Solutions to Practice Test 3

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1 Solutions to Practice Test 3. (a) Find the equation for the plane containing the points (,, ), (, 2, ), and (,, 3). (b) Find the area of the triangle with vertices (,, ), (, 2, ), and (,, 3). Answer: (a) We will label the points A(,, ), B(, 2, ), and C(,, 3). Then, AB (, 2, ) and AC (,, 3). The vector AB AC is normal to the plane containing points A, B, and C. AB AC î ĵ ˆk 2 3 ( 6, 3, 2) Thus, the plane 6(x ) 3y + 2z contains the points A, B, and C. (b) The area of the triangle with vertices A(,, ), B(, 2, ), and C(,, 3) is one-half the area of the parallelogram spanned by the vectors AB and AC. The area of the parallelogram spanned by AB and AC is AB AC. Thus, Area of triangle 2 AB AC Find the distance between the following two parallel lines: l (t) ( 3, 2, 2) + t(3,, ) l 2 (t) (, 2, ) + t(3,, ) Answer: The distance between the two parallel lines is the same as the distance between a point on line and the second line. We will use the method used in Method of Example 7 in Section.5 for finding the distance between a point and a line.

2 A point on line is A( 3, 2, 2), and a point on line 2 is B(, 2, ). A vector in the direction of line 2 is a (3,, ). The distance between the two lines is the length of the vector BA proj aba. BA ( 4,, 2) a BA proj aba a a 2 (3,, ) ( 4,, 2) (3,, ) 9 + (3,, ) ( 3,, ) BA proj a BA ( 4,, 2) ( 3,, ) (,, 3) + 9 Thus, the distance between the two parallel lines is. 3. Evaluate the following double integrals. (a) (b) (c) π sin x 3 3y 2 4 x 2 2 y cos x dy dx cos(x 2 ) dx dy 4 x 2 e (x2 +y2) dy dx 2

3 Answer: (a) π sin x y cos x dy dx π π [ [ ] sin x 2 y2 cos x dx 2 sin2 x cos x dx 6 sin3 x ] π Note that the antiderivative of sin 2 x cos x can be found using the substitution u sin x. (b) Since there is no nice antiderivative of cos(x 2 ), we will start by reversing the order of integration. The region is the triangle bounded by the x-axis, the line y x, and the line x y cos(x 2 ) dx dy Then, we can evaluate the integral: 3 x 3 cos(x 2 ) dy dx 3 x cos(x 2 ) dy dx [ y cos(x 2 ) ] x 3 dx 3 x cos(x2 ) dx [ 6 sin(x2 ) sin(9) 6 ] (c) The region for this integral is a circle of radius 2 centered at the origin. Probably, polar coordinates will be the easiest way to evaluate this integral. We begin by switching to polar coordinates. 2 4 x x 2 e (x2 +y2) dy dx 3 2 re r2 dr dθ

4 Now, we can evaluate the integral: 2 re r2 dr dθ [ ] 2 2 er2 dθ 2 e4 2 dθ π(e 4 ) Consider the solid W that is described by the equations z, z 9 x 2 y 2, and x 2 + y 2 4. Find the volume of this solid. Answer: Cylindrical coordinates work best for finding the volume of this region. If we integrate with respect to z first, the bounds are z 9 r 2. The intersection of the plane z and the paraboloid z 9 x 2 y 2 is the circle x 2 + y 2 9. Thus, the cylinder x 2 + y 2 4 determines the r and θ bounds: r 2 and θ 2π. Thus, Volume 2 9 r r dz dr dθ [zr] 9 r2 dr dθ (9 r 2 )r dr dθ 9r r 3 dr dθ [ 9 2 r2 4 r4 ] 2 dθ 4dθ 28π

5 5. Consider the solid bounded by the paraboloid z x 2 +y 2 and the plane z. Find the moment of inertia of this solid about the z-axis if the density is δ(x, y, z) 2z. Answer: The formula for moment of inertia about the z-axis is on page 367 of the textbook: I z W (x 2 + y 2 )δ(x, y, z) dv For the region given, cylindrical coordinates work well. If we integrate with respect to z first, the bounds are r 2 z. The intersection of the paraboloid z x 2 + y 2 and the plane z is a circle x 2 + y 2. Thus, the bounds for r and θ are r and θ 2π. We are integrating the function (x 2 + y 2 )δ(x, y, z) (x 2 + y 2 )2z, which is 2r 2 z in cylindrical coordinates. Thus, I z 2r 3 z dz dr dθ r 2 [ r 3 z 2] dr dθ r 2 r 3 r 7 dr dθ [ 4 r4 8 r8 ] dθ 8 dθ π Consider the parametric equation x(t) (3t 2, cos(πt), ln t). Find the equation for the line parallel to this curve at time t. Answer: First, we find the velocity: x (t) ( 6t, π sin(πt), ) t 5

6 Next, we evaluate x(t) and x (t) both at t : x() (3,, ) and x () (6,, ). The line tangent to the curve at t will contain the point x() and be in the direction given by the vector x (). Thus, the line has parametric equation: l(t) (3,, ) + t(6,, ) 7. Consider the parametric equation x(t) (cosh t, sinh t, t). Find the length of this curve from t to t. You may find the following facts about the functions cosh x and sinh x helpful: cosh x ex + e x 2 sinh x ex e x 2 d cosh x sinh x dx d sinh x cosh x dx cosh 2 x sinh 2 x Answer: First, we compute the velocity: Next, we compute the speed: x (t) x (t) (sinh t, cosh t, ) sinh 2 t + cosh 2 t + 2 cosh 2 t 2 cosh t In the above computation, we used the fact that sinh 2 t + cosh 2 t. Now, we can compute the length of the curve: 6

7 Length x (t) dt 2 cosh t dt [ 2 sinh t ] 2(sinh sinh ) ( 2 e ) e 8. Consider the parametric equation x(t) (3t, sin 2t, cos 2t). (a) Determine the moving frame {T, N, B} for x(t). (b) Compute the curvature κ for x(t). (c) Compute the torsion τ for x(t). Answer: We can compute all of these quantities using the formulas from Section 3.2. (a) x (t) T x (t) (3, 2 cos 2t, 2 sin 2t) cos 2 2t + 4 sin 2 2t ( ) x (t) 3 x (t) 2 cos 2t 2 sin 2t,, N dt/dt dt/dt (, 4 B T N sin 2t, 4 cos 2t) 6 sin2 2t + 6 cos2 2t (, sin 2t, cos 2t) î ĵ ˆk 3 2 cos 2t 2 sin 2t sin 2t cos 2t ( 2 3, cos 2t, 3 ) sin 2t 7

8 (b) κ dt/dt ds/dt 4/ 4 (c) To compute the torsion τ, we need to compute db ds formula db ds τn: and use the db ds db/dt ds/dt (, 6 sin 2t, 6 cos 2t) (, 6 sin 2t, 6 ) cos 2t Comparing db ds and N, we see that τ Consider the following vector field: F(x, y, z) (cos(yz) x, cos(xz) y, cos(xy) z) (a) Calculate the divergence of the vector field F. (b) Calculate the curl of the vector field F. Answer: (a) div(f) F x + F 2 y + F 3 3 z ( ) F (b) curl(f) 3 F 2, F F 3, F 2 F y z z x x y ( x sin(xy)+x sin(xz), y sin(yz)+y sin(xy), z sin(xz)+z sin(yz)) 8

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