Tuesday, October 30 ∗ Solutions ∗ Transformations of R2.Purpose: In class, we’ve seen several different coordinate systems on R2and R3beyond the usualrectangular ones: polar, cylindrical, and spherical. The lectures on Friday and Monday will cover thecrucial technique of simplifying hard integrals using a change of coordinates (Section 15.9). The pointof this worksheet is to familiarize you with some basic concepts and examples for this process.Starting point: Here we consider a variety of transformations T : R2→ R2. Previously, we have usedsuch functions to describe vector fields on the plane, but we can also use them to describe ways ofdistorting the plane:T1. Consider the transformation T (x, y) =(x −2y, x +2y).(a) Compute the image under T of each vertex in the below grid and make a careful plot ofthem, which should be fairly large as you will add to it later.To speed this up, divide the task up among all members of the group.yx(0,0)(−1,2) (2,2)(−1,−1) (2,−1)SOLUTION:See the image following part (f).(b) For each pair A and B of vertices of the grid joined by a line, add the line segment joiningT (A) to T (B) to your plot. This gives a rough picture of what T is doing.SOLUTION:See the image following part (f).(c) What is the image of the x-axis under T ? The y-axis?SOLUTION:The image of the x-axis is the line y = x. The image of the y-axis is the line y = −x. To seethis, parametrize the x-axis as r(t ) = (t, 0),−∞ < t < ∞. Then T (r(t )) = (t, t), −∞ < t < ∞,which traces out the line y = x. Do the same for the y-axis.(d) Consider the line L given by x + y = 1. What is the image of L under T ? Is it a circle, anellipse, a hyperbole, or something else?SOLUTION:Parametrize L by r(t) = (t, 1 −t),−∞ < t < ∞. T (L) is parametrized by T (r(t )) = (t −2(1 −t), t +2(1 −t)) =(3t −2, −t +2). These are the parametric equations of a line.(e) Consider the circle C given by x2+y2=1. What is the image of C under T ?SOLUTION:Parametrize C by r(t) =(cos t, sin t),0 ≤ t ≤ 2π. Then T (r(t )) = (cos t−2sin t ,cos t+2 sin t), 0 ≤t ≤ 2π. Note that if we let x = cos t −2 sin t, y = cos t +2 sin t , then y − x = 4 sin t andy + x = 2cos t. So the curve T (C) satisfies the equation¡y−x4¢2+¡y+x2¢2= 1. This is theequation of an ellipse.(f) Add T (L), T (C) and T¡ ¢to your picture. Check your answer with the instructor.SOLUTION:Note: The transformation T is a particularly simple sort called a linear transformation.2. Consider the transformation T (x, y) =¡y, x(1+y2)¢. Draw the image of the picture below underT .xy(1,1)SOLUTION:Label the 5 line segments as at left below. The image of the left hand picture is the right handpicture.We can figure this out as follows. First parametrize the line segments:rA(t ) = (0, t ),0 ≤ t ≤ 1 rB(t ) = (t,0), 0 ≤ t ≤ 1rC(t ) = (1, t ),0 ≤ t ≤ 1 rD(t ) = (t,1), 0 ≤ t ≤ 1rE(t ) = (t, t ),0 ≤ t ≤ 1Then compute the image under T of each of these:T (rA(t )) = (t,0), 0 ≤ t ≤ 1 T (rB(t )) = (0, t ),0 ≤ t ≤ 1T (rC(t )) = (t,1 +t2),0 ≤ t ≤ 1 T (rD(t )) = (1,2t), 0 ≤ t ≤ 1T (rE(t )) = (t, t (1 +t2)),0 ≤ t ≤ 1Graphing each of these gives the image above at left.3. In this problem, you’ll construct a transformation T : R2→R2which rotates counter-clockwiseabout the origin by π/4, as shown below.T(a) Give a formula for T in terms of polar coordinates. That is, how does rotation affect r andθ?SOLUTION:T (r,θ) = (r,θ +π/4)(b) Write down T in terms of the usual rectangular (x, y) coordinates. Hint: first convert intopolar, apply part (a) and then convert back into rectangular coordinates.SOLUTION:First convert (x, y) into polar:(r,θ) = (qx2+y2,arctan(y/x))Then apply T in polar coordinates:T (r,θ) = (r,θ +π/4)Then convert the result to rectangular coordinates:T (x, y) = (r cos(θ +π/4),r sin(θ +π/4)), where r =px2+y2,θ = arctan(y/x).Recall the double angle formulas cos(a +b) = cos(a)cos(b) −sin(a)sin(b) and sin(a +b) =sin(a)cos(b) +sin(b)cos(a). Using these we see thatcos(θ +π/4) = cosθ cos(π/4) −sinθ sin(π/4) =p2/2(cosθ −sinθ)andsin(θ +π/4) = sinθ cos(π/4) +sin(π/4)cosθ =p2/2(sinθ +cosθ).Hence we haver cos(θ +π/4) =p2/2(r cosθ −r sin θ) =p2/2(x − y)andr sin(θ +π/4) =p2/2(r sinθ +r cos θ) =p2/2(x − y).So we haveT (x, y) = (p2/2(x − y),p2/2(x −
View Full Document
Unlocking...