MATH H53 : Mid-Term-1

Transcription

1 MATH H53 : Mid-Term-1 22nd September, 215 Name: You have 8 minutes to answer the questions. Use of calculators or study materials including textbooks, notes etc. is not permitted. Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. Question Points Score Total: 5 1

2 1. (1 points) Sketch the curve r = cos θ, θ 2π and find the area of the inner loop. Solution: It is clear that r = when θ = 2π/3 and 4π/3. Hence the equation for the inner loop is r = cos θ, 2π/3 θ 4π/3. So the area is given by A = 1 2 4π/3 2π/3 (1 + 2 cos θ) 2 dθ = 1 2 = 1 2 = = = π/3 π/3 (1 + 2 cos (π + ϕ)) 2 dϕ (where θ = π + ϕ) (1 2 cos ϕ) 2 dϕ (since cos(π + ϕ) = cos ϕ) (1 2 cos ϕ) 2 dϕ (since cos ϕ is even) (1 4 cos ϕ + 4 cos 2 ϕ) dϕ (3 4 cos ϕ + 2 cos 2ϕ) dϕ = 3ϕ 4 sin ϕ + sin 2ϕ = π π/3 2

3 2. (a) (5 points) Find the point of intersection of the lines r 1 (t) = (1,, 2) + t(2, 1, 2) and r 2 (s) = (1, 1, 1) + s( 1, 1, 1) or show that they are skew lines. Solution: If the lines intersect, then we can simultaneously solve 1 + 2t = 1 s t = 1 + s 2 2t = 1 + s From the first two equations we see that t = 1, s = 2. Substituting in the third equation we see that the left hand side is but the right hand side is 1 which is impossible and hence the lines are skew. (b) (5 points) If the lines intersect, find the unique plane containing both the lines. If the lines are skew, find the distance between them. Solution: The common normal to the two lines (which is unique since the lines are not parallel) is given by i j k n = = i + k. Next, points P 1 (1,, 2) and P 2 (1, 1, 1) lie on the two lines, and P 1 P 2 = (, 1, 1). We saw in class that the distance between skew lines is given by the magnitude of the component of P 1 P 2 along n. So, d = comp n P 1 P 2 = n P 1P 2 n =

4 3. (a) (5 points) Show that the points P (1, 2, 2), Q(1, 1, 1), R(, 2, 4) and S(, 2, ) are co-planar. Solution: Let a = PQ = (, 1, 1) b = PR = ( 1,, 2) c = PS = ( 1, 4, 2) Then 1 1 a (b c) = =. Hence the points are co-planar. (b) (5 points) Find the equation of the plane containing all four points. Solution: We can let i j k n = a b = 1 1 = 2i + j k. 1 2 So the equation of the plane is 2x + y z = 2. 4

5 4. (a) (3 points) Find the parametric equation for the curve of intersection of the plane 2z x = 6 with the cone z 2 = x 2 + 3y 2, and identify (circle, ellipse, hyperbola or parabola) the curve. Solution: From the equation of the plane, z = x/ Substituting in the equation on the cone and simplifying we get 3 = x2 4 x + y2 Completing the squares we see that (x/2 1) 2 + y 2 = 4. That is the projection of the curve to the xy-plane is an ellipse which can be parametrized as x = cos t and y = 2 sin t. But then from the equation of the plane z = cos t. So the parametric equation of the curve is r(t) = (2 + 4 cos t, 2 sin t, cos t). Since the projection is bounded, it is clear that the curve will be a circle or an ellipse. Note that r() = (6,, 6) and r(π) = ( 2,, 2). Hence if the curve is a circle then the center will have position vector r c = (2,, 4), and the radius will be R = 2 5. On the other hand r(π/2) = (2, 2, 4), and so r(π/2) r c = 2. If the curve was a circle then this distance should also have been R = 2 5. Hence the curve is not a circle, and is in fact an ellipse! (b) (3 points) Find the curvature at ( 2,, 2). Solution: From he parametric equation above, clearly r(π) = ( 2,, 2). Differentiating the parametric equation, we see that r (π) = (, 2, ), r (π) = (4,, 2) and hence r (π) r (π) = 4( i + 2k). So the curvature κ(π) = r r r 3 = =

6 (c) (4 points) Find the equation for the osculating plane at ( 2,, 2). Solution: A normal to the osculating plane at ( 2,, 2) (or t = π) is given by r (π) r (π) = 4( i + 2k). So we just take n = i + 2k. Then the equation of the plane is x + 2z = 6. Note: This was a silly question. Thanks to Alex for pointing it out. If an entire curve lies in a plane, then the osculating plane of course better be that same plane! A better question might have been to ask for the equation of the osculating circle. Ah...maybe next time! 6

7 5. A particle moves in R 3 so that its position vector r(t) and velocity vector are related by r (t) = b r(t) for some fixed vector b R 3. Suppose also, that r() = 1. (a) (3 points) Show that the particle moves with a constant speed, and that the entire trajectory lies in the unit sphere centered at the origin. Solution: We We compute that d dt r(t) 2 = 2r(t) r (t). But from the equation, r (t) = b r(t), and hence the velocity is orthogonal to the position vector, which in turn implies that the derivative above is zero. Hence r(t) is a constant and the trajectory lies on a sphere centered at the origin. Since r() is a unit vector, the sphere has to be of radius one. To show that the speed is constant we notice that differentiating the equation for velocity, we obtain that r (t) = b r (t). And then the same argument as above shows that the speed is also constant. (b) (4 points) Show that r(t) b is a constant. Using this, show that either the particle is stationary, or it s trajectory is a circle. Solution: Again differentiating d dt r(t) b = r (t) b = (b r(t)) b =. This shows that the trajectory also lies on a plane. Since the (non-empty) intersection of a plane and a sphere is either a point or a circle, the particle is either stationary or the trajectory is a circle. (c) (3 points) If the angle between r() and b is θ, < θ < π show that the particle is not stationary, and calculate the radius of it s circular trajectory in terms of θ. Solution: Since θ, π, the initial velocity is not zero, and the particle is not stationary. The radius can be calculated purely from elementary geometry, or by using our formula for curvature. Method-1 From part (b), b is the normal to the plane containing the trajectory. let the foot of a perpendicular from the origin O to the plane be B (so this will be the center of our trajectory) and denote the tip of r() by P. Then OBP is a right angled triangle with the hypotenuse as OP = 1, and BOP = θ. The radius of the circle is then BP = sin θ. Method-2 Note that, from the above discussion, r (t) = b r (t), r r and r b. Hence κ(t) = r (t) r (t) r (t) 3 = r (t) r (t) r (t) 3 = r (t) b r (t) 2 = b r (t) = b b r(t) sin θ = 1 sin θ. Hence the radius of curvature (which is the radius that we are interested in) is the reciprocal of curvature i.e. R = sin θ. 7