AMS 10/10A, Homework 1
... Problem 4. Express the following complex numbers using exponential notation, i.e., z = reθi . 1. cos α − i sin α; 2. sin α + i cos α; 3. sin α − i cos α. Hint: first use trigonometric identities to write each one in the form of cos θ + i sin θ. ...
... Problem 4. Express the following complex numbers using exponential notation, i.e., z = reθi . 1. cos α − i sin α; 2. sin α + i cos α; 3. sin α − i cos α. Hint: first use trigonometric identities to write each one in the form of cos θ + i sin θ. ...
solns to sample exam
... 8. The air in a small room 10 ft by 10 ft by 10 ft is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon monoxide is blown into the room at a rate of 100 ft3 /min. The air in the room is kept well-mixed, and air flows out of the room through a vent at the same rate. Determine the ...
... 8. The air in a small room 10 ft by 10 ft by 10 ft is 3% carbon monoxide. Starting at t = 0, fresh air containing no carbon monoxide is blown into the room at a rate of 100 ft3 /min. The air in the room is kept well-mixed, and air flows out of the room through a vent at the same rate. Determine the ...
Section 4.3 Line Integrals - The Calculus of Functions of Several
... We need to verify the previous statement in general before we can state our definition of the line integral. Note that in these two examples, ψ(t) = ϕ 2t . In other words, ψ(t) = ϕ(g(t)), where g(t) = 2t for −2 ≤ t ≤ 2. In general, if ϕ(t), for t in an interval [a, b], and ψ(t), for t in an interva ...
... We need to verify the previous statement in general before we can state our definition of the line integral. Note that in these two examples, ψ(t) = ϕ 2t . In other words, ψ(t) = ϕ(g(t)), where g(t) = 2t for −2 ≤ t ≤ 2. In general, if ϕ(t), for t in an interval [a, b], and ψ(t), for t in an interva ...
Page 1 MCV4U0 PROBLEM SET V: Derivatives II In order to receive
... For critical points, we need A ′( x ) = 0 . Hence, A ′( x ) = 8 x − 3x 2 which equals 0 when x = 0 and x = 83 . Since the function is continuous and differentiable over the given interval, then any absolute maximum or minimum must come from the endpoints and/or critical points as guaranteed by the e ...
... For critical points, we need A ′( x ) = 0 . Hence, A ′( x ) = 8 x − 3x 2 which equals 0 when x = 0 and x = 83 . Since the function is continuous and differentiable over the given interval, then any absolute maximum or minimum must come from the endpoints and/or critical points as guaranteed by the e ...
Regularity of minimizers of the area functional in metric spaces
... v ∈ BVf (Ω). We can pick a minimizing √ sequence ui ∈ BVf (Ω) such that F(ui , Ω) → m as i → ∞. Since a ≤ 1 + a2 , we see that kDui k(Ω∗ ) ≤ F(ui , Ω) for every i = 1, 2, . . . , and consequently (kDui k(Ω∗ )) is a bounded sequence of real numbers. Since ui − f ∈ BV0 (Ω), we have Z |ui − f | dµ ≤ C ...
... v ∈ BVf (Ω). We can pick a minimizing √ sequence ui ∈ BVf (Ω) such that F(ui , Ω) → m as i → ∞. Since a ≤ 1 + a2 , we see that kDui k(Ω∗ ) ≤ F(ui , Ω) for every i = 1, 2, . . . , and consequently (kDui k(Ω∗ )) is a bounded sequence of real numbers. Since ui − f ∈ BV0 (Ω), we have Z |ui − f | dµ ≤ C ...
Parametric Curves
... step size for u or…(any ideas?) • …use your Bresenham code to draw short straight lines between the points you generate on the curve (to fill gaps) • There is no trick (that I’m aware of) comparable to the Bresenham approach ...
... step size for u or…(any ideas?) • …use your Bresenham code to draw short straight lines between the points you generate on the curve (to fill gaps) • There is no trick (that I’m aware of) comparable to the Bresenham approach ...
lecture6n
... ( N , M ) order of the equation = # of energy storing devices in the system. Often N M and the order is referred to as N . To solve equations of this kind it is required to have initial values coming from the past (memory) in general for an order N system, N values are required. It is very often u ...
... ( N , M ) order of the equation = # of energy storing devices in the system. Often N M and the order is referred to as N . To solve equations of this kind it is required to have initial values coming from the past (memory) in general for an order N system, N values are required. It is very often u ...
205 13.1 and 13.3
... Ex. Let f x, y xe , find fx and fy and evaluate them at (1,ln 2). x2 y ...
... Ex. Let f x, y xe , find fx and fy and evaluate them at (1,ln 2). x2 y ...
Maths - Nanyang Technological University
... Derivatives of standard functions. Derivative of a composite function. Differentiation of sum, product and quotient of functions and of simple functions defined parametrically. Applications of differentiation to gradients, tangents and normals, stationary points, velocity and acceleration, connected ...
... Derivatives of standard functions. Derivative of a composite function. Differentiation of sum, product and quotient of functions and of simple functions defined parametrically. Applications of differentiation to gradients, tangents and normals, stationary points, velocity and acceleration, connected ...
Here
... Take = /8 and solve the above equation by finite difference for each of the following boundary conditions: a) y(0) = 0, y(/2) = 1; b) y'(0) = 0, y'(/2) = 1.; c) y'(0) + y(0) = 1, y'(/2) + y(/2) = 0. 2. Use Taylor’s method of order two to approximate the solutions for the following initial va ...
... Take = /8 and solve the above equation by finite difference for each of the following boundary conditions: a) y(0) = 0, y(/2) = 1; b) y'(0) = 0, y'(/2) = 1.; c) y'(0) + y(0) = 1, y'(/2) + y(/2) = 0. 2. Use Taylor’s method of order two to approximate the solutions for the following initial va ...
Geodesic ray transforms and tensor tomography
... Need that (M, g ) ⊂⊂ (R × M0 , g ) where (M0 , g0 ) is compact with boundary, and g is conformal to e ⊕ g0 . Here v is related to a high frequency quasimode on (M0 , g0 ). Concentration on geodesics allows to use Fourier transform in the Euclidean part R and attenuated geodesic ray transform in (M0 ...
... Need that (M, g ) ⊂⊂ (R × M0 , g ) where (M0 , g0 ) is compact with boundary, and g is conformal to e ⊕ g0 . Here v is related to a high frequency quasimode on (M0 , g0 ). Concentration on geodesics allows to use Fourier transform in the Euclidean part R and attenuated geodesic ray transform in (M0 ...
Derivatives and Integrals of Vector Functions
... a vector function r is the derivative of r ′, that is, r ″ = (r ′)′. For instance, the second derivative of the function, r(t) = 〈2 cos t, sin t, t〉, is r ″t = 〈–2 cos t, –sin t, 0〉 ...
... a vector function r is the derivative of r ′, that is, r ″ = (r ′)′. For instance, the second derivative of the function, r(t) = 〈2 cos t, sin t, t〉, is r ″t = 〈–2 cos t, –sin t, 0〉 ...
Section 11.6
... Let x = cos t, y = sin t, and z = t. Then x2 + y 2 = cos2 t + sin2 t = 1. Thus, the curve lies on a circular cylinder of radius 1 centered about the z-axis. Since z = t, the curve moves upward along the cylinder as t increases. This curve is called a helix. ...
... Let x = cos t, y = sin t, and z = t. Then x2 + y 2 = cos2 t + sin2 t = 1. Thus, the curve lies on a circular cylinder of radius 1 centered about the z-axis. Since z = t, the curve moves upward along the cylinder as t increases. This curve is called a helix. ...
Solutions for Exam 4
... Instructions. This exam contains seven problems, but only six of them will be graded. You may choose any six to do. Please write DON’T GRADE on the one that you don’t want me to grade. In writing your solution to each problem, include sufficient detail and use correct notation. (For instance, don’t fo ...
... Instructions. This exam contains seven problems, but only six of them will be graded. You may choose any six to do. Please write DON’T GRADE on the one that you don’t want me to grade. In writing your solution to each problem, include sufficient detail and use correct notation. (For instance, don’t fo ...
Problem Set 3 Partial Solutions
... e + C. For the integral of , write = · and use the scalar linearity property ...
... e + C. For the integral of , write = · and use the scalar linearity property ...
The cohomological proof of Brouwer's fixed point theorem
... Figure 1: (a) Brouwer reasoned that at each instant while stirring his morning cup of coffee, there must exist a particle on the surface which is in the same position as it was before the stirring commenced. (b) The vector field V is locally (but not globally!) the gradient of the argument function ...
... Figure 1: (a) Brouwer reasoned that at each instant while stirring his morning cup of coffee, there must exist a particle on the surface which is in the same position as it was before the stirring commenced. (b) The vector field V is locally (but not globally!) the gradient of the argument function ...
Math 165 – worksheet for ch. 5, Integration – solutions
... y = 2t + 5 cos(πt) dt = t2 + sin(πt) + C. π (you could use a substitution u = πt here, with second of the given equations to get 18 = y(2) = 4 + ...
... y = 2t + 5 cos(πt) dt = t2 + sin(πt) + C. π (you could use a substitution u = πt here, with second of the given equations to get 18 = y(2) = 4 + ...
Ph.D. QUALIFYING EXAM DIFFERENTIAL EQUATIONS Spring II, 2009
... 3. Consider the eikonal equation u~ + u~ = u~. (a) Find all solutions of the form u(x, y) = f(x). (b) Use (a) to write down a general solution u = u(x, y, a, b). (Hint: Use the fact that the PDE is invariant under rotations in the xy) (c) Find the solution of the PDE satisfying the condition u(x, x) ...
... 3. Consider the eikonal equation u~ + u~ = u~. (a) Find all solutions of the form u(x, y) = f(x). (b) Use (a) to write down a general solution u = u(x, y, a, b). (Hint: Use the fact that the PDE is invariant under rotations in the xy) (c) Find the solution of the PDE satisfying the condition u(x, x) ...
answers, in pdf - People @ EECS at UC Berkeley
... we have that the curves y = 4x and y = 12 − x2 intersect at x = 2. Letting x = 12 − x2 , we get x2 + x − 12 = 0, i.e., (x + 4)(x − 3) = 0, i.e., x = −4 or x = 3. Since we are dealing with positive x-values only, we have that the curves y = x and y = 12 − x2 intersect at x = 3. Now to find the total ...
... we have that the curves y = 4x and y = 12 − x2 intersect at x = 2. Letting x = 12 − x2 , we get x2 + x − 12 = 0, i.e., (x + 4)(x − 3) = 0, i.e., x = −4 or x = 3. Since we are dealing with positive x-values only, we have that the curves y = x and y = 12 − x2 intersect at x = 3. Now to find the total ...
Document
... If the points P and Q have position vectors r(t) and r(t + h), then represents the vector r(t + h) – r(t), which can therefore be regarded as a secant vector. If h > 0, the scalar multiple (1/h)(r(t + h) – r(t)) has the same direction as r(t + h) – r(t). As h 0, it appears that this vector approac ...
... If the points P and Q have position vectors r(t) and r(t + h), then represents the vector r(t + h) – r(t), which can therefore be regarded as a secant vector. If h > 0, the scalar multiple (1/h)(r(t + h) – r(t)) has the same direction as r(t + h) – r(t). As h 0, it appears that this vector approac ...
HOMEWORK 5 DUE: Fri., Apr. 30 NAME: DIRECTIONS: • STAPLE
... solutions form a fundamental set of solutions? Explain your answer. The characteristic equation is given by m4 − 5m + 4 = 0. This may be factored as m2 − 4 Hence the solution is given by ...
... solutions form a fundamental set of solutions? Explain your answer. The characteristic equation is given by m4 − 5m + 4 = 0. This may be factored as m2 − 4 Hence the solution is given by ...
MATH 2720 Winter 2012 Assignment 4 Questions 1, 2, 6 and 8 were
... For any u, kuk ≥ 0 with kuk = 0 only when u = 0. It follows that V is decreasing except where c is not smooth. Remarks. 1. This property is true if c is piecewise-smooth. Indeed, in more advanced calculus/analysis, it is shown that the definition of a decreasing function can be made more general; f ...
... For any u, kuk ≥ 0 with kuk = 0 only when u = 0. It follows that V is decreasing except where c is not smooth. Remarks. 1. This property is true if c is piecewise-smooth. Indeed, in more advanced calculus/analysis, it is shown that the definition of a decreasing function can be made more general; f ...
Solutions to some problems (Lectures 15-20)
... (b) The integral C F~ · d~r = 0 because F~ is a gradient field, and f is a continuous function, so we can apply the fundamental theorem of calculus and show thatR the circulation of F~ is equal to zero. ~ ·d~r, this integral is equal to zero, since H is a gradient The same for C H field and h is a ...
... (b) The integral C F~ · d~r = 0 because F~ is a gradient field, and f is a continuous function, so we can apply the fundamental theorem of calculus and show thatR the circulation of F~ is equal to zero. ~ ·d~r, this integral is equal to zero, since H is a gradient The same for C H field and h is a ...