Math 138AWinter 2012R. SchultzSOLUTIONS FOR THE TAKE HOME ASSIGNMENT1. (i) Compute the torsion of the curve γ(t) = (t, t2, t4); there is a formula for expressing the torsionin terms of an arbitrary parametrization in one of the exercises.SOLUTION.Since the formula in the exercises is off by a sign, answers that are correct except for the sign will receivefull credit.The torsion depends on γ0, γ00and γ000, so the first step is to compute these:γ0(t) = (1, 2t, 4t3)γ00(t) = (0, 2, 12t2)γ000(t) = (0, 0, 24t)The torsion is given byτ(t) =γ0× γ00· γ000|γ0× γ00|2and the numerator is the determinant of the matrix whose rows are γ0, γ00and γ000. The usual determinantformula shows that the value for the numerator is 48t.Computation of the cross product by the standard rule implies that γ0×γ00is equal to (16t3, −12t2, 2),so that|γ0× γ00|2= 256t6+ 144t4+ 4and if we substitute this and the previously derived formula for [γ0, γ00, γ000] we obtain the torsion as a functionof t:τ(t) =48t256t6+ 144t4+ 4=12t64t6+ 36t4+ 1(ii) Let F be a real valued function of two variables defined on an open region U of the coordinate planesuch that the gradient ∇F is never 0 on U, and let γ(s) be a curve with an arc length like parametrization(tangent vector always has length 1) whose image lies in the set of points of U such that F (x, y) = 0 andwhose curvature is nonzero. Prove that for all values of s the acceleration γ00(s) is a scalar multiple of thegradient of F at γ(s). [Hint: Prove first that γ0(s) is perpendicular to the gradient at γ(s). What can wesay about two nonzero vectors in the plane which are perpendicular to a given nonzero vector, and why isthis true?]SOLUTION.Following the hint, start by showing γ0(s) is perpendicular to the gradient at γ(s). By the hypotheseswe know that F (γ(s)) = 0, and if we differentiate this with respect to s and apply the chain rule we obtainthe equation0 =ddsF (γ(s)) = ∇F (γ(s)) · γ0(s)which means that the tangent vector is perpendicular to the gradient. On the other hand, the assumption ofan arc length like parametrization means that |γ0(s)|2= 1, and as usual if we differentiate this with respectto s we obtain the equation 2 γ0(s) ·γ00(s) = 0 which is equivalent to γ0(s) ·γ00(s) = 0. Hence both γ00(s) and∇F (γ(s)) are perpendicular to the unit vector γ0(s). Since all these vectors lie in a 2-dimensional vectorspace, the set of all vectors perpendicular to γ0(s) is a 1-dimensional subspace. The condition κ(s) 6= 0implies that γ00(s) is nonzero, and since they lie in a 1-dimensional subspace we know that each is a nonzeromultiple of the other.2. (i) Let U be the set of all points in the coordinate plane except (0, 0), and let T (u, v) be thetransformation from U to itself given by(x, y) = T (u, v) =uu2+ v2,−vu2+ v2.Compute the Jacobian of T and solve for u and v as functions of x and y. [Hint: Show that u2+ v2can beexpressed very simply in terms of x2+ y2.]SOLUTION.First compute the partial derivatives of the coordinate functions.∂x∂u=v2− u2(u2+ v2)2,∂y∂u=−2uv(u2+ v2)2,∂x∂v=2uv(u2+ v2)2,∂y∂v=v2− u2(u2+ v2)2This means that the Jacobian is equal to1(u2+ v2)4·v2− u2−2uv2uv v2− u2=(v2− u2)2+ 4uv(u2+ v2)4=(u2+ v2)2(u2+ v2)4=1(u2+ v2)2.Next, solve for u and v as functions of x and y using the hint. One way to start is to express x2+ y2interms of u and v.x2+ y2=u2(u2+ v2)2+v2(u2+ v2)2=1u2+ v2Therefore we also haveu2+ v2=1x2+ y2which in turn implies that x = u(x2+ y2) and y = −v(x2+ y2), and if we solve for u and v we obtain theequationsu =xx2+ y2, v =−yx2+ y2which means that T is equal to its own inverse transformation!Note. One can also approach this using simple facts about complex numbers. From this perspectivethe transformation has the form T (z) = 1/z, and T = T−1corresponds to the identityz =11/z.(ii) If L is the line defined by the equation u + v = 1, then the image of L under T is contained in acircle. Find an equation in x and y which defines that circle.SOLUTION.Substitute the expressions for u(x, y) and v(x, y) into the equation u + v = 1; the resulting equation is1 = u + v =xx2+ y2+−yx2+ y2.If we clear this of fractions and ignore the middle equation we obtain the equation x2+ y2= x − y, whichis equivalent to x2−x + y2+ y = 0. If we complete the squares of the quadratic expressions in x and y thisequation becomesx2− x +14+ y2+ y +14=12which can be rewritten in the formx −122+y +122=12.3. Let T (u, v) be the transformation of the coordinate plane given by(x, y) = T (u, v) =u2− v2, 2uv.(i) If L is the horizontal line defined by the equation v = C for some constant C, then the image of Lunder T is a parabola. Find an equation in x and y which defines this parabola.SOLUTION.We have the following system of three equations in x, y, u, v:x = u2− v2, y = 2uv , v = CWe need to reduce this to a single equation in x and y by eliminating u and v. The third equation quicklyeliminates v, yielding the following system of two equations in x, y, u:x = u2− C2, y = 2uCNow solve the second equation for u in terms of y and substitute this expression into the first equation:uy2C=⇒ x =y24C2− C2This is the equation for the parabola which is the image of the horizontal line v = C.(ii) Suppose now that L is the line defined by the equation u + v = 1. Find a nontrivial equation in xand y which is satisfied by the image of L under T . [Note: A nontrivial equation is one that is not satisfiedby every point in the coordinate plane — for example, 0x + 0y = 0.]SOLUTION.In this case we obtain the following system:x = u2− v2, y = 2uv , u + v = 1The first step is to solve the third equation for v in terms of u and to substitute this into the first and secondequations.v = 1 − u =⇒ x = u2− (1 − u)2= 2u − 1, y = 2u(1 −u) = 2u −2u2Now eliminate u by solving for u in the first equation and substituting the result into the second:x = 2u − 1 =⇒ u =x −12=⇒y = (x − 1) −(x − 1)22= −12x2+ 2x −32(iii) Finally, suppose that L is the line defined by the equation v = u/√3. Then the image of L underT is contained in some line. Find an equation in x and y which defines this line. Are all points on the linedescribable as images of points in L? Give reasons for your answer.SOLUTION.In this case we obtain the following system:x = u2− v2, y = 2uv , v = u/√3Eliminate v from the first two equations by