xx > 0 and f xx f yy (f xy ) 2 > 0. For a saddle point f xx f yy (f xy ) 2 < 0. (Q) What are the special geometric properties of gradient? The gradient of f at a point p points in the direction of greatest increase of f at p. It points in the opposite direction of greatest decrease of f at p. It is orthogonal to the level curve of f at f(p) and therefore, points in the direction of no change. (R) Describe the method of Lagrange Multipliers. To optimize the graph of f under a constraint defined by g(x, y, z) = c, one must solve for x, y, z and λ in the equation f = λ g Problem 3. Find the distance from the point p = (7, 9. 7) to the line L(t) = (3 + 4t, 2 5t, 7 + 2t). Solution 12. The line has direction vector v = (4, 5, 2) and a point on it p o = (3, 2, 7). So the vector p p o = (7 3, 9 2, 7 7) = (4, 7, ) needs to be crossed with (4, 5, 2). Taking the cross product det i j k 4 7 = ( 70, (8 + 56), 20 28) = ( 56, 64, 48) = 8(7, 8, 6) The distance from the point p to the line L is 8(7, 8, 6) (4, 5, 2) = = 8 45 =.557 Problem 4. Find the area of the triangle whose vertices are ( 3, 6), (4, 2) and (1, 5). Solution 13. First we need to pick a point and subtract it from the other two points. The triangle whose vertices are the points above has the same area as the triangle whose vertices are the origin, and the points, (4 3, 2 6) = (7, 4) and (1 3, 5 6) = (4, 1). Therefore, the area of the triangle is half the determinant of the matrix [ ] 7 4 det = = So the area is 9 2.

6 Problem 5. Let P 1 be the plane defined by the equation 3x 10y + 7z = 8 and P 2 be the plane defined by the equation 9 = 4x 5y + 2z. Find the angle between the two planes. Solution. The angle between the planes is the same as the angle between the normal vectors. So the angle is given by cos 1 (3, 10, 7) (4, 5, 2) ( (3, 10, 7) (4, 5, 2) = cos 1 ( ) = cos 1 76 ( ) = o Problem 6. Find the normal and tangential component of acceleration for the curve r(t) = (4 cos(t), 2 sin(t)). Solution 15. To find the normal and tangential component of acceleration, we must first find the speed. r (t) = ( 4 sin(t), 2 cos(t)) = 16 sin 2 (t) + 2 cos 2 (t) = 2 sin 2 (t) + 1 So the tangential component of acceleration is and the normal component of acceleration is a N = a 2 a T 2 = a T = d v(t) dt = d 2 sin 2 (t) + 1 dt 2(28 sin(t) cos(t)) = 2 sin 2 (t) + 1 ( 4 cos(t), 2 sin(t)) 2 ( 2(28 sin(t) cos(t)) 2 sin 2 (t) + 1 )2 = 16 cos 2 (t) + 2 sin 2 (t) 2(282 sin 2 (t) cos 2 (t)) 4( sin 2 (t) + 1) = 16 cos 2 (t) 26 sin 2 (t) 282 sin 2 (t) cos 2 (t) sin 2 (t) + 1 Problem 7. Find the distance a particle travels along the curve r(t) = (3 cos(t), 2 sin(t)) from time t = 0 and t = π 4.

7 Solution 16. You will need a calculator to solve this one. On the test, I will either give you one that you could do by hand, or ask you to set up the integral only. π cos 2 (t) + 9 sin 2 (t)dt When put in the calculator, this is approximately Problem 8. Find the equation for the osculating plane of r(t) = (t, t 2, t 3 ) at the point (1, 1, 1). Solution 17. To find the equation of the osculating plane, you need a normal vector and a point on the plane. To find the vector normal to the plane, we need to find the binormal vector, and to do that, we need the tangent and normal vector. Keep in mind that since we don t need a unit vector to be normal to the plane, then we don t have to worry too much about the length of the vector. We do need to make sure that the tangent vector is unit length because we know that the only way we can ensure that its derivative is orthogonal to it is if we make sure that it has constant length. 1 The tangent vector is T (t) = 1 + 4t 2 + 9t (1, 2t, 4 3t2 ) and (0, 2, 6t) the normal vector is N(t) = 1 + 4t2 + 9t (4t + 18t3 )(1, 2t, 3t 2 ). 4 (1 + 4t 2 + 9t 4 ) 3 2 So at the point (1, 1, 1) the time is t = 1, so T (1) = 1 (1, 2, 3) and N(1) = 1 (0, 2, 6) (1, 2, 3) = ( 11 7, 8 7, 9 7 ) = 1 7 ( 11, 8, 9) So the binormal is the unit vector in the direction of the vector i j k = ( , (9 + 33), ) = (42, 42, ) So we can reduce the vector normal to the osculating plane to be (3,-3,1). Therefore, the equation for the osculating plane is 3x 3y + z = (3, 3, 1) (1, 1, 1) = 1. Problem 9. Show that the limit does not exist lim (x,y) (0,0) y x 2 y Solution 18. To show that the limit doesn t exist, we need to find two paths whose limit a different value. Let y = 2x 2, then Yet if y = x 2, then lim (x,2x 2 ) (0,0) 2x 2 x 2 = 2 lim (x, x 2 ) (0,0) x 2 2x 2 = 1 2

8 Problem 10. Find all critical points and classify them: f(x, y) = x 2 xy 2 + 2x + 2y 4 Solution 19. To find the critical points, we need to get the first and second order partials. f x = 2x y and f y = 2yx + 2 so 2x y = 0 2yx + 2 = 0 The second equation yields 2(yx 1) = 0 which implies yx = 1 The first equation yields x = y2 2 so y( y2 2 ) = So as long as k is a constant, y( y2 2 ) = 0 is the same as solving the equation y 3 2y 2 = 2 0, which unfortunately, due to an inadvertent typo, is very difficult to solve without a calculator. So, using a calculator, we get y = and x = Since, f xx = 2, f yy = 2x, and f xy = 2y = f yx. then f xx f yy (f xy ) 2 = (2)( 2(0.565)) ( 2(1.769)) 2 =.777 < 0 So the critical point (0.565, 1.769) is a saddlepoint. Problem 11. A closed rectangular box is to have volume V. The cost of the material of the box is 4 cents per cm 2 for the top and bottom, 6 cents per cm 2 for the front and back, and 7 cents per cm 2 for the left and right side. What dimensions minimize the total cost of the materials. Solution 20. The top and bottom of the box each have area xy cm 2 and cost 4 cents per cm 2, so the top and bottom cost 8xy cents. The sides of the box each have area xz cm 2 and cost 7 cents per cm 2, so the sides cost xz cents. The front and back each have area yz cm 2 and cost 6 cents per cm 2, so the front and bakc cost 12yz cents. The total cost of making the box is C(x, y, z) = 12yz + 8xy + xz and volume V (x, y, z) = xyz. So to minimize cost under the constraint V = xyz, one needs to find the gradient of both. C = (8y + z, 12z + 8x, 12y + x) and V = (yz, xz, xy). So we need to solve equations (8y + z, 12z + 8x, 12y + x) = λ(yz, xz, xy) From this we get equations: 8y + z = λyz 12z + 8x = λxz 12y + x = λxy By multiplying the first equation by x and the second equation by y and the third equation by z, we get 8xy + xz = λxyz 12yz + 8xy = λxyz 12zy + xz = λxyz

9 and therefore, since x, y, z 0, we get that x = 6 7 y and z = 4 7 y. Using the constraint, we get y( 6 7 y)(4 7 y) = V which implies y3 = 49V. Therefore, the point 24 ( V 7 24, V 49V 7 24, 3 24 ). Problem 12. Find the points on the surface x 2 zy = 4 that is closest to the origin. The problem can be solved by using the functions f(x, y, z) = x 2 +y 2 +z 2 and g(x, y, z) = x 2 zy. So the gradients f = (2x, 2y, 2z) and g = (2x, z, y) yields the equations Solution 21. 2x = λ2x 2y = λz 2z = λy The first equation yields λ = 1 or x = 0. Suppose λ = 1, then 2y = z and 2z = y, which means 4y = y and 5y = 0 or y = 0, which in turn, implies z = 0. This would yield the point (x, 0, 0) on the surface x 2 zy = 4, Therefore, (2, 0, 0) or ( 2, 0, 0) are the point closest or furthest from the origin. The distance from those points to the origin is 2. If x = 0, then the points on the surface 0 yz = 4 are of the form y = 4 z. This would mean 8 z = λz, which would mean z = ± 8. Therefore, y = 4 ± 8, therefore the points are (0, 2 8, ) and (0, 8, 2 ). 2 2