Math 20C Homework 2 Partial Solutions

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3 Math 2C Homework 2 Partial Solutions Problem 1 ( ). Calculate (j k) (j + k). Solution. The basic properties of the cross product are found in Theorem 2 of Section From these properties, we deduce (j k) (j + k) = j (j + k) k (j + k) = j j +j k k j + k }{{}}{{ k } = = = j k k j = j k + j k. Next we recall the following adaptation of Figure 7 (also Figure 14) from Section 12.4: i i j = k k j = j k = i k i = j Therefore we know that j k = i, so from the calculation above we conclude that (j k) (j + k) = i + i = 2i = 2,,. Problem 2 ( ). Find v w, where v and w are vectors of length 3 in the xz-plane, oriented as in the following figure, and θ = π 6. z w v θ y x 1

4 Solution. From Theorem 1 of Section 12.4, we know that v w = v w sin θ = 3 3 sin π 6 = 9 2. Moreover, by the right-hand rule (cf. Theorem 1 of Section 12.4), we know that v w points in the negative y direction: z w v v w θ y x Thus v w =, a, for some positive real number a. From knowing that the length of v w is 9/2, we conclude that a = 9/2, so that v w =, 92,. Problem 3 ( ). Find the area of the parallelogram determined by the vectors a,, and, b, c. Solution. Let u = a,, and v =, b, c. By definition, the area of the parallelogram determined by u and v is u v. By Theorem 1(ii) of Section 12.4, we have u v = u v sin θ, (3.1) where θ is the angle between u and v (and θ π). Next, note that u v = a + b + c =, so u and v are perpendicular by the properties of the dot product (see the bottom box on page 68 in Section 12.3). That is, θ = π. Now from (3.1) we see that 2 u v = u v sin θ = a b 2 + c 2 sin π 2 = a b 2 + c 2, and this is the area of the parallelogram determined by u and v. Note that we could have just computed the cross product u v directly and then taken its magnitude, but that would have been less interesting. 2

5 Problem 4 ( ). Find the equation of the plane passing through (4, 1, 9) and parallel to x + y + z = 3. Solution. Let s call the plane we want to find P. The plane given by the equation x + y + z = 3 has normal vector n = 1, 1, 1 (obtained from just reading off the coefficients of x, y, and z in the equation above). Since P is parallel to this plane, n is also a normal vector for P. Thus, we know a normal vector of P (n) and a point on P (the point (4, 1, 9)). Using Theorem 1 of Section 12.5, we can write the equation of P in the following way: n x, y, z = n 4, 1, 9. This is the vector form of the equation of P. To get the scalar form, we simply expand the equation above to get x + y + z = 14. Problem 5 ( ). Find the equation of the plane passing through (3, 5, 9) and parallel to the xz-plane. Solution. Simply repeat the steps of the previous problem, noting that the xz-plane is given by the equation y =. Problem 6 ( ). Find the equation of the plane containing the lines r 1 (t) = t, 2t, 2t and r 2 (t) = 3t, t, 8t. Solution. To find the equation of this plane, it suffices to find a normal vector to the plane and a point on the plane. For a normal vector n, we can take the cross product of r 1 (1) and r 2 (1) since these vectors lie on the plane (this is not the only option any nonzero choice of t would work) to get n = r 1 (1) r 2 (1) = 1, 2, 3 3, 1, 8 = i j k = 13, 1, 5. Next, a point on the plane can be found by plugging any value of t into either r 1 or r 2. In particular, we can take t = in r 1 to see that (,, ) is on the plane. Thus by Theorem 1 of Section 12.5, an equation of the plane we want in vector form is Expanding this into the scalar form, we get 13, 1, 5 x, y, z = 13, 1, 5,,. 13x + y 5z =. 3

6 Problem 7 ( ). Parametrize the intersection of the surfaces using t = y as the parameter. y 2 z 2 = x 2, y 2 + z 2 = 9 Solution. Right away, set y = t. From the second equation in the problem, we get z 2 = 9 t 2 = z = ± 9 t 2. We see now that we must have 3 t 3. Next, we substitute the formula for z in terms of t into the first equation to get x = y 2 z = t 2 (9 t 2 ) + 2 = 2t 2 7. Thus we have parametrized the intersection of the two surfaces by the following two functions: r 1 (t) = 2t 2 7, t, 9 t 2, r 2 (t) = 2t 2 7, t, 9 t 2, 3 t 3. Problem 8 ( ). Find a parametrization of the curve in the previous problem using trigonometric functions. Solution. From the equation it makes sense to set y 2 + z 2 = 9 = 3 2, y = 3 cos t, z = 3 sin t, t < 2π. Now all we have to do is express x in terms of t. From the equation we see that y 2 z 2 = x 2, x = y 2 z = 9 cos 2 t 9 sin 2 t + 2. There might be a trig identity that simplifies this, but it s not important. Now we finish the problem by giving a trigonometric parametrization of the intersection of our two surfaces: r(t) = 9 cos 2 t 9 sin 2 t + 2, 3 cos t, 3 sin t, t < 2π. Problem 9 ( ). Find a parametrization of the horizontal circle of radius 1 with center (2, 1, 4). Solution. The circle being horizontal means that the z-coordinate of our parametrization is constant. A horizontal circle centered at the origin with radius 1 has the parametrization r 1 (t) = cos t, sin t,, t < 2π. The circle we want is this circle translated so that the center become the point (2, 1, 4). To do this, we simply add the vector 2, 1, 4 to the function r 1 to get the parametrization of the circle we want: r(t) = cos t + 2, sin t 1, 4, t < 2π. 4

7 Problem 1 ( ). Find a parametric equation for the curve ( x ) 2 ( y ) 2 + = Solution. Any time that we see something of the form our first instinct should be to set A 2 + B 2 = C 2, A = C cos t, B = C sin t, t < 2π. For our problem we have A = x, B = y, and C = 1, so we set 5 12 Simplifying, we get x 5 = cos t, y 12 = sin t, t < 2π. x = 5 cos t, y = 12 sin t, t < 2π. These are parametric equations for our curve. 5

8 : Let v(s) =< s 2, 2s, 9s 2 >. Evaluate d ds v(g(s)) at s = 4, assuming g(4) = 3 and g (4) = 9 solution: v (s) =< 2s, 2, 18s 3 > d ds v(g(s)) = v (g(s))g (s) (By the Chain Rule) So at s = 4 this becomes: d ds v(g(4)) = v (g(4))g (4) = v (3)( 9) =< 2(3), 2, 18(3) 3 > ( 9) = 9 < 6, 2, 18/27 >=< 54, 18, 6 > : dr/dt =< 2t 1/2, 6, 8t >, r(1) =< 4, 9, 2 > Find the location and the velicoty of the particle at t=4. solution: velocity at t=4 is the derivative of the position vector evaluated at t=4: velocity(4) = r (4) =< 2(4) 1/2, 6, 8(4) >=< 1, 6, 32 > To find the position vector we need to take the integral of r (t): Integrating element wise gives: (Note: c is the constant of integration) Evaluating for our initial condition: r(t) =< 4t 1/2, 6t, 4t 2 > +c r(1) =< 4(1) 1/2, 6(1), 4(1) 2 > +c =< 4, 9, 2 > c =<, 3, 2 > Plugging in our constant of integration and evaluating at t = 4 yields: r(4) =< 4(4) 1/2, 6(4), 4(4) 2 > + <, 3, 2 >=< 8, 27, 62 > : Evaluate the length of (t 3 + 1, t 2 3), t 1 solution: s = s = b a 1 (x ) 2 + (y ) 2 dt (3t2 ) 2 + (2t ) 2 dt 1

9 s = s = 1 1 (9t4 + 2t 2 dt t (9t 2 + 2dt Using u substitution with u = 9t and dt = du/18 and bounds 4 u 13 s = udu s = 1 18 (2 3 u3/2 ) 13 4 s = 1 27 (133/2 4 3/2 ) : Find the minimum speed of c(t) = (t 3 4t, t 2 + 1) for t. solution: First we need to find the speed of the particle: ds dt = (3t 2 4) 2 + (2t) 2 = 9t 4 2t Since the square root is a strictly increasing function we only need to check the critical points of the function f(t) = 9t 4 2t f (t) = 36t 3 4t = 4t(9t 2 1) Which has critical points at t = and t = 1/9. Evaluating at these points: ds dt () = 16 = 4 ds dt ( 1/9) = 44/ Therefore the minimum speed is 44/ : Find the path that traces a curve in the y = 1 plane with radius 4, and center (2, 1, 3) with constant speed equal to 8. solution: First we need to find start with a basic parametrized curve. r(t) =< 2, 1, 3 > +4 < cos t,, sin t > So we have our center point (2, 1, 3) and our circle of radius 4. Note that the y component of our parametrized curve is constant for all t. (i.e. in the plane y = 1) Now we need to parametrize the curve by substituting t = g(s) so that r(s) = r(g(s)) has the constant speed, r (s) = 8. So we need to find r (s) using the chain rule. r (s) = d ds r(g(s)) = r (g(s))g (s) (1) 2

10 Since r (t) =< 4 sin t,, 4 cos t > we can sub in g(s) for t and get. r (g(s)) =< 4 sin g(s),, 4 cos g(s) > Plugging into equation (1) gives: r (s) =< 4 sin g(s),, 4 cos g(s) > g (s) = 4g (s) < sin g(s),, cos g(s) > r (s) = 4 g (s) (cos g(s)) 2 + (sin g(s)) 2 = 4 g (s) So g (s) must equal 2. Integration give g(s) = 2s. Putting it all together: r(s) = r(g(s)) =< cos (2s), 1, sin (2s) > 3

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29 Problem 1 (15.4.1). Sketch the region D indicated and integrate f(x, y) over D using polar coordinates. f(x, y) = x 2 + y 2 ; x 2 + y 2 2 Solution. The region D indicated is the disk of radius 2 centered at the origin: y D 2 x In polar coordinates D is given by r 2, θ 2π. Moreover, the integrand f(x, y) = x 2 + y 2 becomes f(r, θ) = (r cos θ) 2 + (r sin θ) 2 = r ( 2 cos 2 θ + sin 2 θ ) = r 2 = r after changing to polar coordinates. Therefore our integral is 2π 2 2π f(x, y) da = f(r, θ) r dr dθ = D ( 2π ) ( ) ( 2 = 1 dθ r 2 dr = 2π = 4 2π. 3 2 r 3 3 r 2 dr dθ ) 2 1

30 Problem 2 (15.4.3). Sketch the region D indicated and integrate f(x, y) over D using polar coordinates. f(x, y) = xy; x, y, x 2 + y 2 4 Solution. The region D indicated is the portion of the disk of radius 2 centered at the origin (x 2 + y 2 4) restricted to the first quadrant (x and y ): y D In polar coordinates D is given by 2 r 2, θ π 2. Moreover, the integrand f(x, y) = xy becomes f(r, θ) = (r cos θ)(r sin θ) = r 2 cos θ sin θ = r2 sin(2θ) 2 after changing to polar coordinates and using the trig identity Therefore our integral is D f(x, y) da = π/2 2 = 1 2 ( = 1 2 = 1 2 = 2. ( π/2 sin(2θ) = 2 cos θ sin θ. x π/2 2 f(r, θ) r dr dθ = ) ( 2 ) sin(2θ) dθ r 3 dr π/2 ) ( cos(2θ) 2 ( cos π + cos 2 2 r 4 4 ) ) 2 r 3 sin(2θ) 2 dr dθ 2

31 Problem 3 (15.4.7). Sketch the region of integration and evaluate by changing to polar coordinates. 2 4 x 2 ( x 2 + y 2) dy dx 2 Solution. From the outer integral, we can tell that the x values of the region of integration range from 2 to 2. From the inner integral, we can tell that for a given value of x (with 2 x 2), the y-values range between and 4 x 2. As x ranges from 2 to 2, y = 4 x 2 traces out the upper half of the circle of radius of 2 centered at the origin. Therefore the region of integration looks like the following: y Thus we see that the region is described in polar coordinates as and so our integral is 2 4 x 2 2 r 2, θ π, r 2 {}}{ ( x 2 + y 2) dy dx = π 2 2 r 2 r dr dθ ( π ) ( 2 ) = 1 dθ r 3 dr = π r4 4 x 2 = 4π. Problem 4 (15.4.9). Sketch the region of integration and evaluate by changing to polar coordinates. 1/2 1 x 2 3x x dy dx Solution. From the outer integral, we can tell that the x values of the region of integration range from to 1/2. From the inner integral, we can tell that for a given value of x (with x 1/2), the y-values range between y = 3x (the line with slope 3 and y-intercept ) and y = 1 x 2 (the upper half of the circle of radius 1 centered at the origin). Therefore the region of integration looks like the following: 3

32 y y y 3/2 3/2 1 = 1 = 1 y = 3x x 1/2 1/2 x π/3 1/2 x Thus we see that the region is described in polar coordinates as and so our integral is 1/2 1 x 2 r 1, π/3 θ π/2, π/2 1 x dy dx = r cos θ r dr dθ 3x π/3 ( ) π/2 ( 1 = cos θ dθ = ( π/3 1 ) ) ( r 2 dr = sin θ π/2 π/3 ) ( ) r Problem 5 ( ). Calculate the integral over the given region by changing to polar coordinates. f(x, y) = xy ; x 2 + y 2 1. Solution. The region D over which we re integrating is the unit disk centered at the origin. In polar coordinates, D is given by Moreover, our integrand f(x, y) = xy is r 1, θ 2π. f(r, θ) = (r cos θ) (r sin θ) = r 2 cos θ sin θ = r2 2 sin(2θ) after changing to polar coordinates and using the trig identity sin(2θ) = 2 cos θ sin θ. We can determine the sign of sin 2θ = 2 cos θ sin θ, and hence the value of sin(2θ), in each of the four quadrants in the plane: 4

33 π/2 θ π cos θ sin θ cos θ sin θ π θ 3π/2 cos θ sin θ cos θ sin θ y θ π/2 cos θ sin θ cos θ sin θ x 3π/2 θ 2π cos θ sin θ cos θ sin θ sin(2θ), if θ π/2 sin(2θ), if π/2 θ π = sin(2θ) = sin(2θ), if π θ 3π/2 sin(2θ), if 3π/2 θ 2π Finally, our integral is 2π 1 2π f(x, y) da = f(r, θ) r dr dθ = D = 1 ( 1 ) ( 2π ) r 3 dr sin(2θ) dθ 2 ( = 1 ) ( r 4 1 π/2 π 2 4 sin(2θ) dθ sin(2θ) dθ + π/2 = 1 cos(2θ) π/2 + cos(2θ) π cos(2θ) 3π/ π/2 π 1 r 3 2 3π/2 π sin(2θ) dr dθ + cos(2θ) 2 sin(2θ) dθ 2π 3π/2 2π 3π/2 sin(2θ) dθ = 1 ( cos π + cos + cos(2π) cos π cos(3π) + cos(2π) + cos(4π) cos(3π)) 16 = 1 16 ( ) = 1 2. Problem 6 ( ). Find the volume of the wedge-shaped region contained in the cylinder x 2 + y 2 = 9, bounded above by the plane z = x and below by the xy-plane. Solution. See your text for the picture of the region D. From the picture (and the description of the region) you can conclude that D is given in cylindrical coordinates as Therefore, the volume of D is 3 1 dv = D r 3, π/2 θ π/2, z x = r cos θ. = = 3 ( π/2 r cos θ π/2 r dz dθ dr = 3 ( π/2 ( r z π/2 r cos θ ( ) π/2 ( 3 ) ( π/2 r r cos θ dθ dr = r 2 dr π/2 ) 3 ( ) sin θ = 18. r 3 3 π/2 π/2 ) ) dθ dr cos θ dθ π/2 ) ) 5

34 Problem 7 ( ). Evaluate D x2 + y 2 da where D is the region y 2 x Solution. First, we consider the inner circle. The inner circle D inn is centered at (1, ) and has radius 1. Therefore, in rectangular coordinates it is given by the equation Changing to polar coordinates, we have (x 1) 2 + y 2 = 1. (r cos θ 1) 2 + r 2 sin 2 θ = 1 = r 2 cos 2 θ 2r cos θ r 2 sin 2 θ = 1 = r 2 2r cos θ = = r = 2 cos θ, and π ranges between π/2 and π/2 (since these are the values of θ that make r = ). Thus the integral of the integrand x 2 + y 2 = r over the inner circle D inn is given by D inn x2 + y 2 da = = 8 3 π/2 2 cos θ π/2 π/2 r r dr dθ = π/2 cos 3 θ dθ = 8 3 π/2 π/2 π/2 π/2 ( r cos θ ) dθ (1 sin 2 θ) cos θ dθ. Now let u = sin θ, so that du = cos θ dθ. Also, when θ = π/2, we have u = sin( π/2) = 1, and when θ = π/2, we have u = sin(π/2) = 1. Therefore our integral over the inner circle D inn becomes D inn x2 + y 2 da = (1 u 2 ) du = 8 3 ) (u u3 1 = 8 ( 2 2 ) = This finishes our consideration of the inner circle. Next we consider the outer circle D out. It is given in polar coordinates as r 2, θ 2π. 6

35 Therefore the integral of the integrand x 2 + y 2 = r over D out is D out x2 + y 2 da = 2π 2 r r dr dθ = ( 2π ) ( 2 ) 1 dθ r 2 dr = 16π 3. This finishes our consideration of the outer circle. We can now compute x2 + y D 2 da and finish the problem. Since D is obtained by removing D inn from D out, it follows that x2 + y 2 da = x2 + y 2 da x2 + y 2 da = 16π D D out D inn Problem 8 ( ). Use cylindrical coordinates to calculate f(x, y, z) dv for the W given function and region. f(x, y, z) = x 2 + y 2 ; x 2 + y 2 9, z 5 Solution. In three dimensions, the inequality x 2 + y 2 9 is the filled-in cylinder of radius 3 centered along the z-axis, and in cylindrical coordinates it is given by the inequalities r 3 and θ 2π. Moreover, the inequality z 5 gives us the portion of this cylinder that we need to consider. The integrand f(x, y, z) = x 2 + y 2 is given in cylindrical coordinates as f(r, θ, z) = r 2, and so our integral is W 5 2π 3 f(x, y, z) dv = r 2 r dr dθ dz ( 5 ) ( 2π ) ( 3 = 1 dz 1 dθ ) r 3 dr = 5 2π 34 4 = 45π 2. Problem 9 ( ). Use cylindrical coordinates to calculate f(x, y, z) dv for the W given function and region. f(x, y, z) = y; x 2 + y 2 1, x, y, z 2 Solution. The region W indicated here is the cylinder of radius 1 centered along the z-axis, restricted to the first octant, and with maximum height 2. In polar coordinates, it is given by r 1, θ π/2, z 2. The integrand f(x, y, z) = y is given in cylindrical coordinates as f(r, θ, z) = r sin θ, and so our integral is W 2 π/2 1 f(x, y, z) dv = r sin θ r dr dθ dz ( 2 ) ( ) π/2 ( 1 = 1 dz sin θ dθ ) r 2 dr =

36 Problem 1 ( ). Use cylindrical coordinates to calculate f(x, y, z) dv for the W given function and region. f(x, y, z) = z; x 2 + y 2 z 9. Solution. From the description of the region, we can deduce two inequalities: x 2 + y 2 9, x 2 + y 2 z 9. The first inequality is the cylinder of radius 3 centered along the z-axis; in polar coordinates it is r 3 and θ 2π. The second inequality means that the values of z that we want to integrate over are above the paraboloid z = x 2 + y 2 and below the plane z = 9; in polar coordinates this means that r 2 z 9. Therefore we can already integrate: W f(x, y, z) dv = 2π 3 9 = 2π 3 z r dz dr dθ = r ( 2 ) 81 r 2 r4 dr = π 2 ( 2π 3 ) ( 3 1 dθ r ( z r 2 ( 81r r 5 ) dr = 243π. ) dr ) 8